Prasanna

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 7 Thermodynamics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 7 Thermodynamics

11th Chemistry Guide Thermodynamics Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
The amount of heat exchanged with the surrounding at constant temperature and pressure is given by the quantity
(a) ∆E
(b) ∆H
(c) ∆S
(d) ∆G
Answer:
(b) ∆H

Question 2.
All the naturally occurring processes proceed spontaneously in a direction which leads to
(a) decrease in entropy
(b) increase in enthalpy
(c) increase in free energy
(d) decrease in free energy
Answer:
(d) decrease in free energy

Question 3.
In an adiabatic process, which of the following is true?
(a) q = w
(b) q = 0
(c) ∆E = q
(d) P∆V = 0
Answer:
(b) q = 0

Question 4.
In a reversible process, the change in entropy of the universe is
(a) > 0
(b) ≥ 0
(c) <0
(d) = 0
Answer:
(d) = 0

Question 5.
In an adiabatic expansionof an ideal gas
(a) w = – ∆U
(b) w = ∆U + ∆H
(c) ∆U = 0
(d) w = 0
Answer:
(a) w = – ∆U

Question 6.
The intensive property among the quantities below is
(a) mass
(b) volume
(c) enthalpy
(d) mass/volume
Answer:
(d) mass/volume

Question 7.
An ideal gas expands from the volume of 1 × 10^-3 m³ to 1 × 10^-2 m³ at 300 K against a constant pressure at 1 × 10⁵ Nm^-2. The work done is
(a) -900 J
(b) 900 kJ
(c) 270 kJ
(d) – 900 kJ
Answer:
(a) -900 J

Question 8.
Heat of combustion is always
(a) positive
(b) negative
(c) zero
(d) either positive or negative
Answer:
(b) negative

Question 9.
The heat of formation of CO and CO2 are -26.4 kCal and -94 kCal, respectively. Heat of combustion of carbon monoxide will be
(a) + 26.4 kcal
(b) – 67.6 kcal
(c) – 120.6 kcal
(d) + 52.8 kcal
Answer:
(b) – 67.6 kcal

Question 10.
C(diamond) → C(graphite), ∆H = -ve, this indicates that
(a) graphite is more stable than diamond
(b) graphite has more energy than diamond
(c) both are equally stable
(d) stability cannot be predicted
Answer:
(a) graphite is more stable than diamond

Question 11.
The enthalpies of formation of Al₂O₃ and Cr₂O₃ are – 1596 kJ and – 1134 kJ, respectively.
∆H for the reaction 2Al + Cr₂O₃ → 2Cr + Al₂O₃ is
(a) – 1365 kJ
(b) 2730 kJ
(c) – 2730 kJ
(d) -462 kJ
Answer:
(d) -462 kJ

Question 12.
Which of the following is not a thermodynamic function?
(a) internal energy
(b) enthalpy
(c) entropy
(d) frictional energy
Answer:
(d) frictional energy

Question 13.
If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then
(a) ∆H > ∆U
(b) ∆H – ∆U=0
(c) ∆H + ∆U=0
(d) ∆H < ∆U
Answer:
(d) ∆H < ∆U

Question 14.
Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is
(a) +1 kJ
(b) -5 kJ
(c) +3 kJ
(d) -3 kJ
Answer:
(c) +3 kJ

Question 15.
The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol^-1) reacts with hydrochloric acid in an open beaker at 25°C
(a) – 2.48 kJ
(b) – 2.22 kJ
(c) + 2.22 kJ
(d) + 2.48 kJ
Answer:
(a) – 2.48 kJ

Question 16.
The value of ∆H for cooling 2 moles of an ideal monatomic gas from 1250°C to 250°C at constant pressure will be [given C_p = \(\frac{5}{2}\)R]
(a) – 250 R
(b) – 500 R
(c) 500 R
(d) + 250 R
Answer:
(b) – 500 R

Question 17.
Given that C_(g) + O_2(g) → CO_2(g) ∆H° = – a kJ;
2 CO_(g) + O_2(g) → 2CO_2(g) ∆H° = -b kJ; Calculate the ∆H° for the reaction C_(g) + ½O_2(g) → CO_(g)
(a) \(\frac{b+2 a}{2}\)
(b) 2a – b

(c) \(\frac{2 a-b}{2}\)

(d) \(\frac{b-2 a}{2}\)
Answer:
(d) \(\frac{b-2 a}{2}\)

Question 18.
When 15.68 litres of a gas mixture of methane and propane are fully combusted at 0° C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat of released from this combustion in kJ is (∆H_C(CH4)) = – 890 kJ mol and ∆H_C(C3H8) = – 2220 kJ mol^-1)
(a) -889 kJ
(b) -1390 kJ
(c) -3180 kJ
(d) -632.68 kJ
Answer:
(d) -632.68 kJ

Question 19.
The bond dissociation energy of methane and ethane are 360 kJ mol^-1 and 620 kJ mol^-1 respectively. Then, the bond dissociation energy of C-C bond is
(a) 170 kJ mol^-1
(b) 50 kJ mol^-1
(c) 80 kJ mol^-1
(d) 220 kJ mol^-1
Answer:
(c) 80 kJ mol^-1

Question 20.
The correct thermodynamic conditions for the spontaneous reaction at all temperature is
(a) ∆H < 0 and ∆S > 0
(b) ∆H < 0 and ∆S < 0
(c) ∆H > 0 and ∆S = 0
(d) ∆H > 0 and ∆S > 0
Answer:
(a) ∆H < 0 and ∆S > 0

Question 21.
The temperature of the system, decreases in an
(a) Isothermal expansion
(b) Isothermal Compression
(c) adiabatic expansion
(d) adiabatic compression
Answer:
(c) adiabatic expansion

Question 22.
In an isothermal reversible compression of an ideal gas the sign of q, ∆S and w are respectively
(a) +, -, –
(b) -, +, –
(c) +, -, +
(d) -, -, +
Answer:
(d) -, -, +

Question 23.
Molar heat of vapourisation of a liquid is 4.8 kJ mol^-1 If the entropy change is 16 J mol^-1 K^-1. the boiling point of the liquid is
(a) 323 K
(b) 27°C
(c) 164 K
(d) 0.3 K
Answer:
(b) 27°C

Question 24.
∆S is expected to be maximum for the reaction
(a) Ca(S) + 1/2 O₂(g) → CaO(S)
(b) C(S) + O₂(g) → CO₂(g)
(c) N₂(g) + O₂(g) → 2NO(g)
(d) CaCO₃(S) → CaO(S) + CO₂(g)
Answer:
(d) CaCO₃(S) → CaO(S) + CO₂(g)

Question 25.
The values of ∆H and ∆S for a reaction are respectively 30 kJ mol^-1 and 100 JK^-1 mol^-1. Then the temperature above which the reaction will become spontaneous is
(a) 300 K
(b) 30 K
(c) 100 K
(d) 20°C
Answer:
(a) 300 K

II. Write brief answers to the following questions:

Question 26.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics, known as the law of conservation of energy, states that the total energy of an isolated system remains constant though it may change from one foim to another.
The mathematical statement of the First Law as:
∆U = q + w
where q – the amount of heat supplied to the system; w – work done on the system.

Question 27.
Define Hess’s law of constant heat summation.
Answer:
The heat changes in chemical reactions are equal to the difference in internal energy (∆U) or heat content (∆H) of the products and reactants, depending upon whether the reaction is studied at constant volume or constant pressure. Since ∆U and ∆H are functions of the state of the system, the heat evolved or absorbed in a given reaction depends only on the initial state and final state of the system and not on the path or the steps by which the change takes place.

Question 28.
Explain intensive properties with two examples.
Answer:
The property that is independent of the mass or the size of the system is called an intensive property. Refractive index, Surface tension, density, temperature, Boiling point, Freezing point, molar volume.

Question 29.
Define the following terms:
(a) isothermal process
(b) adiabatic process
(c) isobaric process
(d) isochoric process
Answer:
a. An isothermal process is defined as one in which the temperature of the system remains constant, during the change from its initial to final state. The system exchanges heat with its surrounding and the temperature of the system remains constant. For an isothermal process dT = 0.

b. An adiabatic process is defined as one in which there is no exchange of heat (q) between the system and surrounding during the process. Those processes in which no heat can flow into or out of the system are called adiabatic processes.

c. An isobaric process is defined as one in which the pressure of the system remains constant
during its change from the initial to final state.
For an isobaric process dP = 0.

d. An isochoric process is defined as the one in which the volume of system remains constant during its change from initial to final state. Combustion of a fuel in a bomb calorimeter is an example of an isochoric process.
For an isochoric process, dV= 0.

Question 30.
What is the usual definition of entropy? What is the unit of entropy?
Answer:
Entropy is a measure of the molecular disorder (randomness) of a system.
The thermodynamic definition of entropy is concerned with the change in entropy that occurs as a result of a process.
It is defined as, dS = dq_rev /T

Question 31.
Predict the feasibility of a reaction when
(i) both ∆H and ∆S positive
(ii) both ∆H and ∆S negative
(iii) ∆H decreases but ∆S increases
Answer:
(i) ∆H_r = + ve, ∆S = + ve; non-spontaneous at low temperature spontaneous at high temperature
(ii) ∆H_r = – ve, ∆S = – ve; spontaneous at low temperature non-spontaneous at high temperature
(iii) ∆H decreases but ∆S increases = Spontaneous at all temperature.

Question 32.
Define is Gibb’s free energy.
Answer:
G is expressed as G = H – TS, free energy change of a process is given by the relation
∆G = ∆H – T∆S.

Question 33.
Define enthalpy of combustion.
Answer:
The heat of combustion of a substance is defined as “The change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen”. It is denoted by ∆H_C.

Question 34.
Define molar heat capacity. Give its unit.
Answer:
The heat capacity for 1 mole of substance, is called molar heat capacity (cm). It is defined as “The amount of heat absorbed by one mole of the substance to raise its temperature by 1 kelvin”.
The SI unit of molar heat capacity is JK^-1 mol^-1

Question 35.
Define the calorific value of food. What is the unit of calorific value?
Answer:
The calorific value is defined as the amount of heat produced in calories (or joules) when one gram of the substance is completely burnt. The SI unit of calorific value is J kg^-1. However, it is usually expressed in cal g^-1.

Question 36.
Define enthalpy of neutralization.
Answer:
The heat of neutralisation is defined as “The change in enthalpy when one gram equivalent of an acid is completely neutralised by one gram equivalent of a base or vice versa in dilute solution”.
HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O₁
∆H = – 57.32 kJ

Question 37.
What is lattice energy?
Answer:
Lattice energy is defined as the amount of energy required to completely remove the constituent ions from its crystal lattice to an infinite distance. It is also referred as lattice enthalpy.

Question 38.
What are state and path functions? Give two examples.
Answer:
State function:
A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path (or manner) by which the particular state is reached.
Example:
Pressure (P), Volume (V), Temperature(T), Internal energy (U), Enthalpy (H), free energy (G) etc.
Path functions:
A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.
Example:
Work (w), Heat (q).

Question 39.
Give Kelvin statement of second law of thermodynamics.
Answer:
It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink.

Question 40.
The equilibrium constant of a reaction is 10, what will be the sign of ∆G? Will this reaction be spontaneous?
Answer:
∆G = -2.303 RT logK_eq
Substituting the known values of R, T and K_eq
R = 8.314(JK^-1 mol^-1)
T = 300 K
K_eq = 10
∆G = –2.303 8.314(JK^-1) 300(K) log 10
= -5744.14 J/mol
= -5.744 KJ/mol
∆G < 0, then it is spontaneous.

Question 41.
Enthalpy of neutralization is always a constant when a strong acid is neutralized by a strong base: account for the statement.
Answer:
It is because in dilution solution all strong acids and strong bases are completely ionized. The neutralization of strong acid and strong bases are completely ionized. The neutralization of a strong acid and strong base simply involves the combination of H⁺ ions (from acid) and OH^– ions (from base) to form unionized water molecules with the evolution of 57.1 kJ heat.
H⁺_(aq) + OH^–_(aq) → H₂O(l),
∆₂H° = -57.1 kJ
Since the same reaction takes place during neutralization of all strong acids and strong bases, the value of enthalpy of neutralization is constant.

Question 42.
State the third law of thermodynamics.
Answer:
The third law of thermodynamics states that the entropy of pure crystalline substance at absolute zero is zero. Otherwise it can be stated as it is impossible to lower the temperature of an object to absolute zero in a finite number of steps. Mathematically,
lim_T→0 S = 0 for a perfectly ordered crystalline state.

Question 43.
Write down the Born-Haber cycle for the formation of CaCl₂.
Answer:

Step 1:
Solid calcium is converted to gaseous state (Enthalpy of atomization)
Ca_(S) → Ca_(g) Ca(g)                         ∆H°_a = 178 KJ/mol
Step 2:
The calcium is converted to ionic form
(divalent cation): (Ionisation enthalpy)
Ca → Ca⁺ + e^–                          ∆H°_IE = 590 kJ
Ca → Ca²⁺ + e^–                        ∆H°_IE = 590 KJ
Step 3:
Atomisation of chlorine molecule to chlorine atom: (Atomisation enthalpy)
\(\frac{1}{2}\)Cl₂ → Cl               ∆H°_Cl-Cl = 121 KJ
For two chlorine atoms required, atomistion energy = 2 × 121 = 242 KJ/mol
Step 4:
Convrsion of chlorine atom into ion:(Electron affinity)
Cl + e^– → Cl^–;                           ∆H°_ea = – 364KJ
Step 5:
Finally the two ions join together by lattice energy as: (here two Cl^– ions are involved)
Ca²⁺ + 2Cl^– → CaCl₂

Hence Lattice energy = Heat of formation – heat of atomization – dissociation energy – (sum of ionization energies) – (sum of electron affinities)
Since, heat of formation (i.e standard enthalpy of formation ) of CaCl₂ = – 796KJ/mol
Lattice energy = -796 -178 -242 -(590 + 1145) – (2 × – 364) = -2223KJ/mol

Question 44.
Identify the state and path functions out of the following: (a) Enthalpy (b) Entropy (c) Heat (d) Temperature (e) Work (f) Free energy.
Answer:
State function:
Enthalpy, Temperature, Free energy, Entropy
Path function:
Work, Heat.

Question 45.
State the various statements of second law of thermodynamics.
Answer:
Entropy statement:
The second law of thermodynamics can be expressed in terms of entropy, i.e “the entropy of an isolated system increases during a spontaneous process”.

Kelvin-Planck statement:
It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink.

Clausius statement:
It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing some work.

Question 46.
What are spontaneous reactions? What are the conditions for the spontaneity of a process?
Answer:

1. A reaction that occurs under the given set of conditions without any external driving force is called a spontaneous reaction.
2. The spontaneity of any process depends on three different factors.
3. If the enthalpy change of a process is negative, then the process is exothermic and may be spontaneous. (∆H is negative)
4. If the entropy change of a process is positive, then the process may occur spontaneously. (∆S is positive)
5. The gibbs free energy which is the combination of the above two (∆H – T∆S) should be negative for a reaction to occur spontaneously, i.e. the necessary condition for a reaction to be spontaneous is ∆H – T∆S < 0
6. For spontaneous process,
   ∆S_total > 0, ∆G < 0, ∆S < 0

Question 47.
List the characteristics of internal energy.
Answer:
The internal energy of a system is an extensive property. It depends on the amount of the substances present in the system. If the amount is doubled, the internal energy is also doubled.

The internal energy of a system is a state function. It depends only upon the state variables (T, P, V, n) of the system. The change in internal energy does not depend on the path by which the final state is reached.

The change in internal energy of a system is expressed as ∆U = U_f – U_i.
In a cyclic process, there is no internal energy change. ∆U_(cyclic) = 0

If the internal energy of the system in the final state (U_f) is less than the internal energy of the system in its initial state (U_i), then ∆U would be negative.
∆U = U_f – U_i = -ve(U_f < U_i)

If the internal energy of the system in the final state (U_f) is greater than the internal energy of the system in its initial state (U_i), then ∆U would be positive.
∆U = U_f – U_i = + ve(U_f > U_i).

Question 48.
Explain how heat absorbed at constant volume is measured using bomb calorimeter with a neat diagram.
Answer:
Calorimeter is used for measuring the amount of heat change in a chemical or physical change. In calorimetry, the temperature change in the process is measured which is directly proportional to the heat capacity. By using the expression C = \(\frac{q}{m \Delta T}\), we can calculate the amount of heat change in the process. Calorimetric measurements are made under two different conditions
(i) At constant volume (qV)
(ii) At constant pressure (qp)
(A) ∆U Measurements:
For chemical reactions, heat evolved at constant volume, is measured in a bomb calorimeter.

The inner vessel (the bomb) and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws.

A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurized with excess oxygen. The bomb is immersed in water, in the inner volume of the calorimeter. A stirrer is placed in the space between the wall of the calorimeter and the bomb, so that water can be stirred, uniformly. The reaction is started by striking the substance through electrical heating.

A known amount of combustible substance is burnt in oxygen in the bomb. Heat evolved during the reaction is absorbed by the calorimeter as well as the water in which the bomb is immersed. The change in temperature is measured using a Beckman thermometer. Since the bomb is sealed its volume does not change and hence the heat measurements is equal to the heat of combustion at a constant volume (∆U)_c.

The amount of heat produced in the reaction (∆U)_c is, equal to the sum of the heat abosrbed by the calorimeter and water.
Heat absorbed by the calorimeter
q₁ = k.∆T
where k is a calorimeter constant equal to mc Cc ( me is mass of the calorimeter and Cc is heat capacity of calorimeter)
Heat absorbed by the water q₂ = m_w C_w ∆T
where mw is molar mass of water
C_w is molar heat capacity of water (4,184 kJ K^-1 mol^-1)
Therefore ∆U_c = q₁ + q₂
= k.∆T + m_w C_w ∆T
= (k + m_w C_w)∆T

Calorimeter constant can be determined by burning a known mass of standard sample (benzoic acid) for which the heat of combustion is known (-3227 kJ mol^-1). The enthalpy of combustion at constant pressure of the substance is calculated from the equation
∆H°_c (pressure) = ∆U°_c (Vol) + ∆n_gRT

Question 49.
Calculate the work involved in expansion and 1 compression process.
Answer:
Work involved in expansion and compression processes:
In most thermodynamic calculations we are dealing with the evaluation of work involved in the expansion or compression of gases. The essential condition for expansion or compression of a system; is that there should be difference between external pressure (P_ext) and internal pressure (P_int).

For understanding pressure-volume work, let us consider a cylinder which contains V moles of an ideal gas fitted with a frictionless piston of cross sectional area A. The total volume of the gas inside is V and pressure of the gas inside is P_int. If the external pressure P_ext is greater than P_int, the piston moves inward till the pressure inside becomes equal to P_ext. Let this change be achieved in a single step
and the final volume be V_f.In this case, the work is done on the system (+w). It can be calculated as follows
w = -F.∆x ……….(1)
where dx is the distance moved by the piston during the compression and F is the force acting on the gas.

F = P_extA ……….(2)
Substituting (2) in (1)
w = – P_ext. A. ∆x
A .∆x = change in volume = V_f – V_i
w = -P_ext.{V_f – V_i) ……….(3)
w = -P_ext. (-∆V) …………(4)
= P_ext.∆V

Since work is done on the system, it is a positive quantity.
If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV. In such a case we can calculate the work donp on the gas by therelation
W_rev = – \(\int_{V_{i}}^{v_{f}}\) p_ext dV.
In a compression process, Pext the external pressure is always greater than the pressure of the system.
i.e., P_ext = (P_int + dP)
In an expansion process, the external pressure is always less than the pressure of the system
i.e., P_ext = (P_int – dP)

When pressure is not constant and changes in infinitesimally small steps (reversible conditions) during compression from V_i to V_f. the P-V plot looks like in figure. Work done on the gas is represented by the shaded area.
In general case we can write,
P_ext = (P_int ± dP). Such processes are called reversible processes. For a compression process work can be related to internal pressure of the system under reversible conditions by writing equation
W_rev = – \(\int_{V_{i}}^{v_{f}}\) P_int dV
For a given system with an ideal gas
P_int V = nRT
p_int = \(\frac{n R T}{V}\)
W_rev = – \(\int_{V_{i}}^{v_{f}}\) \(\frac{n R T}{V}\) dV
W_rev = – nRTln(\(\frac{V_{f}}{V_{i}}\))
W_rev= -2.303nRTlog\(\frac{V_{f}}{V_{i}}\) ………….(5)

If V_f > V_i (expansion), the sign of work done by the process is negative.
If V_f < V_i (compression) the sign of work done on the process is positive.

Question 50.
Derive the relation between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.
Answer:
When the system at constant pressure undergoes changes from an initial state with H₁, U₁ and V₁ to a final state with H₂, U₂ and V₂ the change in enthalpy
∆H, can be calculated as follows:
H =U + PV
In the initial state
H₁ = U₁ + PV₁ ………………(1)
In the final state
H₂ = U₂ + PV₂ ……………..(2)
change in enthalpy is (2) – (1)
(H₂ – H₁) = (U₂ – U₁) + P(V₂ – V₁)
∆H = ∆U + P∆V ………………..(3)
As per first law of thermodynamics,
∆U = q + w
Equation (3) becomes
∆H = q + w+P∆V
w = -P∆V
∆H = qp – P∆V + P∆V
∆H = qp ………….(4)
qp – is the heat absorbed at constant pressure and is considered as heat content.
Consider a closed system of gases which are chemically reacting to form gaseous products at constant temperatureand pressure with V_i and V_f,as the total volumes of the reactant and product gases respectively, and niand nfas the number of moles of gaseous reactants and products, then,
For reactants (initial state):
PV_i = n_iRT ………..(5)
For products (final state):
PV_f = n_fRT ……………..(6)
(6) – (5)
P{V_f – V_i) = (n_f – n_i)RT
P∆V = ∆n_(g) RT ………..(7)
Substituting in (7) in (3)
∆H = ∆U+ ∆n_(g)RT …………….(8)

Question 51.
Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride crystal.
Answer:
The formation of NaCl can be considered in five steps. The sum of the enthalpy changes of these steps is equal to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.
Let us calculate the lattice energy of sodium chloride using Bom-Haber cycle

∆H = heat of formation of sodium chloride = -411.3 kJmol^-1
∆H₁ = heat of sublimation of Na_(s) = 108.7 kJmol^-1
∆H₂ = ionisation energy of Na_(s) = 495. kJ mol^-1
∆H₃ = dissociation energy of Cl_2(g) = 244 kJ mol^-1
∆H₄ = Electron affinity of Cl_(g) = -349 kJ mol^-1
U = lattice energy of NaCl
∆H_f = ∆H₁ + ∆H₂ + ∆H₃ + ∆H₄ + U
∴ U = (∆H_f ) – (∆H₁ + ∆H₂ + \(\frac{1}{2}\)∆H₃ + ∆H₄)
=» U = (-411.3) – (108.7 + 495.0 + 122 – 349)
U = (-411.3) – (376.7)
∴ U = -788kJ mol^-1

Question 52.
List the characteristics of Gibbs free energy.
Answer:
(i) G is defined as (H-TS) where H and S are the enthalpy and entropy of the system respectively. T = temperature. Since H and S are state functions, G is a state function.
(ii) G is an extensive property while ∆G = (G₂ – G₁) which is the free energy change between the initial (1) and final (2) states of the system becomes the intensive property when mass remains constant between initial and final states (or) when the system is a closed system.
(iii) G has a single value for the thermodynamic state of the system.
(iv) G and ∆G values correspond to the system only. There are three cases of ∆G in predicting the nature of the process. When, ∆G < 0 (negative), the process is spontaneous and feasible; ∆G = 0. The process is in equilibrium and ∆G > 0 (positive), the process is nonspontaneous and not feasible.
(v) ∆G = ∆H – T∆S. But according to I law of thermodynamics,
∆H = ∆G + P∆V and ∆U = q + w
∴ ∆G = q + w + P∆V – T∆S
But ∆S = \(\frac{T}{q}\) and T∆S = q = heat involved in the process.
∴ ∆G = q + w + P∆V – q = w + P∆V
(or) – ∆G = – w – P∆V = network.
The decrease in free energy – ∆G, accompanying a process taking place at constant temperature and pressure is equal to the maximum obtainable work from the system other than work of expansion.
This quantity is called as the “net work” of the system and it is equal to (- w – P∆V).
∴ Network = – w – P∆V
-∆G represents all others forms of work obtainable from the system such as electrical, chemical or surface work etc other than P-V work.

Question 53.
Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25°C and normal pressure.
Solution:
Given
n = 2 moles
V_i = 500 ml = 0.5 lit
V_f = 2 lit
T = 25°C = 298 K
w = -2303 nRTIog (\(\frac{V_{f}}{V_{i}}\))
w = -2.303 × 2 × 8.314 × 298 × log (4)
w = -2.303 × 2 × 8.314 × 298 × 0.6021
w = -6871 J
w = -6.871 kJ.

Question 54.
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5kJ K^-1. Calculate the enthalpy of combustion of the gas in kJ mol^-1.
Solution: Given
T_i = 298 K
T_f = 298.45 K
k = 2.5 kJ K^-1
m = 3.5g
Mm = 28
heat evolved = k∆T
= k(T_f – T_i)
= 2.5KJ K^-1 (298.45 – 298)K = 1.125 kJ
∆H_c = \(\frac{1.125}{3.5}\) × 28 kJ mol^-1

∆H_c = 9 kJmol^-1

Question 55.
Calculate the entropy change in the system, and surroundings, and the total entropy change in the universe during a process in which 245 J of heat flow out of the system at 77°C to the surrounding at 33°C.
Solution:
Given
T^sys = 77° C = (77 + 273) = 350 K
T^surr = 33°C = (33 + 273) = 360 K
q = 245J

∆S_sys = \(\frac{q}{T_{s y}}=\frac{-245}{350}\) = -0.7 JK^-1

∆S_surr = \(\frac{q}{T_{s \mathrm{sr}}}=\frac{+245}{306}\) = +0.8 JK^-1

∆S_univ = ∆S_sys + ∆S_surr

∆S_univ = -0.7 JK^-1 + 0.8 JK^-1

∆S_univ = 0.1 JK^-1

Question 56.
1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710/and expands to 2 litres. Calculate the entropy change in expansion process.
Solution:
Given
n = 1 mole
P = 4.1 atm
V = 2 Lit
T = ?
q = 3710 J
∆S = \(\frac{q}{T}\)

∆S = \(\frac{q}{\frac{p v}{n R}}\)

∆S = \(\frac{n R q}{P V}\)

∆S = \(\frac{1 \times 0.082 \text { lit atm } K^{-1} \times 3710 J}{4.1 \text { atm } \times 2 \text { lit }}\)

∆S = 37.10 JK^-1

Question 57.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride.
Answer:
Given,
∆H_f(NaCl) = 30.4 kJ = 30400 J mol^-1
∆S_f(NaCl) = 28.4 JK^-1 mol^-1
∆S^-1 = \(\frac{\Delta H_{f}}{T_{f}}\)

T_f = \(\frac{\Delta H_{f}}{\Delta S_{f}}\)

T_f = \(\frac{30400 J m o l^{-1}}{28.4 J K^{-1} m o l^{-1}}\)

T_f = 1070.4 K

Question 58.
Calculate the standard heat of formation of propane, if its heat of combustion is -2220.2 kJ mol^-1. The heats of formation of CO₂(g) and H₂O(1) are -393.5 and -285.8 kJ mol^-1 respectively.
Solution:
Given:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
∆H°C = 2220.2kJ mol^-1 ……………..(1)
C + O₂ → CO₂
∆H°_f = -393.5 kf mol^-1 ……….(2)
H₂ + \(\frac{1}{2}\)O₂ → H₂O
∆H°_f = -285.8 kJ mol^-1 ……………(3)
3C + 4H₂ → C₃H₈
∆H°_c =?
(2) × (3)
⇒ 3C + 3O₂ → 3CO₂
∆H°_f = -1180.5 kJ ………..(4)
(3)× (4)
⇒ 4H₂ + 2O₂ → 4H₂O
∆H°_f = 1143.2 kJ ……..(5)
(4) + (5) – (1)
⇒ 3C + 3O₂ + 4H₂ + 2O₂ + 3CO₂ + 4H₂O → 3CO₂ + 4H₂O + C₃H₈ + 5O₂
∆H°_f = -1180.5 – 1143.2 – (-2220.2) kJ
3C + 4H₂ → C₃H₈
∆H°_f = -103.5 kJ
Standard heat of formation of propane is ∆H°_f(C3H8) = -103.5 kJ

Question 59.
You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below.

Answer:
For ethanol
Given:

Question 60.
For the reaction Ag₂O(s) → 2Ag(s)+12O₂(g) ∆H = 30.56 kJ mol^-1 and ∆S = 6.66JK^-1 mol^-1 (at 1 atm). Calculate the temperature at which G is equal to zero. Also predict the direction of the reaction (I) at this temperature and (ii) below this temperature.
Solution:
Given,
∆H = 30.56 kJ mol^-1
∆S = 6.66 x 10^-3 kJK^-1 mol^-1
T = ? at which ∆G =0
∆G = ∆H – T∆S
0 = ∆H – T∆S
T = \(\frac{\Delta H}{\Delta S}\)

T = \(\frac{30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}}{66.6 \times 10^{-3} \mathrm{~kJ} K^{-1} \mathrm{~mol}^{-1}}\)
T = 4589 K

(i) At 4589K;
∆G = 0 the reaction is in equilibrium.
(ii) at temperature below 4598 K
∆H > T∆S
∆G = ∆H – T∆S > 0, the reaction in the forward direction, is non spontaneous. In other words the reaction occurs in the backward direction.

Question 61.
What is the equilibrium constant K_eq for the following reaction at 400K.
2NOCl(g) ⇌ 2NO(g) + Cl₂(g), given that ∆H° = 77.2 kJ mol^-1 ∆S° = 122 JK^-1 mol^-1
Answer:

Question 62.
Cyanamide (NH₂CN) is completely burnt in excess oxygen in a bomb calorimeter, ∆U was found to be -742.4 kJ mol^-1, calculate the enthalpy change of the reaction at 298K.
NH₂CN(s) + \(\frac{3}{2}\)O₂(g) → N₂(g) + CO₂(g) + H₂O (l) ∆H = ?
Solution:
Given
T = 298 K;
∆U = -742.4 kJ mol^-1
∆H =?
∆H = ∆U + ∆n_(g)RT
∆H = ∆U + (n_p – n_r)RT
∆H = 742.4 + 2 – \(\frac{3}{2}\) × 8.314 × 10^-3 × 298
= -742.4 + (0.5 × 8.314 ×10^-3 × 298)
=-742.4 + 1.24
= -741.16 kJ mol^-1

Question 63.
Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of C – H, C – C, C = C and H – Hare 414, 347, 618 and 435 kJ mol^-1.
Solution:
Given
E_C-H= 414 kJ mol^-1
E_C-C= 347 kJ mol^-1
E_C-C= 6l8 kJ mol^-1
E_H-H = 435 kJ mol^-1

∆H_r = Σ(Bond energy)_r – Σ(Bond energy)_p
∆H_r = (E_C=C + 4E_C-H + E_H-H) – (E_C-C + 6E_C-H)
∆H_r = (618 + (4 × 414) + 315) – (347 + (6 ×414))
∆H_r = 2709 – 2831
∆H_r = -122 kJ mol^-1

Question 64.
Calculate the lattice energy of CaCl2 from the given data
Ca(s) + Cl₂(g) → CaCl₂(s) ∆H°_f = – 795 kJ mol^-1
Atomisation:
Ca(s) → Ca(g) ∆H°₁ = -795 kJ mol^-1

Ionisation:
Ca(g) → Ca²⁺(g) + 2e^-1
∆H°₂ = 2422 kJ mol^-1

Dissociation:
Cl₂(g) → 2Cl(g)
∆H°₃ = + 242.8 kJ mol^-1

Electron Affinity:
Cl(g) + e^– → Cl^-1
∆H°₄ = -355 kJ mol^-1
Answer:
∆H_f = ∆H₁ + ∆H₂ + ∆H₃+∆H₄ + u
-795 = 121 + 2422 + 242.8 + (2 × -355) + u
-795 = 2785.8 – 710 + u
-795 = 2075.8 + u
u = -795 – 2075.8
u = -2870.8 kJ mol^-1

Question 65.
Calculate the enthalpy change for the reaction Fe203 + 3C0 – 2Fe + 3C02 from the following data.
2Fe + \(\frac{3}{2}\)O₂ → Fe₂O₃; ∆H = -741 kJ
C + \(\frac{3}{2}\)O₂ → CO; ∆H=-137kJ
C + O₂ → CO2; ∆H = – 394.5 kJ
Answer:

Question 66.
When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne(A), 95.2% 2-pentyne(B), and 3.5% of 1,2 pentadiene (C) the equilibrium was maintained at 175°C, calculate ∆G° for the following equilibria.
B ⇌ A; ∆G°₁ =?
B ⇌ C; ∆G°₂ =?
Solution:
Given
T = 175°
C = 175 + 273 = 448 K
Concentration of 1-pentyne [A] = 1.3%
Concentration of 2-pentyne [B] = 95.2%
Concentration of 1, 2-pentadiene [C] = 3.5%
At equilibrium
B ⇌ A
95.2% 1.3% ⇒
B ⇌ C
95.2% 3.5% ⇒
K₂ = \(\frac{3.5}{95.2}\) = 0.0367
⇒ ∆G°₁ = -2.303 RTlogK
∆G°₁ = – 2.303 × 8.314 × 448 × log 0.0136
∆G°₁ = +16010 J
∆G°₁ = +16 kJ

⇒ ∆G°₂ = – 2.303 RT log K₂
∆ G°₂ = -2.303 × 8.314 × 448 × log 0.0367
∆ G°₂ = +12312J
∆G°₂ = +12.312 kJ

Question 67.
At 33K, N₂O₄ is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
Solution:
Given
T= 33 K
N₂O₄ ⇌ 2NO₂
Initial concentration: 100%  0
Concentration dissociated 50% – Concentration remaining at equilibrium 50% 100%
K_eq = \(\frac{100}{50}\) = 2
∆G° = -2.303 RT log K_eq
∆G°= -2.303 × 8.314 × 33 × log 2
∆G° = -190.18 Jmol^-1

Question 68.
The standard enthalpies of formation of SO₂ and SO₃ are – 297 kJ mol^-1 and – 396 kJ mol^-1 respectively. Calculate the standard enthalpy of reaction for the reaction: SO +4-0 — SO
Solution:
Given
∆H°_f(SO₂) = – 297 kJ mol^-1
∆H°_f(SO₃) = – 396 kJ mol^-1
SO₂ + 12O₂ → SO₃;
∆H°_r =?
∆H°_r = (∆H°_f)_compound – Σ( ∆H_f)_elements
∆H°_r =∆H°_r(SO₃) – ∆H°_f
∆H°_r = 396 kJ mol^-1– (- 297 kJ mol^-1 + 0)
∆H°_r = – 396 kJ mol^-1 + 297
∆H°_r = -99 kJ mol^-1

Question 69.
For the reaction at 298 K: 2A + B → C
∆H = 400 Jmol^-1; ∆S = 0.2 JK∆ mol^-1 Determine the temperature at which the reaction would be spontaneous.
Solution:
Given,
T = 298 ∆T
∆H = 400 J mol^-1 = 400 J mol^-1
∆S = 0.2 JK^-1 mol^-1
∆G = ∆H – T∆S
if T = 2000 K
∆G = 400 – (0.2 × 2000) = 0
if T > 2000 K
∆G will be negative. The reaction would be spontaneous only beyond 2000 K

Question 70.
Find out the value of equilibrium constant for the following reaction at 298K,
2NH₃ + CO₂ ⇌ NH₂CONH₂ (aq) + H₂O (l) Standard Gibbs energy change, ∆G°_r at the given temperature is -13.6 kJ mol^-1.
Solution:
Given:
T = 298 K
∆G°_r = -13.6 kJ mol^-1
= – 13600 J mol^-1
∆G° = – 2.303 RT log K_eq
log K_eq = \(\frac{-\Delta G^{0}}{2.303 R T}\)

log K_eq = \(\frac{13.6 \mathrm{~kJ} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \times 10^{-3} \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}}\)

log K_eq = 2.38
K_eq = antilog (2.38)
K_eq = 239.88

Question 71.
A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25°C and at 1 atm pressure produce 6.11 lit of carbondioxide. Find out the amount of heat evolved in kJ, during this combustion. (∆H_C(CH4)) = -890 kJmol^-1 and ∆H_C(C2H4) = -1423 kJ mol^-1
Solution:
Given
∆H_C(CH4) = – 890 kJ mol^-1
∆H_C(C2H4) = -1423 kJ mol^-1
Let the mixture contain x lit of CH₄ and (3.67 – x) lit of ethylene.
CH₄ + 2O₂ → CO + 2H₂O
x lit               x lit
C₂H₄ + 3O₂ → 2 CO₂ + 2H₂O
(3.67 -x) lit 2 (3.67 -x) lit
Volume of Carbondioxide formed = x + 2(3.67 – x) = 6.11 lit
x + 7.34 – 2x = 6.11
7.34 – x = 6.11
x = 1.23 lit
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence

11th Chemistry Guide Thermodynamics Additional Questions and Answers

I. Choose the best answer:

Question 1.
_______is a part of universe which is separated from the rest of the universe by real or imaginary boundaries.
(a) Surroundings
(b) boundary
(c) system
(d) matter
Answer:
(c) system

Question 2.
Which one of the following is a closed system?
(a) Hot water in a closed beaker
(b) Hot water in a thermos flask
(c) Hot water in a open beaker
(d) Chemical reactions
Answer:
(a) Hot water in a closed beaker

Question 3.
Which of the following is/are extensive properties?
1. volume
2. Surface tension
3. mass
4. internal energy
(a) 1, 2 and 4
(b) 1, 3 and 4
(c) 1 and 3
(d) 1, 2 and 3
Answer:
(b) 1, 3 and 4

Question 4.
Which of the following is/are intensive properties?
1. refractive index
2. density
3. number of moles
4. molar volume
(a) 1, 2 and 4
(b) 1, 3 and 4
(c) 1, 2 and 3
(d) 2, 3 and 4
Answer:
(b) 1, 3 and 4

Question 5.
For an isothermal process _______.
(a) dT = 0
(b) dV = 0
(c) dq = 0
(d) dP = 0
Answer:
(a) dT = 0

Question 6.
Which of the following is/are state function?
1. Pressure
2. work
3. internal energy
4. Free energy
5. heat
(a) 1, 2 and 4
(b) 1, 3 and 4
(e) 1, 2 and 3
(d)2, 3 and 4
Answer:
(b) 1, 3 and 4

Question 7.
Which of the following is/are path function?
1. pressure
2. work
3. internal energy
4. Free energy
5. heat
(a) 1, 2 and 4
(b) 1, 3 and 4
(c) 2 and 5
(d) 2, 3 and 4
Answer:
(c) 2 and 5

Question 8.
Which is the correct about internal energy, U?
(a) It is an extensive property
(b) It is a state function
(c) For a cyclic process. ∆U = 0.
(d) Change in internal energy is ∆U = U_i – U_f
Answer:
(a) It is an extensive property

Question 9.
Work done by’ a given system with an ideal gas in a reversible process is w_rev =
(a) – nRln (V_f/V_i)
(b) – nRTln (V_i/V_f)
(c) – nRTln (V_f/V_i)
(d) – nTln (V_f/V_i)
Answer:
(c) – nRTln (V_f/V_i)

Question 10.
For a cyclic process involving isothermal expansion of an ideal gas
(a) ∆U = 0
(b) ∆U = q
(c) ∆U = q + w
(d) ∆U = q – w
Answer:
(a) ∆U = 0

Question 11.
Choose the correct statement about enthalpy
(a) Enthalpy is a path function
(b) Enthalpy change ∆H = ∆U + V∆P
(c) In an endothermic process, ∆H = 0.
(d) In an exothermic process. ∆H is negative
Answer:
(d) In an exothermic process. ∆H is negative

Question 12.
Which ofthc following reactions correctly indicates the process of atomization?
(a) CH₄ → CH₃(g) + H(g)
(b) CH₄(g) → CH(g) + 4H(g)
(c) CH(g) + O₂(g) → CO₂(g)
(d) N₂(g) + 3H₂(g) → 2NN₃(g)
Answer:
(b) CH₄(g) → CH(g) + 4H(g)

Question 13.
A reaction has both H and S negative. The rate of reaction
(a) increases with increase of temperature
(b) increases with decrease of temperature
(e) remains unaffected by change of temperature
(d) cannot be predicted for change in temperature
Answer:
(b) increases with decrease of temperature

Question 14.
Evaporation of water is
(a) an exothermic change
(b) an endothermic change
(e) a process where no heat changes occur
(d) a process accompanied by chemical reaction
Answer:
(b) an endothermic change

Question 15.
The entropy change for a non spontaneous reaction is 140 JK^-1 mol at 298 K. The reaction is
(a) reversible
(b) irreversible
(c) exothermic
(d) endothermic
Answer:
(d) endothermic

Question 16.
The enthalpy and entropy change for the reaction: Br₂(l) +Cl₂(g) → 2BrCl(g) are 30 kJ mol^-1 and 105 kJ mol^-1 respectively. The temperature at which the reaction will be in
equilibrium is
(a) 300 K
(b) 285.7 K
(c) 273 K
(d) 450 K
Answer:
(b) 285.7 K

Question 17.
Three moles ofan ideal gas expanded spontaneously into vaccum. The work done will be
(a) Infinite
(b) 3 JouIes
(c) 9 Joules
(d) zero
Answer:
(d) zero

Question 18.
If the enthalpy change for the transition of liquid water to steam is 30 kJ mol^-1 at 27°C, the entropy change for the process would be
(a) 10 J mol^-1K^-1
(b) 1.0 J mol^-1K^-1
(c) 0.1 J mol^-1K^-1
(d) 100 J mol^-1K^-1
Answer:
(d) 100 J mol^-1K^-1

Question 19.
Enthalpy change for the reaction,
4H(g) → 2H₂(g) is – 869.6 kJ
The dissociation energy of H-H bond is
(a) -434.8 kJ
(b) -869.6 kJ
(c) + 434.8 kJ
(d) +217.4 kJ
Answer:
(c) + 434.8 kJ

Question 20.
Which of the following is correct option for free expansion of an ideal gas under adiabatic condition?
(a) q = 0, ∆T ≠ 0, w = 0
(b) q ≠ 0, ∆T = 0, w = 0
(c) q = 0, ∆T = 0, w = 0
(d) q = 0, ∆T < 0, w ≠ 0
Answer:
(b) q ≠ 0, ∆T = 0, w = 0

Question 21.
The total entropy change for a system and its surroundings increases, if the process is
(a) reversible
(b) irreversible
(c) exothermic
(d) endothermic
Answer:
(b) irreversible

Question 22.
∆S is positive for the change
(a) mixing of two gases
(b) boiling of liquid
(c) melting of solid
(d) all of these
Answer:
(d) all of these

Question 23.
If w₁, w₂, w₃ and w₄ are work done in isothermal, adiabatic, isobaric and isochoric reversible processes, the correct order (for expansion) will be
(a) w₁ > w₂ > w₃ > w₄
(b) w₃ > w₂ > w₁ > w₄
(c) w₃ > w₂ > w₄ > w₁
(d) w₃ > w₁ > w₂ > w₄
Answer:
(d) w₃ > w₁ > w₂ > w₄

Question 24.
The subject matter of thermodynamics comprises
(a) energy transformations in a system
(b) mass changes in molecular reactions
(c) total energy of system
(d) rates of chemical reactions
Answer:
(a) energy transformations in a system

Question 25.
A hydrogen bond is very useful in determining the structures and properties of compounds. Its energy varies between
(a) 12 – 20 kJ /mol
(b) 50 – 100 kJ/mol
(c) 10 – 100 kJ/mol
(d) 75 – 200 kJ/mol
Answer:
(c) 10 – 100 kJ/mol

Question 26.
In which of the enlisted cases, Hess’s law is not applicable?
(a) Determination of lattice energy
(b) Determination of resonance energy
(c) Determination of enthalpy of transformation of one allotropic form to another
(d) Dtermination of entropy
Answer:
(d) Dtermination of entropy

Question 27.
The amount of energy required to completely remove the constituent ions from its crystal lattice to an infinite distance is called
(a) lattice energy
(b) ionization energy
(c) internal energy
(d) free energy
Answer:
(a) lattice energy

Question 28.
Statement -1:
The allotropes of carbon, namely, graphite and diamond differ each other.
Statement -2:
They possess different internal energies and have different structures.
In the above statement/s
(a) 1 alone is correct
(b) 2 alone is correct
(c) both 1 and 2 are correct
(d) both 1 and 2 are incorrect
Answer:
(c) both 1 and 2 are correct

Question 29.
Which of the following statement is incorrect? According to thermodynamics, work
(a) is a path function
(b) appears only at the boundary of the system
(c) appears during the change in the state of the system.
(d) surroundings is so large that macroscopic changes occurs in the surroundings
Answer:
(d) surroundings is so large that macroscopic changes occurs in the surroundings

Question 30.
The work done by a force of one Newton through a displacement of one meter is called
(a) joule
(b) calorie
(c) erg
(d) tesla
Answer:
(a) joule

Question 31.
The enthalpy of neutralization of strong acid vs strong base is approximately equal to ________(in kJ).
(a) 57.32
(b) – 57.32
(c) 5.98
(d) – 5.98
Answer:
(b) – 57.32

Question 32.
The maximum efficiency of an automobile engine working between the temperatures 816°C and 21°C is
(a) 73 %
(b) 45%
(c) 67%
(d) 78%
Answer:
(b) 45%

Question 33.
A reaction that occurs under the given set of conditions without any external driving force is called a reaction.
(a) Reversible
(b) spontaneous
(c) irrversible
(d) cyclic
Answer:
(b) spontaneous

Question 34.
Which one of the following spontaneous reaction is endothermic?
(a) combustion of methane
(b) dissolution of ammonium nitrate
(c) acid-base neutralization reaction
(d) none of the above
Answer:
(b) dissolution of ammonium nitrate

Question 35.
The available energy in the system to do work is called
(a) Gibbs free energy
(b) internal energy
(c) potential energy
(d) kinetic energy
Answer:
(a) Gibbs free energy

Question 36.
Which one of the following is incorrect about Gibbs free energy?
(a) Extensive property
(b) path function
(c) ∆G < 0 for a spontaneous process (d) ∆G > 0 for a non-spontaneous process
Answer:
(b) path function

Question 37.
Match the following:

(a) A – iii, B – iv, C – i, D – ii
(b) A – ii, B – iv, C – iii, D – i
(c) A – iv, B – iii, C – i, D – ii
(d) A – i, B – iv, C – ii, D – iii
Answer:
(a) A – iii, B – iv, C – i, D – ii

Question 38.
For a cyclic process involving isothermal expansion of an ideal gas, q =
(a) 0
(b) P∆V
(c) w
(d) – w
Answer:
(d) – w

Question 39.
In Calorimeter, the expression used to calculate the amount of heat change in the process is
(a) C = qm∆T
(b) C = m/q∆T
(c) C = q/m∆T
(d) C = q∆T/m
Answer:
(c) C = q/m∆T

Question 40.
“It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing some work”. This statement is given by
(a) Clausius
(b) Kelvin
(c) Gibbs
(d) Joule
Answer:
(a) Clausius

II. Very short question and answers (2 Marks):

Question 1.
Define isolated system.
Answer:
A system which can exchange neither matter nor energy with its surroundings is called an isolated system. Here boundary is sealed and insulated.

Question 2.
What is a homogeneous system? Give an example.
Answer:
A system is called homogeneous if the physical state of all its constituents is the same. Example: a mixture of gases.

Question 3.
Illustrate Closed system with an example.
Answer:
Hot water contained in a closed beaker is an example for a closed system. In this system energy (heat) is transferred to the surroundings but no matter (water vapour) can escape from this system. A gas contained in a cylinder fitted with a piston constitutes a closed system.

Question 4.
Define irreversible process.
Answer:
The process in which the system and surrounding cannot be restored to the initial state from the final state is called an irreversible process. All the processes occurring in nature are irreversible processes.

Question 5.
Define standard entropy change.
Answer:
The absolute entropy of a substance at 298 K and one bar pressure is called the standard entropy S°.

Question 6.
What is Zeroth law of thermodynamics?
Answer:
The law states that’ If two systems are separately in thermal equilibrium with a third one, then they tends to be in thermal equilibrium with themselves’.

Question 7.
Define enthalpy.
Answer:
The enthalpy (H), is a thermodynamic property of a system, is defined as the sum of the internal energy (U) of a system and the product of pressure and volume of the system.
H = U + PV

Question 8.
Define specific heat capacity of a system.
Answer:
The heat absorbed by one kilogram of a substance to raise its temperature by one Kelvin at a specified temperature.

Question 9.
What is a reversible process?
Answer:
The process in which the system and surrounding can be restored to the initial state from the final state without producing any changes in the thermodynamic properties of the universe is called a reversible process.

Question 10.
What is a thermodynamic process? Give two examples.
Answer:
The method of operation which can bring about a change in the system is called thermodynamic process.
Examples: Heating, Cooling, expansion.

Question 11.
Distinguish between state function and path function.
Answer:
A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path by which the particular state is reached. A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.

Question 12.
Write any two characteristics of internal energy.
Answer:

1. The internal energy of a system is an extensive property.
2. The internal energy of a system is a state function.

Question 13.
Write the importance of internal energy.
Answer:
The internal energy possessed by a substance differentiates its physical structure. For example, the allotropes of carbon, namely, graphite C (graphite) and diamond C (diamond), differ from each other because they possess different internal energies and have different structures.

Question 14.
Write notes on heat.
Answer:
The heat (q) is regarded as an energy in transit across the boundary separating a system from its surrounding. Heat changes lead to temperature differences between system and surrounding. Heat is a path function.

Question 15.
Write the thermodynamic significance of work.
Answer:
The work

1. is a path function.
2. appears only at the boundary of the system.
3. appears during the change in the state of the system.
4. in thermodynamics, surroundings is so large that macroscopic changes to surroundings do not happen.

Question 16.
What is meant by pressure-volume work?
Answer:
In elementary thermodynamics the only type of work generally considered is the work done in expansion (or compression) of a gas. This is known as pressure-volume work, PV work or expansion work.

Question 17.
What is specific heat capacity of a system?
Answer:
The heat absorbed by one kilogram of a substance to raise its temperature by one Kelvin at a specified temperature is called specific heat capacity of a system.

Question 18.
Distinguish between Cv and Cp.
Answer:
The molar heat capacity at constant volume (Cv) is defined as the rate of change of internal energy with respect to temperature at constant volume. The molar heat capacity at constant pressure (Cp) can be defined as the rate of change of enthalpy with respect to temperature at constant pressure.

III. Short Question and Answers (3 Marks):

Question 1.
What is internal energy?
Answer:
The internal energy of a system is equal to the energy possessed by all its constituents namely atoms, ions and molecules. The total energy of all molecules in a system is equal to the sum of their translational energy (U_t), vibrational energy (U_v), rotational energy (U_r), bond energy (U_b), electronic energy (U_e) and energy due to molecular interactions (U_i).
Thus:
U = U_t + U_v + U_r + U_b + U_e + U_i
The total energy of all the molecules of the system is called internal energy.

Question 2.
Give applications of bomb calorimeter.
Answer:

1. Bomb calorimeter is used to determine the amount of heat released in combustion reaction.
2. It is used to determine the calorific value of food.
3. Bomb calorimeter is used in many industries such as metabolic study, food processing, explosive testing etc.

Question 3.
What is Molar heat of fusion?
Answer:
The molar heat of fusion is defined as “the change in enthalpy when one mole of a solid substance is converted into the liquid state at its melting point”.
For example, the heat of fusion of ice
H₂O(s) → H₂O(l)
∆H(fusion) = +5.98 kJ

Question 4.
Write any three statements of first law of thermodynamics.
Answer:

1. Whenever an energy of a particular type disappears, an equivalent amount of another type must be produced.
2. The total energy of a system and surrounding remains constant (or conserved)
3. “Energy can neither be created nor destroyed, but may be converted from one form to another”.

Question 5.
Describe the need for the second law of thermodynamics.
Answer:
From the first law of thermodynamics, the energy of the universe is conserved. Let us consider the following processes:
A glass of hot water over time loses heat energy to the surrounding and becqmes cold.
When you mix hydrochloric acid with sodium hydroxide, it forms sodium chloride and water with evolution of heat.

In both these processes, the total energy is conserved and are consistent with the first law of thermodynamics. However, the reverse process i.e. cold water becoming hot water by absorbing heat from surrounding on its own does not occur spontaneously even though the energy change involved in this process is also consistent with the first law. However, if the heat energy is supplied to cold water, then it will become hot. i.e. the change that does not occur spontaneously and an be driven by supplying energy.

Question 6.
Write notes on entropy statement of second law of thermodynamics.
Answer:
The second law of thermodynamics can be expressed in terms of entropy, i.e “the entropy of an isolated system increases during a spontaneous process”.
For an irreversible process such as spontaneous expansion of a gas,
S_total > 0
S_total > S_system + S_surrounding
i.e., S_universe > S_system + S_surrounding
For a reversible process such as melting of ice,
S_system = S_surrounding
S_universe = 0

IV. Long Question and Answers(5 Marks):

Question 1.
Explain steps to write thermochemical equations.
Answer:
A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (∆H). The following conventions are adopted in thermochemical equations:

1. The coefficients in a balanced thermochemical equation refer to number of moles of reactants and products involved in the reaction.
2. The enthalpy change of the reaction ∆Hr has to be specified with appropriate sign and unit.
3. When the chemical reaction is reversed, the value of ∆H is reversed in sign with the same magnitude.
4. The physical states (gas, liquid, aqueous, solid in brackets) of all species are important and must be specified in a thermochemical reaction, since ∆H depends on the physical state of reactants and products.
5. If the thermochemical equation is multiplied throughout by a number, the enthalpy change is also multiplied by the same number.
6. The negative sign of ∆H_r indicates that the reaction is exothermic and the positive sign of ∆H_r indicates an endothermic reaction.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 13 Photosynthesis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis

11th Bio Botany Guide Photosynthesis Text Book Back Questions and Answers

Part-I

Questions 1.
Assertion (A): Increase in Proton gradient inside lumen responsible for ATP synthesis.
Reason (R): Oxygen evolving complex of PS I located on the thylakoid membrane facing Stroma, releases H⁺ ions.
(a) Both Assertion and Reason are True.
(b) Assertion is True and Reason is False.
(c) Reason is True and Assertion is False.
(d) Both Assertion and Reason are False.
Answer:
(a) Both Assertion and Reason are True.

Question 2.
Which chlorophyll molecule does not have a phytol tail?
a) Chl – a
b) Chl – b
c) Chl – c
d) Chl – d
Answer:
(c) Chl – c

Question 3.
The correct sequence of flow of electrons in the light reaction is:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.
(b) PS I, plastoquinone, cytochrome, PS II ferredoxin.
(c) PS II, ferredoxin, plastoquinone, cytochrome, PS I.
(d) PS I, plastoquinone, cytochrome, PS II, ferredoxin.
Answer:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.

Question 4.
For every CO₂ molecule entering the C₃ cycle, the number of ATP & NADPH required
a) 2ATP + 2NADPH
b) 2ATP + 3NADPH
c) 3ATP + 2NADPH
d) 3ATP + 3NADPH
Answer:
(c) 3ATP + 2NADPH

Question 5.
Identify true statement regarding light reaction of photosynthesis?
(a) Splitting of water molecule is associate with PS I.
(b) PS I and PS II involved in the formation of NDPH + H⁺.
(c) The reaction center of PS I is Chlorophyll a with absorption peak at 680 nm.
(d) The reaction center of PS II is Chlorophyll a with absorption peak at 700 nm.
Answer:
(b) PS I and PS II involved in the formation of NDPH + H⁺.

Question 6.
Two groups (A&B) of bean plants of similar size and same leaf area were placed in identical conditions. Group A was exposed to light of wavelength 400 – 450 nm & Group B to light of wavelength of 500 – 550nm. Compare the photosynthetic rate of the 2 groups giving reasons.
Answer:
‘A’ group of plants exposed to light of 400 – 450nm. Chlorophyll a shows maximum absorption peak at 450nm (blue region). Hence rate of photosynthesis was high.
‘B’ group of plants exposed to light of 500 – 550nm. This wavelength refers to green region of the spectrum. Chlorophyll does not absorb light in the green region but reflects green. So plants appear green rate of photosynthesis was negligible in these plants.

Question 7.
A tree is believed to be releasing oxygen during nighttime. Do you believe the truthfulness of this statement? Justify your answer by giving reasons?
Answer:
Yes, a tree is believed to be releasing O₂ during nighttime because at night CAM plants fix CO₂ with the help of phospho Enol Pyruvic acid and produce oxala acetic acid, which is converted into malic acid-like C₄ cycle.

Question 8.
Grasses have an adaptive mechanism to compensate for photorespiratory losses – Name and describe the mechanism.
Answer:
The photorespiratory losses are checked by certain grasses by having physiological adaptation. The process of photosynthesis occurs in mesophyll cells and bundle sheath cells.

Mesophyll cells:

• Initially, CO₂ is taken up by Phosphoenolpyruvate (PEPA) (3C) and changed to oxaloacetate (4C) in the presence of PEP carboxylase.
• Oxaloacetate is reduced to Malate/Aspartate. The product formed reaches the bundle sheath.

Bundle Sheath:

• The oxidation of Malate and Aspartate occurs with the release of carbon dioxide and the formation of Pyruvate (3C)
• Due to increased CO₂ concentration RUBISCO functions as a carboxylase and not as Oxygenase.
• The photosynthetic losses are prevented.
• RUBP operates now under the Calvin cycle and pyruvate transported back to Mesophyll cells is changed into Phosphoenolpyruvate to keep the cycle going.

Question 9.
In Botany class, teacher explains, Synthesis of one glucose requires 30 ATPs in C₄ plants and only 18 ATPs in C3 plants. The same teacher explains C₄ plants are more advantageous than C₃ plants. Can you identify the reason for this contradiction?
Answer:
C₄ Plants are more advantageous than C₃ plants because of the following reasons:

Question 10.
When there is plenty of light and higher concentration of O₂, what kind of pathway does the plant undergo? Analyse the reasons.
Answer:
The rate of photosynthesis decreases when there is an increase in oxygen concentration. This Inhibitory effect of oxygen was first discovered by Warburg (1920) using green algae, Chlorella.

Part-II.

11th Bio Botany Guide Photosynthesis Additional Important Questions and Answers

I. Choose the Correct Answers

Question 1.
Photosynthesis is the major:
(a) endothermic reaction
(b) exothermic reaction
(c) endergonic reaction
(d) exergonic reaction
Answer:
(c) endergonic reaction

Question 2.
The physiological unit of photosynthesis is
a) 150-250 Chlorophyll molecules
b) 200-300 chlorophyll molecules
c) 440-660 chlorophyll molecules
d) 450- 650 chlorophyll molecules
Answer:
b) 200-300 chlorophyll molecules

Question 3.
How many million tonnes of dry matter produced annually by photosynthesis?
(a) 1700 million tonnes
(b) 1900 million tonnes
(c) 1400 million tonnes
(d) 2000 million tonnes
Answer:
(a) 1700 million tonnes

II. Match Correctly & Choose The Right Answer

Question 4.
I) Black Mann – A) the importance of Chlorophyll
II) Warburg – B) Law of limiting factor
III) Dustrochet – C) C4 cycle
IV) Hatch & Slack – Chlorella

Answer:
c) B-D-A-C

Question 5.
Thylakoid disc diameter is:
(a) 0.35 to 0.75 microns
(b) 0.25 to 0.8 microns
(c) 0.45 to 0.8 microns
(d) 0.50 to 0.9 microns
Answer:
(b) 0.25 to 0.8 microns

Question 6.
The pigment responsible for the yellowing of leaves during autumn season is
a) Violaxanthin
b) Fucoxanthin
c) Phycobillin
d) Lycopene
Answer:
d) Lycopene

Question 7.
The no of quanta of light required for the release of one oxygen molecule
a) 18 quanta
b) 8 quanta
c) 81 quanta
d) 19 quanta
Answer:
b) 8 quanta

Question 8.
Each pyrrole ring comprises of:
(a) six carbons and one nitrogen atom
(b) three carbons and one nitrogen atom
(c) four carbons and one nitrogen atom
(d) four carbons and two nitrogen atom
Answer:
(c) four carbons and one nitrogen atom

Question 9.
RUBISCO – Constitute …………. of chloroplast protein
a) 17%
b) 20%
c) 18%
d) 16%
Answer:
d) 16%

Question 10.
Pheophytin resembles chlorophyll ‘a’ except that it lacks:
(a) Fe atom
(b) Mn atom
(c) Mg atom
(d) Cu atom
Answer:
(c) Mg atom

Question 11.
According to Emerson the fall in quantum yield about 680 nm is called
a) Phoisynthtic drop
b) Emerson drop
c) Airburg effect
d) Red drop
Answer:
d) Red drop

Question 12.
Which one of the photosynthetic pigments is called shield pigment:
(a) carotenes
(b) chlorophyll ‘b’
(c) pheophytin
(d) carotenoids
Answer:
(d) carotenoids

Question 13.
Which of the following equation correctly sums up photosynthesis

Answer:

Question 14.
Photosynthetic rate of red light (650 nm) is equal to:
(a) 42.5
(b) 10.0
(c) 43.5
(d) 40.8
Answer:
(c) 43.5

Question 15.
Which photosystem is found to be located on the outer surface of the thylakoid?
a) PS I
b) P.S II
c) P. 890
d) Both (a) and (b)
Answer:
a) P.S I

Question 16.
Phosphorylation taking place during respiration is called as:
(a) Photophorylation
(b) Oxidative phosphorylation
(c) Reductive phosphorylation
(d) None of the above
Answer:
(b) Oxidative phosphorylation

Question 17.
The term Quantosome was coined by
a) Emerson
b) Liebig
c) Calvin & Melvin
d) Park & Biggins
Answer:
d) Park & Biggins

Question 18.
In bioenergetics of light reaction, to release one electron from pigment system it requires:
(a) two quanta of light
(b) four quanta of light
(c) one quantum of light
(d) eight quanta of light
Answer:
(a) two quanta of light

Question 19.
Photosynthesis produces
a) 1700 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year
b) 7100 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year
c) 1600 million tonnes of dry matter/week by fixing 56 x 12¹⁰ kg of carbon every week
d) 6100 million tonnes of dry matter/month by fixing 100 x 10¹² kg of carbon every month
Answer:
a) 1700 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year

Question 20.
In C₄ plants, how many ATPs and NADPH + H⁺ are utilised for the release of one oxygen molecule:
(a) 3 ATPs and 2 NADPH + H⁺
(b) 4 ATPs and 3 NADPH + H⁺
(c) 2 ATPs and 2 NADPH + H⁺
(d) 5 ATPs and 2 NADPH + H⁺
Answer:
(d) 5 ATPs and 2 NADPH + H⁺

Question 21.
Products of light reaction in photosynthesis are
a) ATP & NADPH₂
b) ADP & glucose
c) Ferredoxin and cytochrome b6
d) Cytochrome
Answer:
a) ATP & NADPH₂

Question 22.
In the sugarcane plant, the dicarboxylic acid pathway was first discovered by:
(a) Hatch and Slack
(b) Kortschak, Hart and Burr
(c) Calvin and Benson
(d) Mitchell and Root
Answer:
(b) Kortschak, Hart and Burr

Question 23.
Photosynthetic pigments in chloroplasts lie embedded in
a) Chloroplast envelope
b) Plastogloblue
c) matrix
d) thylakoids
Answer:
d) thylakoids

Question 24.
Indicate the correct answer:
(a) C₄ plants are adapted to only rainy conditions
(b) C₄ plants are partially adapted to drought condition
(c) C₄ plants are exclusively adapted to desert condition
(d) C₄ plants are adapted to aquatic condition
Answer:
(b) C₄ plants are partially adapted to drought condition

Question 25.
Carotenoids and Xanthophylls are also known as
a) Respiratory pigments
b) Accessory pigments
c) Photosynthetic pigments.
d) Photolytic pigments
Answer:
b) Accessory pigments

Question 26.
Which metal ion is a
a) Iron
b) cobalt
c) Magnesium
d) Zinc
Answer:
c) Magnesium

Question 27.
The important external factors affecting photosynthesis are:
(a) light, chlorophyll, temperature
(b) light, stomatal opening, oxygen
(c) light, protoplasmic factor, oxygen
(d) light, CO₂ and oxygen
Answer:
(d) light, CO₂ and oxygen

Question 28.
The process of photophosphorylation was discovered by
a) Priestly
b) Calvin
c) Amon
d) Warburg
Answer:
c) Arnon

Question 29.
Which of the following is a C4 plant
a) Potato
b) Sugarcane
c) Pea
d) Papaya
Answer:
b) Sugarcane

Question 30.
Splitting of water molecule (photolysis) produces:
(a) hydrogen and oxygen
(b) electrons, protons and oxygen
(c) electrons and oxygen
(d) hydrogen, carbon dioxide and oxygen
Answer:
(b) electrons, protons and oxygen

Question 31.
Dimorphism in chloroplasts is seen in
a) C4 plants
b) C₂ plants
c) CAM – plants
d) C3 plants
Answer:
a) C4 plants

Question 32.
Energy required for ATP synthesis in PSII comes from
a) Proton gradient
b) Electron gradient
c) Reduction of glucose
d) Oxidation of glucose
Answer:
a) Proton gradient

Question 33.
The by-product of Photosynthesis is
a) O₂
b) CO₂
c) Carbohydrate
d) H₂O
Answer:
a) O₂

Question 34.
Which of the following process is called reverse of Glycolysis?
a) CO₂ reduction
b) RUBP carboxylation
c) RUBP regeneration
d) ATP synthesis
Answer:
a) CO₂ reduction

Question 35.
The Dark reaction of Photosynthesis occurs in
a) Matrix
b) Grana
c) Stroma
d) Cytoplasm
Answer:
c) Stroma

Question 36.
A granal chloroplasts are characteristics of
a) Mesophyll of pea leaves
b) Bundle sheath of Mango leaves
c) Mesophyll of maize leaves
d) Bundle sheath of sugar cane leaves
Answer:
d) Bundle sheath of Sugar cane leaves

Question 37.
Which of the following plant is a better photosynthesis?
a) Mango
b) Sugarcane
c) Wheat
d) Rice
Answer:
b) Sugarcane

Question 38.
The enzyme that is not found in a C3 plant is
a) RUBP carboxylase
b) PEP carboxylase
c) NADP reductase
d) ATP synthase
Answer:
b) PEP carboxylase

Question 39.
Which of the following factors affect the rate of photosynthesis?
I. Light
II. Protoplasmic factor
III. Hormones Codes
IV. Haemoglobin
a) only III
b) I and II
c) only IV
d) I, II, and III
Answer:
d) I, II and III

Question 40.
Match the following columns

Question 41.
One complete light reaction involves light energy.
a) 30 quanta
b) 48 quanta
c) 40 quanta
d) 25 quanta
Answer:
b) 48 quanta

Question 42.
During photosynthesis which of the following event does not take place?
a) oxidation of CO₂
b) Reduction of CO₂
c) oxidation of H₂O
d) Light absorption
Answer:
a) Oxidation of CO₂

Question 43.
The site of light trapping in the chloroplast is
a) Thylakoid membrane
b) Stroma
c) Plasma fluid
d) Stromal lamellae
Answer:
a) Thylakoid membrane

Question 44.
Kranz anatomy is traced in the Leaves of
a) Wheat
b) Potato
c) Mustard
d) Sugarcane
Answer:
d) Sugarcane

Question 45.
The intermediate got from Kreb’s cycle that is used for chlorophyll synthesis is
a) Citric acid
b) Isocitric acid
c) Succinic acid
d) Fumaric acid
Answer:
c) Succinic acid

Question 46.
The existence of light and dark reaction of photosynthesis was proved by
a) Blackman
b) Emerson
c) Warburg
d) Arnon
Answer:
a) Blackman

Question 47.
Choose the wrong match.
a) Hatch & Slack – Dicarboxylic acid pathway
b) Decker – PCO cycle
c) Ruben, Kamen – CAM cycle
d) Calvin Benson – PCR cycle
Answer:
c) Ruben, Kamen – CAM cycle

Question 48.
I) Primary CO, acceptor – PEPA
II) 4C compound produced 1 st – OAA
III) 1st carboxylation occurs in – Bundle sheath cells
IV) 2nd carboxylation occurs in – Mesophy 11 cells
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
c) III & IV

Question 49.
Say True or False with respect to C₂ cycle
I) RUBISCO has the most abundant protein on earth.
II) Photorespiration does not yield any free energy in the form of ATP
III) The end product is a 2 – c compound. So the cycle is known as C₂ cycle
IV) Under certain conditions 50% of the photosynthetic potential is lost because of photorespiration.

Answer:
b) True – True – False – True

Question 50.
Say True or False and choose the right option from the given choice.
I) PAR is between – 400 – 700 mil.
II) Heliophytes (Beans) require higher light intensity than sccophytes (oxalis)
III) Red light induces lowest rate of photosynthesis.
IV) Green light induces highest rate of photosynthesis.

Answer:
c) True – True – False – False

Question 51.
Choose the wrongly matched pair
a) Stephen hales – Father of plant physiology
b) Lavoisier – Purifying gas oxygen is produced in sun light
c) Vonmayer – Green plants convert solar energy into chemical energy
d) Emerson &Amoid – C4 cycle
Answer:
d) Emerson & Arnold – C4 cycle

Question 52.
Choose the rightly matched pair
a) Chlorophyll a – Accessory pigments and trap solar energy
b) Chlorophyll b – Differs from Chlorophyll a in having CH3 instead of CHO – at 3rd C atom
c) Chlorophyll c – Differs from Chlorophyll a by lacking a phytol tail
d) Chlorophyll d – It has CHO at 3rd at the 3rd carbon atom at 11 – pyrrole ring
Answer:
c) Chlorophyll c – Differs from chlorophyll a by lacking a phytol tail.

Question 53.
Choose the right matched pairs from the given options.
I) Green non sulphur bacteria – Clostridium & Lynbya
II) Green sulphur bacteria – Chlorobacterium & Chlorobium
III) Purple sulphur bacteria – Thiospirillum & Chromatium
IV) Purple non sulphur bacteria – Rhodopseudomonas & Rhodospirillum
a) I, II, & III
b) II, III & IV
C) I, II & IV
d) I, III & IV
Answer:
b) II, III & IV

II. Assertion (A) & Reason (R)

a) Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion
b) Both Assertion (A) and Reason (R) are true – Reason is not the correct explanation of Assertion.
c) Assertion (A) is false but Reason (R) is true.
d) Both Assertion (A) and Reason (R) are false.

Question 1.
Assertion (A): Chlorophyll appears green.
Reason (R): It absorbs light mainly in the region of green part of light spectrum.
Answer:
c) Assertion (A) is False but Reason (R) is true.

Question 2.
Assertion (A): Red of spectrum contains high energy.
Reason (R): Green light of Visible spectrum contain low energy than red light.
Answer:
c) Assertion (A) is False but Reason (R) is true.

Question 3.
Assetion (A): Non cyclic photo phosphorylation occurs in the stroma of chloroplasts
Reason (R): There is a continuous flow of electrons in this process.
Answer:
d) Both Assertion (A) and Reason (R) are false.

Question 4.
Assetion (A): Carotenes and Xanthophylls are soluble in either
Reason (R): These are accessory pigments of photosynthesis
Answer:
b) Both Assertion (A) and Reason (R) are true, Reason is not the correct explanation of Assertion

Question 5.
Assetion (A): Carotenoids are accessory pigments
Reason (R): Absorbed light energy is transferred to reaction centre by carotenoids.
Answer:
a) Assertion (A) and Reason (R) True and Reason is the correct explanation of Assertion.

III. 2 Marks Questions

Question 1.
What is the function of the plant in the universe?
Answer:
Plants are the major machinery which produces organic compounds like carbohydrates,lipids, proteins, nucleic acids and other biomolecules.

Question 2.
What is PAR?
Answer:
It refers to Photosynthetically Active Radiation, which is between 400 – 700 nm photosynthetic rate is maximum in blue and red light – Green light induces lowest rate of photosynthesis.

Question 3.
What is the site of photosynthesis?
Answer:
Chloroplasts are the main site of photosynthesis and both the energy-yielding process (Light reaction) and fixation of carbon dioxide (Dark reaction) that takes place in the chloroplast.

Question 4.
Name the photosynthetic pigments of Algae?
Answer:

• Chlorophyll b – Green Algae
• Chlorophyll c – Dianoflagellates, Diatoms & Brown Algae
• Chlorophyll d – Red Algae
• Chlorophyll e – Xantho phycean Algae.

Question 5.
Endosymbiotic hypothesis says that chloroplasts evolved from bacteria. Substantiate the statement.
Answer:
Presence of 70S ribosome and DNA gives them status of semi-autonomy and proves endosymbiotic hypothesis which says chloroplast evolved from bacteria.

Question 6.
Write down the significance of photorespiration.
Answer:

• Glycine and Serine synthesized during this process are precursors of many biomolecules like Chlorophyll, Proteins, Nucleotides.
• It consumes excess NADH + H⁺ generated.
• Glycolate protects cells from Photooxidation.

Question 7.
What are the conclusions of Hill’s Reaction?
Answer:

• During Photosynthesis oxygen is evolved from water.
• Electrons for the reduction of CO₂ are obtained from water.
• A reduced substance produced, later helps to reduce CO₂
• 2H₂O + 2A→ 2 AH₂ + O₂

Question 8.
What are Xanthophylls?
Answer:
Yellow (C₄₀H₅₆O₂) pigments are like carotenes but contain oxygen. Lutein is responsible for yellow colour change of leaves during autumn season. Examples: Lutein, Violaxanthin and Fueoxanthin.

Question 9.
Notes on Phycobillins.
Answer:

• They are proteinaceous pigments.
• They are soluble in water.
• Lack ‘Mg’ and phytol tail.
• There are 2 forms 1. Phycocyanin 2. Phycoerythrin.
• Phycocyanin occur in Cyanobacteria.
• Phyco erythrin occur in Rhodophycean Algae.

Question 10.
Define absorption spectrum.
Answer:
Pigments absorb different wavelengths of light. A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called its absorption spectrum.

Question 11.
Why do we call carotenoids shield pigments?
Answer:

• Carotenoids are yellow to orange pigments mostly tetraterpens and absorb light strongly in the blue to violet region of the visible spectrum.
• These pigments protect chlorophyll from photosynthetic oxidative damage.

Question 12.
What is known as substrate-level phosphorylation?
Answer:
Phosphorylation taking place during respiration is called oxidative phosphorylation and ATP produced by the breakdown of substrate is known as substrate-level phosphorylation.

Question 13.
What are Quantosomes?
Answer:

• They are physiological photosynthetic units, located on the inner membrane of thylakoid lamellae of size 180A X 160 A length & breadth.
• It was named by Park &Pickins( 1964).
• One quantosome contains about 230 chlorophyll molecules.
• It constitutes a photosynthetic unit responsible for the production of one O₂ molecule or reduction of one CO₂ molecule.

Question 14.
What is Bioluminescence?
Answer:
It is the special aspect of few living organism, in which there are some biochemical substances production is responsible for the emission of light by a living organism.

Question 15.
What is the significance of photorespiration?
Answer:
Significance of photorespiration:

1. Glycine and Serine synthesized during this process are precursors of many biomolecules like chlorophyll, proteins, nucleotides.
2. It consumes excess NADH + H⁺ generated.
3. Glycolate protects cells from Photooxidation.

Question 16.
Explain water oxidizing clock or S state mechanism.
Answer:
The splitting of water molecule, mechanism was studied by KoK et, al (1970).
It consists of a series of 5 states so, s₁, s₂, s₃, s₄.
Each sate acquires positive charge by a photon (hv) and after the state s₄ – if acquires 4 positive charges 4 electron and evolution of oxygen.

Two molecules of water go back to the so.
At the end of photolysis 4H⁺, 4e^– and O₂ are evolved from water.
4H₂O → 4H⁺+ + 40H^–
40H^{–  →} 2H₂O+O₂+4e^–
2H₂O → 4H⁺+ O₂ + 4e^–.

Question 17.
Distinguish between photophosphorylation and oxidative phosphorylation.
Answer:

Question 18.
What are the air pollutants, that affect the rate of photosynthesis?
Answer:
Pollutants like SO₂, NO₂, O₃ (Ozone) and Smog affect the rate of photosynthesis.

Question 19.
What are the conditions for the occurrence of Non-cyclic photophosphorylation?
Answer:

• Non-Cyclic Photo Phosphorylation occurs, when.
• There is the availability of NADP⁺ for reduction.
• Two molecules of water go back to the so.
• When there is the splitting of water molecules.
• When both PSI and PS II are activated.

Question 20.
Name any three photosynthetic bacteria.
Answer:
Three photosynthetic bacteria:

1. Chlorobacterium
2. Thiospirillum
3. Rodhospirillum

Question 21.
What are the 2 phases of light reaction?
Answer:
The light reaction has 2 phases

1. Photooxidation phase
2.  Photochemical phase.

I) Photooxidation phase (POP):

•  Absorption of light energy.
• Transfer of energy from accessory pigments to the reaction centre.
• Activation of chlorophyll ‘a’ -molecule.

II) Photochemical phase (PCO):

• Photolysis of water and evolution of oxygen.
• Electron transport and synthesis of assimilatory power.

Question 22.
Greenlight induces lowest rate of photosynthesis justifies.
Answer:

• Yes green light induces the lowest rate of photosynthesis because it is not coming under photosynthetically
• Active reduction – (400 – 700 nm) known as PAR.
• PAR – (Photosynthetic rate is maximum in blue and red light not in green Light.

Question 23.
What will be the quanta requirement for the complete light reaction which releases 6 oxygen molecules?
Answer:

• Complet light reaction releases 6 oxygen molecules.
• If one molecule of oxygen evolution requires 8 quanta means for 6 oxygen molecules (6 x 8 = 48) quanta of light reaquired for a complete light reaction.

Question 24.
Give the balance sheet of Calvin or C3 cycle.
Answer:
One molecule of CO₂ is fixed in one turn of the Calvin or C³ cycle.
So 6 turns of cycle will be required to fix 6 molecules of C0₂ – (i.e) to form one molecule of Glucose C6H₁₂O6

IV. 3 Mark Questions

Question 1.
Mention any three significance of photosynthesis.
Answer:
Three significance of photosynthesis:

1. Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly.
2. It is the only natural process that liberates oxygen in the atmosphere and balances the oxygen level.
3. Photosynthesis balances the oxygen and carbon cycle in nature.

Question 2.
What are the properties of light.
Answer:

• Light is a transverse electromagnetic wave.
• It consists of ocillating electric and magnetic fields that are perpendicular to each other and perpenticular to the direction of propagation of the light.
• Light moves at a speed of 3 x 108 ms^-1
• Wave length is the distance between successive crests of the wave.
• Light as a particle is called photon. Each photon contains an amount of energy known as quantum.
• The energy of a photon depends on the frequency of the light.

Question 3.
Distinguish between Absorption spectrum & Action Spectrum.
Answer:

Question 4.
Distinguish between fluorescence and phosphorescence.
Answer:

Question 5.
What is meant by the ground state?
Answer:
The action of photon plays a vital role in the excitation of pigment molecules to release an electron. When the molecules absorb a photon, it is in an excited state. When the light source turned off, the high-energy electrons return to their normal low-energy orbitals as the excited molecule goes back to its original stable condition known as the ground state.

Question 6.
Write down the significance of Photosynthesis.
Answer:

• Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly All other organism depend on them for energy.
• It liberates oxygen in the atmosphere and balances.
• Fuels such as coal, petroleum, and other fossil fuels are preserved forms got only from photosynthetic plants.
• It also provides fodder, fibre, firewood, timber useful medicinal products and these sources come by the act of photosynthesis.

Question 7.
What is DCMU?
Answer:

• It is a chemical herbicide having an inhibiting effect on photosynthesis.
• It is Dichloro phenyl D₁ Methyl Urea.
• It can inhibit electron flow during light reactions of photosynthesis.
• It is a herbicide that blocks the plastoquinone binding site of P.S II and inhibits electron flow from plastoquinone to cytochrome.

Question 8.
State black man’s law of limiting factor.
Answer:
It is a modified law proposed by Liebig’s law of minimum.

Definition:
According to Blackman at any given point of time, the lowest factor among essentials will limit the rate of Photosynthesis.

Example:
When in a condition, if light intensity is low also C0₂ concentration low, in this situation among the two factors which ever is the lowest is considered as the limiting factor here among the essentials CO₂ concentration is the limiting factor.

Question 9.
What is meant by dicarboxylic acid pathway?
Answer:
C₄ pathway is completed in two phases, first phase takes place in the stroma of mesophyll cells, where the CO₂ acceptor mblecule is 3 – Carbon compound, phosphoenolpyruvate (PEP) to form 4 – carbon Oxalo acetic acid (OAA). The first product is a 4 – carbon and so it is named as C₄ cycle. Oxalo acetic acid is a dicarboxylic acid and hence this cycle is also known as a dicarboxylic acid pathway.

Question 10.
State some interesting facts about C4 cycle.
Answer:

• C4 cycle is an alternative path way for CO2 fixation.
• It occur in nearly 1000 plant species 300 dicots but mostly 700 monocots (tropical and sub tropical grasses)
• It represent about 5% earths biomass and 1% of its known plants
• 30% terrestrial carbon fixation on earth is due to C4 So, if C4 plants on earth is increased, then by carbon sequestration by thus strategy severe climate change would be avoided in the near future.

Question 11.
What is Kranz Anatomy or what is meant by Dimorphism of Chloropiasts in C4 plants.
Answer:

Question 12.
what is the significance of the CAM cycle?
Answer:
The significance of the CAM cycle:

1. It is advantageous for succulent plants to obtain CO₂ from malic acid when stomata are closed.
2. During daytime, stomata are closed and CO₂ is not taken but continues their photosynthesis.
3. Stomata are closed during the daytime and help the plants to avoid transpiration and water loss.

Question 13.
Compare the C3 and C4 on the basis of ATP production.
Answer:

Question 14.
What will be the quanta requirement for the complete light reaction which releases 6 oxygen molecules?
Answer:
The complete light reaction releases 6 oxygen molecules if one molecule of oxygen evolution, requires 81 quanta means for 6 oxygen molecules 6 x 8 = 48 quanta of light required for the complete light reaction.

Question 15.
Draw the Graphical representation of any 3 factors affecting photosynthesis.
Answer:

1. Light Intensity
2. CO₂ Concentration
3. Temperature

V. 5 Mark Questions

Question 1.
Distinguish between Photosystem – I and photosystem – II
Answer:
Photosystem – I:

1. The reaction centre is P700.
2. PS I is involved in both cyclic and non – cyclic.
3. Not involved in the photolysis of water and evolution of oxygen.
4. It receives electrons from PS II during non – cyclic photophosphorylation.
5. Located in unstacked region granum facing chloroplast stroma.
6. Chlorophyll and Carotenoid ratio is 20 to 30 : 1.

Photosystem – II:

1. Reaction centre is P680.
2. PS II participates in Non – cyclic pathway.
3. Photolysis of water and evolution of oxygen take place.
4. It receives electrons by photolysis of water.
5. Located in stacked region of thylakoid membrane facing lumen of thylakoid.
6. Chlorophyll and Carotenoid ratio is 3 to 7 : 1.

Question 2.
Structure of Chloroplast.
Answer:

Question 3.
Notes on photosystem and Reaction centre.
Answer:

• Thylakoid membrane contains photosystem I (PSI) and photosystem II (PSII)
• PS I is unstacked region of granum tàcing stroma ofchÍoroplast.
• PS II is found in stacked region of thylakoid membrane facing lumen of thylakoid.
• Each photosystem consists of central core complex (CC) and light harvesting complex (LHC) or Antenna molecules.
• The core complex consists of respective reaction centre associated with proteins, electron donors and acceptors.
• PSI – CCI consists of reaction centre P 700 and LHC – I
• PS II- CC II consists of reaction centre P680 and LHC – II
• Light harvesting complex consists of several chiorophylls, carotenoids and xanthophyll molecules.
• The main function of LHC is to harvesting light energy and transfer it to their respective reaction centre.

Question 4.
Difference between photosystem I and Photosystem II.
Answer:

Question 5.
Differences between Cyclic Photophosphorylation and Non – cyclic photophosphorylation.
Answer:

Question 6.
Compare and contrast the photosynthetic processes in C₃ and C₄ plants.
Answer:
Contrast the photosynthetic processes in C₃ and C₄ plants:
C₃ Plants:

• CO₂ fixation takes place in mesophyll cells only.
• CO₂ acceptor is RUBP only.
• First product is 3C – PGA.
• Kranz anatomy is not present.
• Granum is present in mesophyll cells.
• Normal Chloroplast.
• Optimum temperature 20° to 25° C.
• Fixation of CO₂ at 50 ppm.
• Less efficient due to higher photorespiration.
• RUBP carboxylase enzyme used for fixation.
• 18 ATPs used to synthesize one glucose.
• Efficient at low CO₂.
• eg: Paddy, Wheat, Potato and so on.

C₄ Plants:

• CO₂ fixation takes place mesophyll and bundle sheath.
• PEP in mesophyll and RUBP in bundle sheath cells.
• First product is 4C – OAA.
• Kranz anatomy is present.
• Granum present in mesophyll cells and absent in bundle sheath.
• Dimorphic chloroplast.
• Optimum temperature 30° to 45° C.
• Fixation of CO₂ even less than 10 ppm.
• More efficient due to less photorespiration.
• PEP carboxylase and RUBP carboxylase used.
• 30 ATPs to produce one glucose.
• Efficient at higher CO₂.
• eg: Sugar cane, Maize, Sorghum, Amaranthus and so on.

Question 7.
Explain Non cyclic photophosphorylation.
Answer:
When PS II (P680) gets activated, electrons from a high energy state pass through a series of electron carriers like pheophytin, plastoquinone cytochrome complex, plastocyanin, and finally accepted by PS I (P700).

During this flow ATP is generated:

• PS. I (P 700) is activated by light electrons moved to high energy state and accepted by electron acceptor (FRS) Ferredoxin Reducing Substance, during downhill passes through Ferredoxin. During this process NADPH is reduced by H+ formed during photolysis.
• Electrons released from PS II are not cycled back but used in the reduction of NADPH+ into NADPH+H⁺.
• During electron transport, it generates ATP and this type of Phophorylation is called.

Non Cyclic Photophosphorylation:
The electron flow looks like the letter ‘Z’ so known as Z scheme. It has 3 stages.

1. Electron transport from water to P 680: Electrons lost by the PS II are replaced by electrons from splitting of the water molecule, producing electron, protons, and oxygen.
2. Electron transport from P680 to P 700: The flow, through various electron carrier molecules, like pheophytin, plastoquinone (PQ) cytochrome b6 – F complex, plastocyanin (PC) finally reaches P 700 (P.S.I)
3. Electron transport from P 700 to NADP⁺: PSI (P700) is excited now and the electrons pass through ferredoxin, NADP is reduced to NADPH + H⁺

Question 8.
Explain Calvin cycle or C₃ cycle?
Answer:

• It follows a light reaction.
• Utilises ATP and NADPH + H⁺ produced during the light reaction, and reduce carbon dioxide carbohydrate.
• These reactions do not require light so named as Dark reactions.
• The first formed product is a 3 carbon compound (Phospho Glyceric Acid) and so-known as C₃ cycle.
• It was found by Melvin, Calvin, and Benson – so known as the Calvin cycle.
• Occur in the stroma of the chloroplast.
• It is temperature-dependent, and so it is also called a thermochemical reaction.

Phase I carboxylation (Carbon fixation):

• The 5 C compound Ribulose 1 – 5 Bisphosphate (RUBP) with the help of (RUBISCO) enzyme accepts one molecule of carbon dioxide → 6 carbon compound (unstable)
• The 6c compound is broken into → 2 molecules of 3 c compound.

Phase II – Glycolytic Reversai/Reduction :

Phase III – Regeneration:

• The regeneration of RUBP involves several intermediate compounds of 6c, 5c, 4c, and 7c compounds.
• Fixation of one CO₂ require 3 ATPs + 2NADPH⁺ + H⁺
• Fixation of six CO₂ require I8 ATPs+ 12NADPH⁺ + H⁺
• 6 c compound is the net gain to form hexose sugar.

Question 9.
Draw the flow Chart of C₄ pathway?
Answer:

Question 10.
Explain CAM Cycle?
Answer:

• It is one of the carbon path ways in succulent plants growing in semi arid or xerophytic condition.
• The stomata are closed during day (scoto active) and open during night.
• This reverse rhythm help to conserve water loss through transpiration and will stop the fixation of CO₂ during day.
• At night time CAM plants fix CO, with help of (PEP) phospho Enol Pyruvic acid and produce (OAA) Oxalo Acetic Acid.
• Subsequently OAA is converted into a Malic acid-like C4 cycle and get accumulated in the vacuole, increasing the acidity.
• During daytime stomata, are closed and Malic acid is decarboxylated into pyruvic acid resulting in the decrease of acidity.
• CO₂ thus formed enters into the Calvin cycle and produces carbohydrates.

Question 11.
Give the flow chart of Photo respiration or C₂ cycle.
Answer:

Question 12.
Differentiate Photorespiration and Dark respiration.
Answer:

Question 13.
Give any 5 External factors affecting photosynthesis.
Answer:
1. Carbon dioxide:
330ppm or 0.3% of CO₂ is available in the atomsphere If there is an increase in CO₂ concentration the rate of Photosynthesis increases -If it increases beyond 500 PPm rate of photosynthesis will be inhibited.

2. Oxygen:
When there is increase in oxygen concentration there is unhibition of photosynthesis Warburg – studied this in chlorellain 1920. This effect is known as Warburg effect.

3. Temperature:

• Optimum temperature for photosynthesis vary from plant to plant
• Normally it is 25°C to 3 5°C
• In Opuntia, it is 55°C
• In Lichens it is 20°C
• In Algae growing in hot spring it is 75°C
• At high and low temperature the stomata will close also the enzymes get inactivated.

4. Water:

• Pholysis of water provide electrons and protons for the reduction of NADP – directly.
• Affect stomatal movement and hydration of protoplasm – indirectly.
• During water stress, supply of NADPH + H⁺ affected

5. Minerals:

Question 14.
Explain the test tube funnel experiment.
Answer:
AIM: To proove that oxygen is evolved during Photosynthesis.
Procedure:
Take some hydrilla plant and place them at the bottom of a beaker containing water – Add, little NaHCO3 in to the water. Cover plant with an inverted funnel Invert a test tube over the funnel keep this set up in sun light.

Observation: Air bubbles are released from Hydrilla plant and collected in the test tube by downward displace ment of water. Take the test tube carefully by closing with a finger and then introduce a burning match stick, it bum brightly.
Inference: Hydrilla plant perform photosynthesis and oxygen is liberated during photosynthesis.

Question 15.
Explain the experiment to determine rate of photosynthesis by Witmott’s bubbler.
Answer:
Procedure:

• Wilmott’s bubbler consists of a wide mouth bottle fitted with a single holed cork, a glass tube with lower and having wider opening to insert hydrilla plant.
• The upper end is fitted to a narrow bottle with water.
• Fill the bottle with water and insert hydrilla living into wider part of the tube.
• Hydrilla plant should be cut inside the water to avoid entry of
  air bubbles.
• Fix the tube with jar which acts as water reservoir.
• Keep the apparatus in sunlight count the bubbles when they are in same size.

Question 16.
Differentiate photosynthesis in plants and Bacterial photosynthesis.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 15 Polymorphism Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 15 Polymorphism

11th Computer Science Guide Polymorphism Text Book Questions and Answers

Part I

Choose The Correct Answer :

Question 1.
Which of the following refers to a function having more than one distinct meaning?
a) Function Overloading
b) Member overloading
c) Operator overloading
d) Operations overloading
Answer:
a) Function Overloading

Question 2.
Which of the following reduces the number of comparisons in a program?
a) Operator overloading
b) Operations overloading
c) Function Overloading
d) Member overloading
Answer:
c) Function Overloading

Question 3.
void dispchar(char ch=’$’,int size=10) ‘
{
for(int i=1;i<=size;i++)
cout<<ch;
}

How will you invoke the function dispchar() the following input?
To print $ for 10 times
a) dispchar();
b) dispchar(ch,size);
c) dispchar($,10);
d) dispcharC$’, 10 times);
Answer:
a) dispchar();

Question 4.
Which of the following is not true with respect to function overloading?
a) The overloaded functions must differ in their signature.
b) The return type is also considered for overloading a function.
c) The default arguments of overloaded functions are not considered for Overloading.
d) Destructor function cannot be overloaded.
Answer:
b) The return type is also considered for overloading a function.

Question 5.
Which of the following is an invalid prototype for function overloading?
a) void fun (int x);
void fun (char ch);.
b) void fun (int x);
void fun (int y);
c) void fun (double d);
void fun (char ch);
d) void fun (double d);
void fun (int y);
Answer:
b) void fun (int x);
void fun (int y);

Question 6.
Which of the following function(s) combination cannot be considered as overloaded function(s) in the given snippet?
void print(char A,int B); // F1
void printprint(int A, float B); // F2
void Print(int P=10); // F3
void print(); // F4
a) F1,F2,F3,F4
b) F1,F2,F3
c) F1,F2,F4
d) F1,F3,F4
Answer:
b) F1,F2,F3

Question 7.
Which of the fallowing operator is by default overloaded by the compiler?
a) *
b) +
c) +=
d) ==
Answer:
c) +=

Based on the following program answer the questions (8) to (10)
#include<iostream>
using namespace std;
class Point
{
private:,
int x, y;
public:
Point(int x1,int y1)
{
x=x1;
y=y1;
}
void operator+(Point &pt3);
void show()
{
cout << “x = ” << x << “, y = “<<y;
}
};
void Point: :operator+(Point &pt3)
{
x += pt3.x;
y += pt3.y;
}
int main()
{
Point pt1(3,2),pt2(5,4);
pt1+pt2;
pt1.show();
return 0;
}

Question 8.
Which of the following operator is overloaded?
a) +
b) operator
c) ::
d) =
Answer:
a) +

Question 9.
Which of the following statement invoke operator overloading?
a) pt1+pt2;
b) Point pt1(3,2),pt2(5,4);
c) pt1.show();
d) return 0;
Answer:
a) pt1+pt2;

Question 10.
What is the output for the above program?
a) x=8, y=6
b) x=14, y=14
c) x=8, y=6
d) x=5, y=9
Answer:
a) x=8, y=6

Part – II

Very Short Answers

Question 1.
What is function overloading?
Answer:
The ability of the function to process the message or data in more than one form is called function overloading. In other words, function overloading means two or more functions in the same scope share the same name but their parameters are different.

Question 2.
List the operators that cannot be overloaded.
Answer:
The following operators can not be overloaded:

• scope operator::
• sizeof
• member selector.
• member pointer selector *
• ternary operator ?:

Question 3.
class add
{
int x;
public:
add(int);
};

Write an outline definition for the constructor.
Answer:
OUTLINE CONSTRUCTOR DEFINITION
add :: add(int a)
{
x = a;
cout<<“\nParameterized constructor”;
}

Question 4.
Does the return type of function help in overloading a function?
Answer:
No. The return type of overloaded functions is not considered for overloading the same data type.

Question 5.
What is the use of overloading a function?
Answer:

• Function overloading is not only implementing polymorphism but also reduces the number of comparisons in a program and makes the program to execute faster.
• It also helps the programmer by reducing the number of function names to be remembered.
• Program complexity is reduced.

Part – III

Short Answers

Question 1.
What are the rules for function overloading?
Answer:
Rules for function overloading:

1. The overloaded function must differ in the number of its arguments or data types.
2. The return type of overloaded functions is not considered for overloading the same data type.
3. The default arguments of overloaded functions are not considered as part of the parameter list in function overloading.

Question 2.
How does a compiler decide as to which function should be invoked when there are many functions? Give an example.
Answer:
The number and types of a function’s parameters are called the function’s signature. When we call an overloaded function, the compiler determines the most appropriate definition to use, by comparing the argument types we have used to call the function with the parameter types specified in the definitions. The process of selecting the most appropriate overloaded function or operator is called overload resolution.

Example:
float area (float radius);
float area (float half, float base, float height);
float area (float length , float breadth);
According to the input passed, the respective function is called.

For example:
x= area(5); // calls area() with one input.
x= area(5,6); // calls area() with two input.
x= area(0.5,5,6);// calls area() with three input.

Question 3.
What is operator overloading? Give some example of operators which can be overloaded.
Answer:
The term operator overloading refers to giving additional functionality to the normal C++ operators.
Example:
The following operators can be overloaded +,++,-,—,+=,-=,*.<,>, etc.

Question 4.
Discuss the benefit of constructor overloading.
Answer:
Function overloading can be applied for constructors, as constructors are special functions of classes. A class can have more than one constructor with a different signature. Constructor overloading provides the flexibility in creating multiple types of objects for a class.

1. Memory is allocated for the objects.
2. Initialisation for the objects.

Question 5.
class sale
{
int cost, discount;
public:
sale(sale &);
};
Write a non-inline definition for the constructor specified;
Answer:
Non-inline definition for constructor:
sale :: sale(sale &s)
{
cost = s.cost;
discount = s.discount;
}

Explain In Detail

Question 1.
What are the rules for operator overloading?
Answer:
Following are some rules to be followed while implementing operator overloading.

1. The precedence and associativity of an operator cannot be changed.
2. No new operators can be created, only existing operators can be overloaded.
3. Cannot redefine the meaning of an operator’s procedure. You cannot change how integers are added. Only additional functions can be to an operator.
4. Overloaded operators cannot have default arguments.
5. When binary operators are overloaded, the left-hand object must be an object of the relevant class.

Question 2.
Answer the question (i) to (v) after going through the following class.
Answer:
classBook
{
int BookCode ;
char Bookname[20];
float fees;
public:
Book() //Function 1
{
fees =1000;
BookCode=1;
strcpy (Bookname,”C++”);
}
void display(float C) //Function 2
{
cout< < BookCode < < “:”< < Boo kname<<“:”<<fees<<endl;
}
~Book() //Function 3
{
cout<<“End of Book
Object”<<endl;
}
Book (int BC,char S[ ],float F); //
Function 4
};

i) In the above program, what are Function 1 and Function 4 combined together referred to?
ii) Which concept is illustrated by Function3? When is this function called/ invoked?
iii) What is the use of Function3?
iv) Write the statements in main to invoke function1 and function2
v) Write the definition for Function4
Answer:
i) Constructor
ii) Destructor. It will be executed automatically when object goes out of scope.
iii) To remove the memory space of the object allocated at the time of creation.
iv) a) Book b; // object b automatically call the constructor function Book(); (Function 1)
b) display(4.5); // Invokes the display function by passing 4.5. (Function 2)
v) Definition of Function 4:
Book (int BC,char S[ ],float F)
{
fees=F;
BookCode=BC;
strcpy (Bookname,S);
}

Question 3.
Write the output of the following program.
#include<iostream>
using namespace std;
class Seminar
{
int Time;
public:
Seminar()
{
Time=30;cout<<“Seminar starts now”<<endl;
}
void Lecture()
{
cout<<“Lectures in the seminar on”<<endl;
}
Seminar(int Duration)
{
Time=Duration;cout <<“Welcome to Seminar “<<endl;
}
Seminar(Seminar &D)
{
Time=D.Time;cout<<“Recap of Previous Seminar Content “<<endl;
}
~Semina r()
{
cout<<“Vote of thanks”<<endl;
}
};
int main()
{
Seminar s1,s2(2),s3(s2);
s1.LectureQ;
return 0;
}
Output:

Question 4.
Debug the following program.
#include<iostream>
using namespace std;
class String
{
public:
charstr[20];
public:
void accept_string
{
cout<<“\n Enter String :
cin>>str;
}
display_string()
{
cout<<str;
}
String operator *(String x) //Concatenating String
{
String s;
strcat(str,str);
strcpy(s.str,str);
gotos;
}
}
int main()
{
String str1, str2, str3;
str1.accept_string();
str2 .accept_string();
Cout<<“\n\n First String is : “;
str1.display_string();
cout<<“\n\n Second String is : “;
str2.display_string();
str3=str1+str2;
cout<<“\n\n Concatenated String is :”;
str3.display_string();
return 0;
}

Correct Program :

using namespace std;
#include<iostream>
#include<string.h>
class String
{
public:
char str[20];
public:
void accept_string ()
{
cout<<“\n Enter String :”;
cin>>str;
}
void display__string()
{
cout<<str;
}
String operator+(String x)//
Concatenating String
{
Strings;
strcat(str,x.str);
strcpy(s.str,str);
return(s);
}
};
int main()
{
String str1, str2, str3;
str1.accept_string();
str2 .accept_stri ng();
Cout<<“\n\n First String is : “;
str1.display_string();
cout<<“\n\n Second String is : “;
str2.display_string();
str3=str1+str2;
cout<<“\n\n Concatenated String is :”;
str3.display_string();
return 0;
}

Output:

Question 5.
Answer the questions based on the following program.
#include<iostream>
#include<string.h>
using namespace std;
class comp
{
public:
chars[10];
void getstring(char str[10])
{
strcpy(s,str);
}
void operator==(comp);
};
void comp::operator==(comp ob)
{
if(strcmp(s,ob.s)==0)
cout<<“\nStrings are Equal”;
else
cout<<“\nStrings are not Equal”;
}
int main()
{
comp ob, ob1;
char stringl[10], string2[10];
cout<<“Enter First String:”;
cin>>string1;
ob.getstring(stringl);
cout<<“\nEnter Second String:”;
cin>>string2;
ob1.getstring(string2);
ob==ob1;
return 0; ‘
}

i) Mention the objects which will have the scope till the end of the program.
ii) Name the object which gets destroyed in between the program.
iii) Name the operator which is overloaded and write the statement that invokes it.
iv) Write out the prototype of the overloaded member function
v) What types of operands are used for the overloaded operator?
vi) Which constructor will get executed? Write the output of the program
Answer:
i) Objects ob and obi in the main( )will have the scope till the end of the program.
ii) Object ob in the operator== function will be destroyed when exit from it.
iii) The operator overloaded is ==.
The statement invoke the operator
overloaded function is ob == ob1;
iv) Prototype of the overloaded member function is as follows:
void comp::operator==(comp);
v) User-defined data type class (comp) objects are used.
vi) No explicit constructor is not defined in the class. So, a compiler-generated default constructor is created and executed.
Output:

11th Computer Science Guide Polymorphism Additional Questions and Answers

Choose The Correct Answer: (1 Mark)

Question 1.
The number and types of a function’s parameters are called the …………………
(a) overload resolution
(b) function’s signature
(c) function overloading
(d) operator overloading
Answer:
(b) function’s signature

Question 2.
In C++, polymorphism is achieved through _________ overloading.
a) Function
b) Operator
c) Operand
d) Either A or B
Answer:
d) Either A or B

Question 3.
The return type of overloaded functions is not considered for overloading same …………………
(a) polymorphism
(b) prototype
(c) data type
(d) overloading
Answer:
(c) data type

Question 4.
The number and types of a function’s parameters are called the function’s________.
a) Signature
b) Syntax
c) Either A or B
d) None of these
Answer:
a) Signature

Question 5.
The mechanism of giving special meaning to an operator is known as …………………
(a) operator overloading
(b) parameter
(c) function overloading
(d) polymorphism
Answer:
(a) operator overloading

Question 6.
______ overloading is not only implementing polymorphism but also reduces the number of comparisons in a program and makes the program execute faster.
a) Function
b) Operator
c) Operand
d) Either A or B
Answer:
a) Function

Question 7.
The overloaded operator is given using the keyword ………………… followed by an operator symbol.
(a) operator
(b) data type
(c) object
(d) function
Answer:
(a) operator

Question 8.
The ______ of overloaded functions are not considered as part of the parameter list in function overloading.
a) arguments
b) default arguments
c) data
d) None of these
Answer:
b) default arguments

Question 9.
The ______ of overloaded functions are not considered for overloading the same data type.
a) return type
b) arguments
c) data
d) None of these
Answer:
a) return type

Question 10.
The overloaded function must differ in ______
a) the number of its arguments
b) data types
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 11.
_________ overloading provides the flexibility of creating multiple types of objects for a class,
a) Constructor
b) Destructor
c) Member function
d) None of these
Answer:
a) Constructor

Question 12.
Compiler identifies a given member function is a constructor by its ________
a) name
b) return type
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 12.
Compiler Identifies a given member function is a constructor by its ______
a) name
b) return type
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 13.
The term overloading refers to giving additional functionality to the normal C++ operators.
a) Function
b) Operator
c) Operand
d) Either A or B
Answer:
b) Operator

Question 14.
__________ operator can not be overloaded.
a) scope operator::
b) sizeof
c) member selector.
d) All the above
Answer:
d) All the above

Question 15.
________ operator can not be overloaded.
a) Member pointer selector
b) Ternary operator ?:
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 16.
can be Overloaded.
a) User-defined types (objects)
b) Literals
c) Identifiers
d) None of these
Answer:
a) User-defined types (objects)

Very Short Answers (2 Marks)

Question 1.
Give the syntax for operator overloading.
Answer:
RetumType classname :: Operator Operator Symbol (argument list)
{

\\ Function body

}
Example: Deposit Deposit: : operator + (Deposit dl);

Question 2.
How polymorphism is applied in C++?
Answer:
In C++, polymorphism is achieved through function overloading and operator overloading.

Question 4.
What is a function signature?
Answer:
The number and types of a function’s parameters are called the function’s signature.

Short Answers (3 Marks)

Question 1.
How does the compiler determine the appropriate function in overloading?
Answer:
When you call an overloaded function, the compiler determines the most appropriate definition to use, by comparing the argument types you have used to call the function with the parameter types specified in the definitions. The process of selecting the most appropriate overloaded function or operator is called overload resolution.

Question 2.
Explain overload resolution.
Answer:
When you call an overloaded function, the compiler determines the most appropriate definition . to use, by comparing the argument types you have used to call the function with the parameter types specified in the definitions. The process of selecting the most appropriate overloaded function or operator is called overload resolution.

Question 3.
Write operator overloading syntax.
Answer:
Operator Overloading Syntax

Question 4.
Write a program to implement function overloading.
Answer:
Program:
#include <iostream>
using namespace std;
void print(int i)
{
cout<< ” It is integer” << i <<endl;
}
void print(double f)
{
cout<< ” It is float” << f <<endl;
void print(string c)
{
cout<< ” It is string ” << c <<endl;
}
int main()
{
print(10);
print(10.10);
print(‘Ten”);
return 0;
}
Output:
It is integer 10
It is float 10.1
It is string Ten

Explain in Detail (5 Marks)

Question 1.
Write a program to implement function overloading.
Program
#include <iostream>
using namespace std;
long add(long, long);
long add(long,long,long);
float add(float, float);
intmain()
{
long a, b, c,d;
float e, f, g;
cout << “Enter three integers\n”;
cin >> a >> b>>c;
//number of arguments different but same
data type
d=add(a,b,c);
cput << “Sum of 3 integers: ” << d << endl;
cout << “Enter two integers\n”;
cin >> a >> b;
//two arguments data type same with above function call and different with below function call
c = add(a, b);
cout << “Sum of 2 integers: ” << c << endl;
cout << “Enter two floating point numbers\n”;
cin >> e >> f;
//two arguments similar to the above function call but data type different
g = add(e, f);
cout << “Sum of floats: ” << g << endl;
}
long add(long c, long g)
{
long sum;
sum = c + g;
return sum;
}
float add(float c, float g)
{
float sum;
sum = c + g;
return sum;
}
long add(long c, long g,long h)
{
long sum;
sum = c + g+h;
return sum;
}

Output:
Enter three integers
3 4 5
Sum of 3 integers: 12 Enter two integers
4 6
Sum of 2 integers: 10
Enter two floating-point numbers
2.1 3.1
Sum of floats: 5.2

Question 2.
Write the coding for the following output using constructor overloading.
Output:
Constructor without parameters..
Parameterized constructor…
Copy Constructor…
Enter data… 20 30
Object a:
The numbers are..20 30
The sum of the numbers are.. 50
Object b:
The numbers are..10 20
The sum of the numbers are.. 30
Object c:
The numbers are..10 20
The sum of the numbers are.. 30
Answer:
#include
using namespace std;
class add
{
int num1, num2, sum;
public:
add()
{
cout << “\n Constructor without parameters…”;
num1 = 0;
num2 = 0;
sum = 0;
}
add (int s1, int s2 )
{
cout << “\n Parameterized constructor…”;
num1= s1;
num2=s2;
sum=0;
}
add (add &a)
{
cout << “\n Copy Constructor…”; ‘
num1 = a.num1;
num2 = a.num2;
sum = 0;
}
void getdata()
{
cout << “\n Enter data …”; cin>>num 1 >> num2;
}
void addition()
{
sum=num 1 + num2;
}
void putdata()
{
cout << “\n The numbers are..”;
cout < cout << “\n The sum of the numbers are..” << sum; }
};
int main()
{
add a, b (10, 20), c(b);
a. getdata();
a. addition();
b. addition();
c. addition();
cout << “\n Object a : “;
a. putdata();
cout << “\n Object b : “;
b. putdata();
cout << “\n Object c..”;
c. putdata();
return 0;
}

Question 3.
Write a program to find complex number addition and subtraction using binary operator overloading.
Answer:
Program:
//Complex number addition and subtraction
#include<iostream>
using namespace std;
class complex
{
int real,img;
public:
void read()
{
cout<<“\nEnter the REAL PART :”;
cin>>real;
cout<<“\nEnter the IMAGINARY
PART : “;
cin>>img;
}
complex operator+(complex c2)
{
complex c3;
c3.real=real+c2.real;
c3.img=img+c2.img;
return c3;
}
complex operator-(complex c2)
{
complex c3;
c3.real=real-c2.real;
c3.img=img-c2.img;
return c3;
}
void display()
{
cout<<real<<“+”<<img<<“i”;
}
};
int main()
{
complex c1,c2,c3;
int choice, cont;
do
{
cout<<“\t\tCOMPLEX NUMBERS\n\
n1.ADDITION\n\n2.SUBTRACTION\n\n”;
cout<<“\nEnter your choice :”;
cin>>choice;
if(choice==1|| choice==2)
{
cout<<“\n\nEnter the First Complex
Number”;
cl.read();
cout<<“\n\nEnter the Second Complex
Number”;
c2.read();
}
switch(choice)
{
// binary + overloaded
case 1 : c3=c1+c2;
cout<<“\n\nSUM = “;
c3.display();
break;
case 2 : c3=c1-c2; // binary -overloaded
cout<<“\n\nResult = “;
c3.display();
break;
default: cout<<“\n\nUndefined Choice”;
}
cout<<“\n\nDo You Want to Continue?(1-Y,0-N)”;
cin>>cont;
}while(cont==1);
return 0;
}

Output :
COMPLEX NUMBERS
1.ADDITION
2.SUBTRACTION
Enter your choice : 1
Enter the First Complex Number
Enter the REAL PART : 3
Enter the IMAGINARY PART: 4
Enter the Second Complex Number
Enter the REAL PART: 5
Enter the IMAGINARY PART: 8
SUM = 8+12i,
Do You Want to Continue?(1-Y,0-N)1
COMPLEX NUMBERS
1.ADDITION
2.SUBTRACTION
Enter your choice: 2
Enter the First Complex Number
Enter the REAL PART: 8
Enter the IMAGINARY PART: 10
Enter the Second Complex Number
Enter the REAL PART: 4
Enter the IMAGINARY PART: 5
Result = 4+5i
Do You Want to Continue? (1-Y,0-N)0

Question 4.
Write a program for concatenation of string using operator overloading.
Answer:
program:
#include<string.h>
# i ncl ude < iostream >
using namespace std;
class strings
{
public:
char s[20];
void getstring(char str[])
{
strcpy(s,str);
}
void operator+(strings);
};
void strings: :operator+(strings ob)
{
strcat(s,ob.s);
cout<<“\nConcatnated String is:”<<s;
>
int main()
{
strings ob1, ob2;
char stringl[10], string2[10];
cout<<“\nEnter First String:”;
cin>>string1;
ob1.getstring(string1);
cout<<“\nEnter Second String:”;
cin>>string2;
ob2.getstring(string2);
//Calling + operator to Join/Concatenate strings
ob1+ob2;
return 0;
}

Output :
Enter First String: COMPUTER
Enter Second String: SCIENCE
Concatenated String is: COMPUTER SCIENCE

Case Study

Suppose you have a Kitty Bank with an initial amount of Rs.500 and you have to add some more amount to it. Create a class ‘Deposit’ with a data member named ‘amount’ with an initial value of Rs.500. Now make three constructors of this class as follows:

1. without any parameter – no amount will be added to the Kitty Bank
2. has a parameter which is the amount that will be added to the Kitty Bank
3. whenever an amount is added an additional equal amount will be deposited automatically
Create an object of the ‘Deposit’ and display the final amount in the Kitty Bank.
Program:
using namespace std;
#include<iostream>
class Deposit
{
public:
int amount;
Deposit()
{
amount = 500;
}
Deposit(int a)
{
amount = 500 + a;
}
Deposit(int a, int b)
{
amount = 500 + a + b;
}
void display()
{
cout<<amount;
}
};
int main()
{
Deposit D1;
int amt;
cout<<“\nEnter amount to deposit”;
cin>>amt;
cout<<“\nInitial Amount in the Bank
Rs.”<<D1.amount;
Deposit D2(amt);
cout<<“\nAmount in the Bank after deposit the amount is Rs.”<<D2.amount;
Deposit D3(amt,amt);
cout<<“\nAmount in the Bank after deposit with addition equal amount deposit
RS.”<<D3. amount;
}
Output:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 16 Inheritance Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 16 Inheritance

11th Computer Science Guide Inheritance Text Book Questions and Answers

Part I

Choose The Correct Answer:

Question 1.
Which of the following is the process of creating new classes from an existing class?
a) Polymorphism
b) Inheritance
c) Encapsulation
d) superclass
Answer:
b) Inheritance

Question 2.
Which of the following derives a dass student from the base class school?
a) school: student
b) class student: public school
c) student: public school
d) class school: public student
Answer:
b) class student: public school

Question 3.
The type of inheritance that reflects the transitive nature is
a) Single Inheritance
b) Multiple Inheritance
c) Multilevel Inheritance
d) Hybrid Inheritance
Answer:
c) Multilevel Inheritance

Question 4.
Which visibility mode should be used when you want the features of the base class to be available to the derived class but not to the classes that are derived from the derived dass?
a) Private
b) Public
c) Protected
d) All of these
Answer:
a) Private

Question 5.
Inheritance is the process of creating new class from
a) Base class
b) abstract
c) derived class
d) Function
Answer:
a) Base class

Question 6.
A class is derived from a class which is a derived class itself, then this is referred to as
a) multiple inheritances
b) multilevel inheritance
c) single inheritance
d) double inheritance
Answer:
b) multilevel inheritance

Question 7.
Which amongst the following is executed in the order of inheritance?
a) Destructor
b) Member function
c) Constructor
d) Object
Answer:
b) Member function

Question 8.
Which of the following is true with respect to inheritance?
a) Private members of base class are inherited to the derived class with private
b) Private members of base class are not inherited to the derived class with private accessibility
c) Public members of base class are inherited but not visible to the derived class
d) Protected members of base class are inherited but not visible to the outside class
Answer:
b) Private members of base class are not inherited to the derived class with private accessibility

Question 9.
Based on the following dass decoration answer the questions (from 9.1 o 9.5 )

class vehicle
{
int wheels;
public:
void input_data(float,float);
void output_data();
protected:
int passenger;
};
class heavy_vehicle: protected vehicle
{
int diesel_petrol;
protected:
int load;
protected:
int load;
public:
void read_data(float,float)
void write_data(); };
class bus: private heavy_vehicle
{
char Ticket[20];
public:
void fetch_data(char);
void display_data(); >;
};

Question 9.1.
Which is the base class of the class heavy, vehicle?
a) Bus
b) heavy_vehicle
c) vehicle
d) both (a) and (c)
Answer:
c) vehicle

Question 9.2.
The data member that can be accessed from the function displaydata()
a) passenger
b) load
c) Ticket
d) All of these
Answer:
d) All of these

Question 9.3.
The member function that can be accessed by an objects of bus Class is
a) input_data()
b) read_data() ,output_data()write_data()
c) fetch_data(),display_data()
d) All of these
Answer:
c) fetch_data(),display_data()

Question 9.4.
The member function that is inherited as public by Class Bus
a) input_data()
b) read_data(),output_data(),write_data()
c) fetch_data(), display_data()
d) None of these
Answer:
d) None of these

Question 10.
class x
{
int a;
public :
x()
{}
};
class y
{
x x1;
public:
y()
{}
};
class z : public y,x
{
int b;
public:
z()
{}
}z1;

What is the order of constructor for object z to be invoked?
a) z,y,x,x
b) x,y,z,x
c) y,x,x,z
d) x,y,z
e) x,y,x,z
Answer:
e) x,y,x,z

Part – II

Very Short Answers

Question 1.
What is inheritance?
Answer:
Inheritance is one of the most important features of Object-Oriented Programming. In object-oriented programming, inheritance enables a new class and its objects to take on the properties of the existing classes.

Question 2.
What is a base class?
Answer:
The class to be inherited is called a base class or parent class.

Question 3.
Why derived class is called a power-packed class?
Answer:

• Multilevel Inheritance: In multilevel inheritance, the constructors will be executed in the order of inheritance.
• Multiple Inheritance: If there are multiple base classes, then it starts executing from the leftmost base class.

Question 4.
In what multilevel and multiple inheritances differ though both contains many base class?
Answer:
In case of multiple inheritance derived class have more than one base classes (More than one parent). But in multilevel inheritance derived class have only one base class (Only one parent).

Question 5.
What is the difference between public and private visibility mode?
Answer:
Private visibility mode:
When a base class is inherited with private visibility mode the public and protected members of the base class become ‘private’ members of the derived class.

Public visibility mode:
When a base class is inherited with public visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.

Part – III

Short Answers

Question 1.
What are the points to be noted while deriving a new class?
Answer:

The following points should be observed for defining the derived class:

1. The keyword class has to be used.
2. The name of the derived class is to be given after the keyword class.
3. A single colon.
4. The type of derivation (the visibility mode), namely private, public or protected. If no visibility mode is specified, then by default the visibility mode is considered private.
5. The names of all base classes (parent classes) separated by a comma.

class derivedclass_name :visibility_mode
base_class_name
{
// members of derived class
};

Question 2.
What is differences between the members present in the private visibility mode and the members present in the public visibility mode?
Answer:
Private visibility mode:
When a base class is inherited with private visibility mode the public and protected members of the base class become ‘private’ members of the derived class.

Private visibility members can not be inherited further. So, it can not be directly accessed by its derived classes.
Public visibility mode:
When a base class is inherited with public visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.

Public visibility members can be inherited by its child and can access in it.

Question 3.
What is the difference between polymorphism and inheritance though are used for the reusability of code?
Answer:
Polymorphism:

• Reusability of code is implemented through functions (or) methods.
• Polymorphism is the ability of a function to respond differently to different messages.
• Polymorphism is achieved through overloading.

Inheritance:

• Reusability of code is implemented through classes.
• Inheritance is the process of creating derived classes from the base class or classes.
• Inheritance is achieved by various types of inheritances namely single, multiple, multilevel, hybrid and hierarchical inheritances.

Question 4.
What do you mean by overriding?
Answer:
When a derived class member function has the same name as that of its base class member function, the derived class member function shadows/hides the base class’s inherited function. This situation is called function overriding and this can be resolved by giving the base class name followed by :: and the member function name.

Question 5.
Write some facts about the execution of constructors and destructors in inheritance.
Answer:

1. Base class constructors are executed first, before the derived class constructors execution.
2. Derived class cannot inherit the base class constructor but it can call the base class constructor by using Base_class name: :base_class_constructor() in the derived class definition
3. If there are multiple base classes, then it starts executing from the leftmost base class
4. In multilevel inheritance, the constructors will be executed in the order of inheritance The destructors are executed in the reverse order of inheritance.

IV. Explain In Brief (Five Marks)

Question 1.
Explain the different types of inheritance.
Answer:
Types of Inheritance;
There are different types of inheritance viz., Single Inheritance, Multiple inheritance, Multilevel inheritance, hybrid inheritance and hierarchical inheritance.

1. Single Inheritance:
When a derived class inherits only from one base class, it is known as single inheritance.

2. Multiple Inheritance;
When a derived class inherits from multiple base classes it is known as multiple inheritance.

3. Hierarchical inheritance:
When more than one derived classes are created from & single base class known as Hierarchical inheritance.

4. Multilevel Inheritance
The transitive nature of inheritance is itself reflected by this form of inheritance. When a class is derived from a class which is a derived class – then it is referred to as multilevel inheritance.

5. Hybrid inheritance:
When there is a combination of more than one type of inheritance, it is known as hybrid inheritance. Hence, it may be a combination of Multilevel and Multiple inheritances or Hierarchical and Multilevel inheritance or Hierarchical, Multilevel and Multiple inheritances.

Question 2.
Explain the different visibility modes through pictorial representation.
Answer:
Private visibility mode:
When a base class is inherited with private visibility mode the public and protected members of the base class become ‘private’ members of the derived class.

Protected visibility mode:
When a base class is inherited with protected visibility mode the protected and public members of the base class become ‘protected members’ of the derived class.

Public visibility mode:
When a base class is inherited with public visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.

Question 3.
#include<iostream>
#include<string.h>
#include<stdio.h>
using name spacestd;
class publisher
{
char pname[15];
char hoffice[15];
char address[25];
double turnover;
protected:
char phone[3][10];
void register();
public:
publisher();
publisher();
void enter data();
void disp data();
};
class branch
{
char bcity[15];
char baddress[25];
protected:
int no_of_emp;
public:
char bphone[2][10];
branch();
~branch();
void havedata();
void givedata();
};
class author: public branch, publisher
{
int aut_code;
charaname[20];
float income;
public:
author();
~author();
void getdata();
void putdata();
};

Answer The Following Questions Based On The Above Given Program:

3.1. Which type of Inheritance is shown in the program?
3.2. Specify the visibility mode of base classes.
3.3 Give the sequence of Constructor/Destructor Invocation when object of class author is created.
3.4. Name the base class(/es) and derived class (/es).
3.5 Give number of bytes to be occupied by the object of the following class:
(a) publisher
(b) branch
(c) author
3.6. Write the names of data members accessible from the object of class author.
3.7. Write the names of all member functions accessible from the object of class author.
3.8 Write the names of all members accessible from member functions of class author.
Answer:
3.1 Multiple Inheritance
3.2 public
3.3 Constructors branch, publisher and author are executed.
Destructors author, publisher and branch will be executed.
3.4 Base classes : branch and publisher Derived class : author
3.5 a) publisher class object requires 93 bytes
b) branch class object requires 64 bytes
c) author class object requires 181 bytes

Question 4.
Consider the following C++ code and answer the questions.
class Personal
{
int Class, Rno;
char Section;
protected:
char Name[20];
public:
personal();
void pentry();
void Pdisplay();
};
class Marks:private Personal
{
float M{5};
protected:
char Grade[5];
public:
Marks();
void Mentry();
void Mdisplay();
};
class Result:public Marks
{
float Total,Agg;
public:
char FinalGrade, Commence[20];
Result();
void Rcalculate();
void Rdisplay();
};

4.1. Which type of Inheritance is shown in the program?
4.2. Specify the visibility mode of base classes.
4.3 Give the sequence of Constructor/Destructor Invocation when object of class Result is created.
4.4. Name the base class(/es) and derived class (/es).
4.5 Give number of bytes to be occupied by the object of the following class:
(a) Personal
(b) Marks
(c) Result
4.6. Write the names of data members accessible from die object of class Result.
4.7. Write the names of all member functions accessible from the object of class Result.
4.8 Write the names of all members accessible from member functions of class Result.
Answer:
4.1 Multilevel Inheritance
4.2 For Marks class – private visibility
For Result class – public visibility
4.3 Constructors Personal, Marks and Result be executed.
Destructors Result, Marks and Personal will be executed.
4.4 Base classes : Personal and Marks
Derived classes : Marks and Result

4.5 a) Personal class object requires 28 bytes (using Dev C++)
b) Marks class object requires 53 bytes (using Dev C++)
c) Result class requires 82 bytes (using Dev C++)

4.6 Data members FinalGrade, Commence(Own class members) alone can be accessed.
No members inherited under public, so base class members can not be accessed.

4.7 Member functions
Rcalculate( ), Rdisplay (own class member functions)
Mentry, Mdisplay (derived from Marks class) alone can be accessed.
Personal class public member functions can not be accessed because Marks class inherited under private visibility mode.

4.8 1) Data members

• Total, Agg, Final Grade and Commence of its own class
• M, Grade from Marks class can be accessed.

Personal class data members can not be accessed because Marks class inherited under private visibility mode.

2) Member functions
Mentry and Mdisplay from Marks class can be invoked from Result class member
Personal class member-functions can not be accessed because Marks class inherited under private visibility mode.

Question 5.
Write the output of the following program.
#include<iostream>
using namespace std;
class A
{
protected:
int x;
public:
void show()
{
cout<<“x = “<<x<<endl;
}
A()
{
cout<<endl<<” I am class A “<<endl;
}
~A()
{
cout<<endl<<” Bye”;
}
};
class B : public A
{
protected:
int y;
public:
B(int x, int y)
{
//this -> is used to denote the objects datamember this->x = x;
//this -> is used to denote the objects datamember this->y = y;
}
B()
{
cout<<endl<<“I am class B”<<endl;
}
~B()
{
cout<<endl<<” Bye”;
}
void show()
{
cout<<“x = “<<x<<endl;
cout<<“y = “<<y<<endl;
}
};
int main()
{
A objA;
B objB(30, 20);
objB.show();
return 0;
}
Output:

Question 6.
Debug the following program.
Output:
—————
15
14
13

Program :
%include(iostream.h)
#include<conio.h>
Class A
{
public;
int al,a2:a3;
void getdata[]
{
a1=15;
a2=13;a3=13;
}
}
Class B:: public A()
{
PUBLIC
voidfunc()
{
int b1:b2:b3;
A::getdata[];
b1=a1;
b2=a2;
a3=a3;
cout<<b1<<‘\t'<<b2<<‘t\'<<b3;
}
void main()
{
clrscr()
B der;
derl:func();
getch();
}
Answer:
Modified Error Free Program :
using namespace std;
#include<iostream>
#include<conio.h>
class A
{
public:
int a1,a2,a3;
void getdata()
{
a1=15;
a2=14;
a3=13;
}
};
class B : public A
{
public:
void func()
{
int b1,b2,b3;
A::getdata();
b1=a1;
b2=a2;
b3=a3;
cout<<b1<<‘\n'<<b2<<‘\n'<<b3;
}
};
int main()
{
B der;
der.func();
getch();
return 0;
}

11th Computer Science Guide Inheritance Additional Questions and Answers

Choose The Correct Answer:

Question 1.
When a derived class inherits only from one base class, it is known as ………………
(a) multiple inheritances
(b) multilevel inheritance
(c) hierarchical inheritance
(d) single inheritance
Answer:
(d) single inheritance

Question 2.
_____ enables new class and its objects to take on the properties of the existing classes.
a) Inheritance
b) Encapsulation
c) Overriding
d) None of these
Answer:
a) Inheritance

Question 3.
When more than one derived classes are created from a single base class, it is called ………………
(a) inheritance
(b) hybrid inheritance
(c) hierarchical inheritance
(d) multiple inheritances
Answer:
(c) hierarchical inheritance

Question 4.
A class that inherits from a superclass is called a _______ class.
a) Sub
b) Base class
c) Derived
d) Sub or Derived
Answer:
d) Sub or Derived

Question 5.
The ……………… are invoked in reverse order.
(a) constructor
(b) destructor
(c) pointer
(d) operator
Answer:
(b) destructor

Question 6.
There are ________ types of inheritance,
a) Two
b) Three
c) Four
d) Five
Answer:
d) Five

Question 7.
_________ inheritance is a type of inheritance.
a) Single or Hybrid
b) Multilevel / Hierarchical
c) Multiple
d) All the above
Answer:
d) All the above

Question 8.
When a derived class inherits only from one base class, it is known as ________ inheritance.
a) Single
b) Multilevel / Hierarchical
c) Multiple
d) Hybrid
Answer:
a) Single

Question 9.
When a derived class inherits from multiple base classes it is known as _________ inheritance.
a) Single
b) Multilevel / Hierarchical
c) Multiple
d) Hybrid
Answer:
c) Multiple

Question 10.
When more than one derived classes are created from a single base class, it is known
as_________inheritance.
a) Single
b) Hierarchical
c) Multiple
d) Hybrid
Answer:
b) Hierarchical

Question 11.
The transitive nature of inheritance is itself reflected by _____ form of inheritance.
a) Single
b) Multilevel
c) Multiple
d) Hybrid
Answer:
b) Multilevel

Question 12.
When a class is derived from a class which is a derived class – then it is referred to as ______ inheritance.
a) Single
b) Multilevel
c) Multiple
d) Hybrid
Answer:
b) Multilevel

Question 13.
When there is a combination of more than one type of inheritance, it is known as ________ inheritance.
a) Single
b) Multilevel
c) Multiple
d) Hybrid
Answer:
d) Hybrid

Question 14.
Hybrid inheritance may be a combination of ________inheritance.
a) Multilevel and Multiple
b) Hierarchical and Multilevel
c) Hierarchical, Multilevel and Multiple
d) All the above
Answer:
d) All the above

Question 15.
The order of inheritance by derived class, to inherit the base class is ________
a) Left to Right
b) Right to Left
c) Top to Bottom
d) None of these
Answer:
a) Left to Right

Question 16.
In ________ inheritance the base classes dc not have any relationship between them,
a) Single
b) Multilevel
c) Hybrid
d) Multiple
Answer:
d) Multiple

Question 17.
In ________inheritance a derived class itself acts as a base class to derive another class.
a) Single
b) Multilevel
c) Multiple
d) Multiple
Answer:
b) Multilevel

Question 18.
_________ inheritance is similar to relation between grandfather, father and child,
a) Single
b) Multilevel
c) Multiple
d) Multiple
Answer:
b) Multilevel

Question 19.
A class without any declaration will have ________byte size.
a) 1
b) 0
b) 2
d) 10
Answer:
a) 1

Question 20.
class x{}; x occupies
a) 1
b) 0
b) 2
d) 10
Answer:
a) 1

Question 21.
In inheritance, which member of the base class will be acquired by the derived class is done by using ________ .
a) Visibility modes
b) Data members
c) Member functions
d) None of these
Answer:
a) Visibility modes

Question 22.
The accessibility of base class by the derived class is controlled by ________
a) Visibility modes
b) Data members
c) Member functions
d) None of these
Answer:
a) Visibility modes

Question 23.
_______ is a visibility modes.
a) private
b) public
c) protected
d) All the above
Answer:
d) All the above

Question 24.
The default visibility mode is ________
a) private
b) public
c) protected
d) All the above
Answer:
a) private

Question 25.
When a base class is inherited with ________ visibility mode the public and protected members of the base class become ‘private’ members of the derived class.
a) private
b) public
c) protected
d) All the above
Answer:
a) private

Question 26.
When a base class is inherited with ________ visibility mode the protected and public members of the base class become ‘protected members’ of the derived class,
a) private
b) public
c) protected
d) All the above
Answer:
c) protected

Question 27.
When a base class is inherited with ________ visibility mode, the protected members of the base class will be inherited as protected members of the derived class and the public members of the base class will be inherited as public members of the derived class.
a) private
b) public
c) protected
d) All the above
Answer:
b) public

Question 28.
When classes are inherited with ________the private members of the base class are not inherited they are only visible.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
d) Either A or B or C

Question 29.
When classes are inherited with ________ the private members of the base class are continue to exist in derived classes, and cannot be accessed.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
d) Either A or B or C

Question 30.
________inheritance should be used when you want the features of the base class to be available to the derived class but not to the classes that are derived from the derived class.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
a) private

Question 31.
________ inheritance should be used when features of base class to be available only to the derived class members but not to the outside world.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
c) protected

Question 32.
_____ inheritance can be used when features of base class to be available the derived class members and also to the outside world.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
b) public

Question 33.
When an object of the derived class is created, the compiler first call the ________class constructor.
a) Base
b) Derived
c) Either Base or Derived
d) None of these
Answer:
a) Base

Question 34.
When the object of a derived class expires first the ________class destructor is invoked.
a) Base
b) Derived
c) Either Base or Derived
d) None of these
Answer:
b) Derived

Question 35.
The ________ are executed in the order of inherited class.
a) Constructors
b) Destructors
c) Either A or B
d) None of these
Answer:
a) Constructors

Question 36.
The ________ are executed in the reverse order.
a) Constructors
b) Destructors
c) Either A or B
d) None of these
Answer:
b) Destructors

Question 37.
If there are multiple base classes, then it starts executing from the ________base class.
a) Leftmost
b) Rightmost
c) Compiler decided
d) None of these
Answer:
a) Leftmost

Question 38.
________ members of the base class can be indirectly accessed by the derived class using the public or protected member function of the base class.
a) private
b) public
c) protected
d) Either A or B or C
Answer:
a) private

Question 39.
________ member function has the access privilege for the private members of the base class.
a) public
b) protected
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 40.
________ functions can access the private members.
a) Member
b) Non-member
c) Destructor
d) None of these
Answer:
a) Member

Question 41.
In case of inheritance there are situations where the member function of the base class and derived classes have the same name. The ________operator resolves this problem.
a) Conditional
b) Membership
c) Scope resolution
d) None of these
Answer:
c) Scope resolution

Question 42.
When a derived class member function has the same name as that of its base class member function, the derived class member function ________the base class’s inherited function.
a) Shadows
b) Hides
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 43.
When a derived class member function has the same name as that of its base class member function, the derived class member function shadows/hides the base class’s inherited function is called function ________
a) Overriding
b) Overloading
c) Shadowing
d) Either A or C
Answer:
d) Either A or C

Question 44.
________ pointer is a constant pointer that holds the memory address of the current object.
a) this
b) void
c) new
d) None of these
Answer:
a) this

Question 45.
________pointer is useful when the argument variable name in the member function and the data member name are same.
a) this
b) void
c) new
d) None of these
Answer:
a) this

Very Short Answers (2 Marks)

Question 1.
Write a short note on hierarchical inheritance.
Answer:
When more than one derived class is created from a single base class, it is known as Hierarchical inheritance.

Question 2.
What are the types of inheritance?
Answer:
There are different types of inheritance viz., Single Inheritance, Multiple inheritance, Multilevel inheritance, hybrid inheritance and hierarchical inheritance.

Question 3.
Give the syntax of deriving a class.
Answer:
Syntax:
class derived_dass_name :visibility_mode base_dass_name
{
// members of derivedclass
};

Question 4.
Write note on this pointer.
Answer:
‘this’ pointer is a constant pointer that holds the memory address of the current object. It identifies the currently calling object. It is useful when the argument variable name in the member function and the data member name are same. To identify the data member it will be given as this->data member name.

Short Answers (3 Marks)

Question 1.
What are inheritance and access control?
Answer:
When you declare a derived class, a visibility mode can precede each base class in the base list of the derived class. This does not alter the access attributes of the individual members of a base class but allows the derived class to access the members of a base class with restriction.

Classes can be derived using any of the three visibility modes:

1. In a public base class, public and protected members of the base class remain public and protected members of the derived class.
2. In a protected base class, public and protected members of the base class are protected members of the derived class.
3. In a private base class, public and protected members of the base class become private members of the derived class.
4. In all these cases, private members of the base class remain private and cannot be used by the derived class.
5. However, it can be indirectly accessed by the derived class using the public or protected member function of the base class since they have the access privilege for the private members of the base class.

Question 2.
Write a program for the working of constructors and destructors under inheritance.
Program
#include<iostream>
using namespace std;
class base
{
public:
base()
{
cout<<“\nConstructor of base class…”;
}
~base()
{
cout<<“\nDestructor of base class….”;
}
};
class derived:public base
{
public :
derived()
{
cout << “\nConstructor of derived …”;
}
~derived()
{
cout << “\nDestructor of derived…”;
}
};
class derived 1 :public derived
{
public :
derived 1()
{
cout << “\nConstructor of derived! //, …”;
}
~derived1()
{
cout << “\nDestructor of derived …”;
}
};
int main()
{
derivedl x;
return 0;
}

Output:
Constructor of base class…
Constructor of derived …
Constructor of derived …
Destructor of derived …
Destructor of derived …
Destructor of base class….

Question 3.
What about access control in a publicly derived class?
Answer:
From a publicly derived class, public and protected members of the base class remain public and protected members of the derived class. The public members can be accessed by the object of the derived class similar to its own members in public.

Question 4.
What about access control in the privately derived class?
Answer:
From a privately derived class, public and protected members of the base class become private members of the derived class. Hence it is not possible to access the derived members using the object of the derived class. The Derived members are invoked by calling it from the publicly defined members.

Explain in Detail

Question 1.
Write a program to implement single Inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno;
public:
void acceptnameO
{
cout<<“\n Enter roll no and name ..”;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name :-“<<name<<endl;
}
};
class exam : public student
//derived class with single base class
{
public:
int markl, mark2 ,mark3,mark4,marks, mark6, total;
void acceptmark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks..
cin>>mark1>>mark2>>mark3>>
mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t, Marks Obtained
cout<<“\n Language.. “<<mark1;
cout<<“\n English .. “<<mark2;
cout<<“\n Physics .. “<<mark3;
cout<<“\n Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout«”\n Maths .. “<<mark6;
}
};
int main()
{
exam e1;
el.acceptname();
//calling base class function using derived
class object
e1.acceptmark();
e1.displayname();
//calling base class function using derived
class object
e1l.displaymark();
return 0;
}
Output
Enter roll no and name . . 1201
KANNAN
Enter lang,eng,phy,che,esc,mat
marks.. 100 100 100 100 100 100
Roll no:-1201
Name:-KANNAN
Marks Obtained
Language.. 100
English .. 100
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100

Question 2.
Write a program to implement multiple inheritance.
Program :
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno;
public:
void acceptname()
{
cout<<“\n Enter roll no and name.. “;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name:-” << name << endl;
}
};
class detail //Base class
{
int dd,mm,yy;
char cl[4];
public:
void acceptdob()
{
cout<<“\n Enter date,month,year in digits and class ..”;
cin>>dd>>mm>>yy>>cl;
}
void displaydob()
{
cout<<“\n class:-“<<cl;
cout<<“\t\t DOB : “<<dd«” –
“<<mm<<“-” <<yy<<endl;
}
};
//derived class with multiple base class
class exam: public student, public detail
{
public:
int mark1, mark2 ,mark3,mark4,mark5, mark6, total;
void acceptmark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks..”;
cin>>mark1>>mark2>>mark3>> mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained
cout<<“\n Language.. “<<markl;
cout<<“\n English .. “<<mark2;
cout<<“\n Physics .. “<<mark3;
cout<<“\n Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout<<“\n Maths .. “<<mark6;
}
};
int main()
{
exam e1;
//calling base class function using derived
class object
e1.acceptname();
//calling base class function using derived
class object
e1.acceptdob();
e1.acceptmark();
//calling base class function using derived
class object
e1.displayname();
//calling base class function using derived
class object
e1.displaydob();
e1.displaymark();
return 0;
}
Output:
Enter roll no and name . . 1201 MEENA
Enter date, month, year in digits and
class .. 7 12 2001 XII
Enter lang, eng, phy, che, esc, mat
marks.. 96 98 100 100 100 100
Roll no:-1201
Name MEENA
class:-XII
DOB : 7 – 12 -2001
Marks Obtained
Language..96
English .. 98
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100

Question 3.
Write a program to implement multilevel inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno; public:
void acceptname()
{
cout<<“\n Enter roll no and name.. “;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name << name <<
endl;
}
};
//derived class with single base class class
exam: public student
{
public:
int mark1, mark2, mark3, mark4, mark5, mark6;
void accept mark()
{
cout<<“\n Enter
lang,eng,phy,che,esc,mat marks..’; cin>>mark1>>mark2>>mark3>> mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained
cout<<“\n Language… “<<markl;
cout<<“\n English… “<<mark2;
cout<<“\n Physics… “<<mark3;
cout<<“\n Chemistry… “<<mark4;
cout<<“\n Comp.sci… “<<mark5;
cout<<“\n Maths… “<<mark6;
}
};
class result: public exam
{
int total;
public:
void showresult()
{
total=markl+mark2+mark3+mark 4+mark5+mark6;
cout<<“\nTOTAL MARK SCORED : “<<total;
}
};
int main()
{
result r1;
//calling base class function using derived
class object
r1.acceptname();
//calling base class function which itself is a derived
r1.acceptmark();
// class function using its derived class object
r1.displayname();
//calling base class function
using derived class //object
//calling base class function which itself is a derived
r1.displaymark();
//class function using its derived class object
r1.showresult();
//calling the child class
function
return 0;
}

Output :
Enter roll no and name .. SARATHI
Enter lang,eng,phy,che,csc,mat
marks.. 96 98 100 100 100 100
Roll no:-1201
Name:-SARATHI
Marks Obtained
Language… 96
English… 98
Physics… 100
Chemistry… 100
Comp.sci… 100
Maths… 100
TOTAL MARK SCORED: 594

Question 4.
Write a program to implement hierarchical inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno;
public:
void acceptname()
{
cout<<“\n Enter roll no and name..”;
cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cout<<“\n Name :-” <<name <<
endl;
}
};
//derived class with single base class
class qexam: public student
{
public:
int mark1, mark2, mark3, mark4, marks, mark6;
void accent mark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks for quarterly exam”;
cin>>markl>>mark2>>mark3>>mark4>>mark5>>mark6;
}
void displaymark()
cout< <“\n\t\t Marks Obtained in quarterly”;
cout< <“\n Language.. “<<marki;
cout<<”\n English .. “<<mark2;
cout< <“\n Physics .. “<<mark3;
cout< <“\fl Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout< <“\n Maths .. “<<mark6;
}
};
//derived class with single base class
class hexam : public student
{
public:
int mark1, mark2, mark3, mark4, marks, mark6;
void acceptmark()
{
cout<<“\n Enter lang, eng, phy, che, esc, mat marks for half/early exam..”;
cin>>markl>>mark2>>mark3>>mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained in Halfyearly”;
cout<<“\n Language., “<< mark1;
coutcc”\n English .. “<< mark2;
coutcc”\n Physics .. “<< mark3;
coutcc”\n Chemistry.. “<< mark4;
coutcc”\n Comp.sci.. “<< mark5;
coutcc”\n Maths .. “<< mark6;
}
};
int main()
{
qexam q1;
hexam hi;
//calling base class function using derived class object
q1.acceptname();
//calling base class function
q1. acceptmark();
//calling base class function using derived
class object
h1.acceptname();
//calling base class function using derived class object
h1.displayname(); .
h1,acceptmark();
//calling base class- function using its // derived class object
h1.displaymark();
return 0
}
Output:
Enter roll no and name . . 1201
KANNAN
Enter lang,eng,phy,che,esc,mat
marks for quarterly exam. .
95 96 100 98 100 99
Roll no :-1201
Name :-KANNAN
Marks Obtained in quarterly
Language.. 95
English .. 96
Physics .. 100
Chemistry.. 98
Comp.sci.. 100
Maths .. 99
Enter roll no and name . . 1201
KANNAN
Enter lang,eng,phy,che,esc, mat marks for the half-yearly exam.
96 98 100 100 100 100
Roll no:-1201
Name:-KANNAN
Marks Obtained in Halfyearly
Language.. 96
English .. 98
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100

Question 5.
Write a program to implement hybrid inheritance.
Program
# include <iostream>
using namespace std;
class student //base class
{
private :
char name[20];
int rno; public:
void acceptname()
{
cout<<“\n Enter roll no and name.. cin>>rno>>name;
}
void displayname()
{
cout<<“\n Roll no :-“<<rno;
cou<<“\n Name name <<
endl;
}
};
//derived class with the single base class
class exam: public student
{
public:
int mark1, mark2, mark3, mark4, marks, mark6;
void accept mark()
{
cout<<“\n Enter larig, eng, phy, che, esc,mat marks..”;
cin>>markl>>mark2>>mark3>> mark4>>mark5>>mark6;
}
void displaymark()
{
cout<<“\n\t\t Marks Obtained “;
cout<<“\n Language.. “<<markl;
cout<<“\n English .. “<<mark2;
cout<<“\n Physics .. “<<mark3;
cout<<“\n Chemistry.. “<<mark4;
cout<<“\n Comp.sci.. “<<mark5;
cout<< “\n Maths .. “<<mark6;
}
};
class detail //base classs 2
{
int dd,mm,yy;
char cl[4];
public:
void acceptdob()
{
cout<<“\n Enter date,month,year in digits and class ..”;
cin>>dd>>mm>>yy>>cl;
}
void displaydob()
{
cou<<“\n class :-“<<cl;
cou<<“\t\t DOB : “<<dd<<” – ”
<<mm<<“-” <<yy<<endl;
}
};
//inherits from the exam, which itself is a //derived
class and also from class detail
class result: public exam, public detail
{
int total;
public:
void showresuit()
{
total=markl+mark2+mark3+ mark4+mark5+mark6;
cout<<“\nTOTAL MARK SCORED: ” <<total;
}
};
class detail //base classs 2
{
int dd,mm,yy;
char cl[4];
public:
void acceptdob()
{
cout<<“\n Enter date,month,year in digits and class ..
cin>>dd>>mm>>yy>>cl;
}
void displaydob()
{
cout<<“\n class :-“<<cl;
cout<<“\t\t DOB : “<<dd<<” – ”
<<mm<<“-” <<yy<<endl;
}
};
//inherits from the exam, which itself is a //derived class and also from class detail class result: public exam, public detail
{
int total;
public:
void showresuit()
{
total = markl + mark2-i-mark3 + mark4+mark5+mark6;
cout<<“\nTOTAL MARK SCORED : ” <<total;
}
};
int main()
{
result r1;
//calling base class function using derived class object
r1.acceptname();
//calling base class which itsel is a derived class function using its derived class object
r1.acceptmark();
r1.acceptdob();
cout<<“\n\n\t\t MARKS STATEMENT”; //calling base class function using derived class object
r1.displayname();
r1.displaydob();
//calling base class which itsel is a derived class function using its derived class object
r1.displaymark();
//calling the child class function
r1.showresuit();
return 0;
>
Output:
Enter roll no and name .. 1201 RAGU
Enter lang,eng,phy,che,esc,mat
marks.. 96 98 100 100 100 100
Enter date,month,year in digits and class .. 7 12 2001 XII
MARKS STATEMENT
Roll no :-1201
Name :-RAGU
class : -XII
DOB : 7 – 12 -2001
Marks Obtained
Language..96
English .. 98
Physics .. 100
Chemistry.. 100
Comp.sci.. 100
Maths .. 100
TOTAL MARK SCORED: 594

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 6 Gaseous State Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

11th Chemistry Guide Gaseous State Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the intermolecular interactions become significant
Answer:
(d) at high pressure the intermolecular interactions become significant

Question 2.
Rate of diffusion of a gas is
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?
(a) [P + \(\frac{a}{n^{2} V^{2}}\)](V – nb) = nRT

(b) [P + \(\frac{n a}{n^{2} V^{2}}\)](V – nb) = nRT

(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

(d) [P + \(\frac{n^{2} a^{2}}{V^{2}}\)(V – nb) = nRT]
Answer:
(c) [P + \(\frac{a n^{2}}{V^{2}}\)](V – nb) = nRT

Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) collide without loss of energy
Answer:
(b) exert no attractive forces on each other

Question 5.
Equal weights of methane and oxygen are mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{3}\) × 273 × 298
Answer:
(a) \(\frac{1}{3}\)

Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature

Question 7.
In a closed room of 1000 m³ a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle

Question 9.
The value of universal gas constant depends upon
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Question 10.
The value of the gas constant R is
(a) 0.082 dm³atm
(b) 0.987 cal mol^-1K^-1
(c) 8.3J mol^-1K^-1
(d) 8erg mol^-1K^-1
Answer:
(c) 8.3J mol^-1K^-1

Question 11.
Use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Question 12.
The table indicates the value of vanderWaals constant ‘a’ in (dm³)² atm. mol^-2. The gas which can be most easily liquefied is

(a) O₂
(b) N₂
(c) NH₃
(d) CH₄
Answer:
(c) NH₃

Question 13.
Consider the following statements
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises
Select the correct statement
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Answer:
(b) II and III

Question 14.
Compressibility factor for CO₂ at 400 K and 71.0 bar is 0.8697. The molar volume of CO₂ under these conditions is
(a) 22.04 dm³
(b) 2.24 dm³
(c) 0.41 dm³
(d) 19.5 dm³
Answer:
(c) 0.41 dm³

Question 15.
If temperature and volume of an ideal gas is increased to twice its valuesthe initial pressure P becomes
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(b) 2P

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3 times that of a hydrocarbon having molecular formula C_nH_{2n – 2}. What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(d) 1

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape.
(a) \(\frac{3}{8}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{1}{4}\)
Answer:
(c) \(\frac{1}{8}\)

Question 18.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion i.e., α = 1\(\left[\frac{\partial V}{\delta T}\right]\)Vp. For an ideal gas α is equal to
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(a) T

Question 19.
Four gases P, Q, R, and S have almost same values of ‘b’ but their a’ values (a. h are Vander Waals Constants) are in the order Q < R < S < p. At a particular temperature, a’nong the four gases the most easily liquelìable one is
(a) P
(b) Q
(c) R
(d) S
Answer:
(c) R

Question 20.
Maximum deviation from ideal gas is expected
(a) CH₄(g)
(b) NH₃(g)
(c) H₂ (g)
(d) N₂ (g)
Answer:
(b) NH₃(g)

Question 21.
The units of Vander Waals constants ‘b’ and ‘a’ respectively
(a) mol L^-1 and L atm² mol^-1
(b) mol L and L atm mol²
(c) mol^-1 L and L² atm mol^-2
(d) none of these
Answer:
(c) mol^-1 L and L² atm mol^-2

Question 22.
Assertion:
Critical temperature of CO₂ is 304 K it can be liquefied above 304 K.
Reason:
For a given mass of gas, volume is to directly proportional to pressure at constant temperature
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 23.
What is the density of N₂ gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K^-1 mol^-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas? (T is measured in K)

Answer:

Question 25.
25g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI

II. Answer these questions briefly:

Question 26.
State Boyle’s law.
Answer:
At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
Mathematically, the Boyle’s law can be written as
V ∝ \(\frac{1}{P}\)     …………..(1)
(T and n are fixed, T-temperature, n-number of moles)
V = k × \(\frac{1}{P}\) …………(2)
k – proportionality constant
PV = k (at constant temperature and mass)

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law.
Answer:
Yes, a balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law. Volume of balloon decrease when the temperature reduced from room temperature to low temperature. When cooled, the kinetic energy of the gas molecules decreases, so that the volume of the balloon also decreases.

Question 28.
Name two items that can serve as a model for ‘Gay Lusaac’ law and explain.
Answer:
Firing a bullet:
When gunpowder burns, it creates; a significant amount of superheated gas. The high pressure of the hot gas behind the bullet forces it out, of the barrel of the gun.
Heating food in an oven:
When you keep food in an oven for heating, the air inside the oven is heated, thus pressurized.

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what j the mathematical expression means.
Answer:
The mathematical expression that relates gas volume and moles is Avogadro’s hypothesis. It may be expressed as
V ∝ n,
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = constant
where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂ are a different set of values of volume and number of moles of the same gas at same temperature and pressure.

Question 30.
What are ideal gases? In what way real gases differ from ideal gases?
Answer:
The kinetic theory of gases which is the basis for the gas equation (PV = nRT), assumes that the individual gas molecules occupy negligible volume when j compared to the total volume of the gas and there is no attractive force between the gas molecules. Gases whose behaviour is consistent with these assumptions under all conditions are called ideal gases.

But in practice both these assumptions are not valid under all conditions. For example, the fact that gases can be liquefied shows that the attractive force exists among molecules. Hence, there is no gas which behaves ideally under all conditions. The non-ideal gases are called real gases. The real gases f tend to approach the ideal behaviour under certain conditions.

Question 31.
Can a Vander Waals gas with a = 0 be liquefied? Explain.
Answer:
If the vander Waals constant (a) = 0 for a gas, then it behaves ideally, (i.e.,) there is no intermolecular forces of attraction. So it cannot be liquefied. Moreover,
P_c = \(\frac{a}{27 b^{2}}\)
If a = 0, then P_c = 0; therefore it cannot he liquefied.

Question 32.
Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:
Gaseous pressure is developed by the continuous bombardment of the molecules of the gas among themselves and also with the walls of the container. When the molecules hit the sticky area of the container, the number of molecules decreases and hence, the pressure decreases. Therefore, pressure is less than the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept under water during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The tyre of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude.
Answer:
(a) Aerated water bottles contains excess dissolved oxygen and minerals which dissolved under certain pressure and if this pressure suddenly decrease due to change in atmospheric pressure, the bottle will certainly burst with decrease the amount of dissolved oxygen in water , this tends to change the aerated water into normal-water.

(b) The vapour pressure of ammonia at room temperature is very high and hence the ammonia will evaporate unless the vapour pressure is decreased. On cooling the vapour pressure decreases so that the liquid remains in the same state. Hence, the bottle is cooled ’ before opening.

(c) In Summer due to hot weather condition, the air inside the tyre expands to large volume due to heat as compared to winter therefore inflated to lesser pressure in summer.

(d) As we move to higher altitude, the atmospheric pressure decreases, therefore balloon can J easily expand to large volume.

Question 34.
Give suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container.
Answer:
According to kinetic theory, gas molecules are moving continuous at random. Hence, they do not settle at the bottom of a container.

(b) Gases diffuse through all the space available to them.
Answer:
Gases have a tendency to occupy all the available space. The gas molecules migrate from region of higher concentration to a region of lowrer concentration. Hence, gases diffuse through all the space available to them.

Question 35.
Suggest why there is no hydrogen in our atmosphere. Why does the moon have no atmosphere?
Answer:
Hydrogen has tendency to combine with oxygen to from water vapour. Hence, the presence of hydrogen is negligible in the atmosphere. The gravitational pull in the moon is very less and hence, there is no atmosphere in the moon.

Question 36.
Explain whether a gas approaches ideal behavior or deviates from ideal behaviour if
(a) it is compressed to a smaller volume at f constant pressure.
Answer:
When the gas is compressed to a smaller volume, the compressibility factor (Z) decreases. Hence, the gas deviates from ideal behavior

(b) the temperature is raised at while keeping the volume constant.
Answer:
When the temperature is increased, the compressibility factor approaches unity. Hence, the gas behaves ideally.

(c) More gas is introduced into the same volume and at the same temperature.
Answer:
When more gas is introduced into a container of the same volume and at the same temperature, the compressibility factor tend to unity. Hence, the gas behaves ideally.

Question 37.
Which of the following gases would you expect to deviate from ideal behavior under conditions of low temperature F₂, Cl₂ or Br₂? Explain.
Answer:
Bromine has greater tendency to deviate from ideal behavior at low temperature. The compressibility factor tends to deviate from unity for bromine.

Question 38.
Distinguish between diffusion and effusion.
Answer:

Question 39.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
If aerosol cans are heated, then they will produce more vapour inside the can, which will make the pressure rise very quickly. The rises of temperature can double the pressure inside. Even though the cans are tested, they will burst if the pressure goes up too far. A bursting can could be dangerous.

Question 40.
When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
Answer:
Helium floats because it is buoyant; its molecules are lighter than the nitrogen and oxygen molecules of our atmosphere and so they rise above it. In the car, it’s the air molecules that are actually getting pulled and pushed around by gravity as the result of the accelerating frame.

That means it moves in a direction opposite to the force on the surrounding air. Normally, air is pulled downwards due to gravity, which pushes the balloon upwards. In this case, the surrounding air in the car is pulled forward by the deceleration of the car, which pushes the helium balloon backwards.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It would be harder on the top of the mountain; because the external pressure pushing on the liquid to force it up the straw is less.

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and Volume.
Answer:
Vander Waals equation for a real gas is given by
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Where n is the number of moles of gas and V is the volume of the container
⇒ P ∝ \(\frac{n^{2}}{V^{2}}\)
⇒ P = \(\frac{a n^{2}}{V^{2}}\)
Where a is proportionality constant and depends on the nature of gas
Therefore,
P = P + \(\frac{a n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\) π(2r)³ = 8V_m
where V_m is a volume of a single molecule.
Excluded volume for single molecule = \(\frac{8 V_{m}}{2}\) = 4V_m
Excluded volume for n molecule
= n(4V_m) = nb
Where b is van der waals constant which is equal to 4V_m
⇒ V’ = nb
V_ideal = V – nb
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas.

Question 43.
Derive the values of van der Vaals equation constants in terms of critical constants.
Answer:
The Van der walls equation for n moles is
(p + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT ……… (1)
For 1 mole
(P + \(\frac{n^{2}}{V^{2}}\)) (V – b) = RT ……………(2)

From the equation we can derive the values of critical constants P_c, V_c, and T_c in terms of a and b, the van der Waals constants, On expanding the above equation,
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0 ………….(3)

Multiply equation (3) by \(\frac{V^{2}}{P}\)
\(\frac{V^{2}}{P}\) (PV +\(\frac{a}{V}\) – pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V³ + \(\frac{a}{P}\)V – bV² – \(\frac{a b}{P}\) – \(\frac{R T V^{2}}{P}\) = 0 ………..(4)
When the above equation is rearranged in powers of V

V³ – [\(\frac{R T}{P}\) + b]V² + \(\frac{a}{P}\)V – \(\frac{a b}{P}\) = 0 ……………..(5)

The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V_c. The pressure and temperature becomes P_c and T_c respectively
V = V_c
V – V_c = 0
(V – V_c)³ = 0
V³ – 3V_cV² + 3V_c²V – V_c³ = 0 …………..(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V₂, V and constant terms in (5) and (6).
-3V_cV² = –[\(\frac{R T_{c}}{P_{c}}\) + b] V²

3V_c = \(\frac{R T_{c}}{P_{c}}\) + b ………………(7)
3V_c² = \(\frac{a}{P_{c}}\) ……………(8)
V_c³ = \(\frac{ab}{P_{c}}\) …………….(9)

The critical constants can be calculated using the values of van der waals constant of a gas and vice versa.
a = 3V_c² P_c and
b = \(\frac{V_{c}}{3}\)

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of moon?
Answer:
Astronauts must wear spacesuits whenever they leave a spacecraft and are exposed to the environment of moon. In moon, there is no air to breath and no air pressure. Moon is extremely cold and filled with dangerous radiation. Without protection, an astronaut would quickly die in space. Spacesuits are specially designed to protect astronauts from the cold, radiation and low pressure in space. They also provide air to breathe. Wearing a spacesuit allows an astronaut to survive and work in moon.

Question 45.
When ammonia combines with HCl, NH₄Cl is formed as white dense fumes. Why do more fumes appear near HCl?
Answer:
HCl and NH₄Cl molecules diffuse through the air towards each other. When they meet, they reactto
form a white powder called ammonium chloride, NH₄Cl.
HCl_(g) + NH_3(g) ⇌ NHC1

Hydrogen chloride + ammonia ⇌ ammonium chloride.
The ring of white powder is closer to the HCl than
the NH₃. This is because the NH₃ molecules are lighter (smaller) and have diffùsed more quickly through the air in the tube.

Question 46.
A sample of gas at 15°C at 1 atm. has a volume of 2.58 dm³. When the temperature is raised to 38°C at 1 atm does the volume of the gas Increase? If so, calculate the final volume.
Answer:
T₁ = 15°C + 273;
T₂ = 38 + 273
T₁ = 228;
T₂ = 311K
V₁ = 2.58dm³;
V₂ = ?
(P = 1 atom constant)
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

V₂ = \(\left[\frac{V_{1}}{T_{1}}\right]\) × T₂

= \(\frac{2.58 d m^{3}}{288 K}\) × 311K
V₂ = 2.78 dm³ i.e, volume increased from 2.58 dm³ to 2.78 dm³

Question 47.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen In a vessel of volume of 37.6 dm³ at 298K, and the sample B Is in a vessel of volume 16.5 dm³ at 298K. Calculate the number of moles in sample B.
Answer:
n_A = 1.5mol; n_B = ?
V_A = 37.6 dm³; V_B = 16.5 dm³
(T = 298K constant)
\(\frac{V_{A}}{n_{A}}=\frac{V_{B}}{n_{B}}\)

n_A = \(\left[\frac{n_{A}}{n_{B}}\right] V_{B}\)

Question 48.
Sulphur hexafluoride Is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas In a steel vessel of volume 5.43 dm3 at 69.5°C, assumIng Ideal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm³
T = 69.5 + 273 = 342.5
P =?
PV = nRT
P = nRT/V
P =
P = 94.25 atm

Question 49.
Argon is an Inert gas used In light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C Is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P₁ = 1.2 atm
T₁ = 180°C + 273 = 291 K
T₂ = 850°C + 273 = 358 K
P₂ =?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P₂ = \(\left[\frac{P_{1}}{T_{1}}\right] \times T_{2}\)

P₂ = \(\frac{1.2 a t m}{291 K}\) × 358 K
P₂ = 1.48 atm

Question 50.
A small bubble rises from the bottom of a lake where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure Is I arm. Calculate the final volume in (mL) of the bubble, If its initial volume 1.5 mL.
Answer:
T₁ = 6°C + 273 = 279 K
P₁ = 4 atm; V₂ = 1.5 ml
T₂ = 25°C + 273 = 298 K
P₂ = 1 atm; V₂ =?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

= \(\frac{4 a t m \times 1.5 m l \times 298 K}{279 K \times 1 a t m}\) = 6.41 mol

Question 51.
Hydrochloric acid Is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 × 10^-3 dm³ of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 × 10^-3 dm³
P = 742 mm of Hg
T = 298K;
m =?
m = \(\frac{P V}{R T}\)

= \(\frac{742 m m H g \times 154.4 \times 10^{-3} L}{62 m m H g L K^{-1} m o l^{-1} \times 298 K}\)

n = \(\frac{\text { Mass }}{\text { Molar Mass }}\)

Mass = n × Molar Mass
= 0.0006 × 2.016
= 0.0121 g = 12.1 mg

Question 52.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What Is the molar mass of the unknowa gas?
Answer:

Question 53.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure In the tanks Is 9.21 atm. calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:

Question 54.
A combustible gas is stored in a metal tank at a pressure of 2.98 atm at 25°C. The tank can withstand a maximum pressure of 12 atm after which It will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal 1100 K).
Answer:
Pressure of the gas in the tank at its melting point
T = 298 K;
P₁ = 2.98 atom;
T₂ = 1100 K;
P₂ =?
\(\frac{P_{1} P_{2}}{T_{1} T_{2}}\) =????
⇒ P₂ = \(\frac{P}{T_{1}}\) × T₂
= \(\frac{2.98 \text { atm }}{298 K}\) × 1100 K = 11 atm

At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

11th Chemistry Guide Gaseous State Additional Questions and Answers

I. Choose the best Answer:

Question 1.
The approximate volume percentage of nitrogen and oxygen in the atmosphere of air are _______ and ______ respectively.
(a) 21, 68
(b) 21, 78
(c) 78, 2
(d) 80, 21
Answer:
(c) 78, 2

Question 2.
The SI unit of pressure is
(a) pascal
(b) atmosphere
(c) bar
(d) torr
Answer:
(a) pascal

Question 3.
The compound widely used in the refrigerator as coolant is
(a) Freon-2
(b) Freon-12
(c) Freon-13
(d) Freon-14
Answer:
(b) Freon-12

Question 4.
“For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature”. This statement is
(a) Boyle’s law
(b) Gay-Lussac law
(c) Avogadro’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 5.
Hydrogen is placed in the ______ of the periodic table.
(a) Group -1
(b) group-17
(c) group-18
(d) group-2
Answer:
(a) Group -1

Question 6.
The plot of the volume of the gas against its temperature at a given pressure is called
(a) isotone
(b) isobar
(c) isomer
(d) isotactic
Answer:
(b) isobar

Question 7.
At constant temperature for a given mass, for each degree rise in temperature, all gases expand by of their volume at 0°C.
(a) 273
(b) 298
(c) \(\frac{1}{273}\)

(d) \(\frac{1}{298}\)
Answer:
(c) \(\frac{1}{273}\)

Question 8.
The precise value of temperature at which the volume of the gas becomes zero is
(a) -273.15 °C
(b) -273 °C
(c) -298.15°C
(d) -298 °C
Answer:
(a) -273.15 °C

Question 9.
‘At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature”. This statement is
(a) Charle’s law
(b) Boyle’s law
(c) Gay-Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay-Lussac’s law

Question 10.
The value of gas constant, R, in terms of JK^-1 mol^-1 is
(a) 8.314
(b) 4.184
(c) 0.0821
(d) 1.987
Answer:
(a) 8.314

Question 11.
A mixture of gases containing 4 mole of hydrogen and 6 mole of oxygen. The partial pressure of hydrogen, if the total pressure is 5 atm, is
(a) 10 atm
(b) 0.4 atm
(c) 20 atm
(d) 2 atm
Answer:
(d) 2 atm

Question 12.
A mixture of gases containing 1 mole of He, 4 mole of Ne and 5 mole of Xe. The correct order of partial pressure of the gases, if the total pressure is 10 atm is
(a) Xe < Ne < He
(b) He < Ne < Xe
(c) Xe < Ne < He
(d) He < Xe < Ne
Answer:
(b) He < Ne < Xe

Question 13.
The property of a gas which involves the movement of the gas molecules through another gases is called
(a) effusion
(b) dissolution
(c) difusion
(d) expansion
Answer:
(c) difusion

Question 14.
The process in which agas escapes from a container through a very small hole is called
(a) diffusion
(b) effusion
(c) occlusion
(d) dilution
Answer:
(a) diffusion

Question 15.
The rate of diffusion of a gas is inversely proportional to the
(a) square of molar mass
(b) square root of density
(c) square root of molar mass
(d) square of density
Answer:
(c) square root of molar mass

Question 16.
An unknown gas X diffuses at a rate of 2 times of oxygen at the same temperature and pressure. The molar mass (in g mol^-1) of the gas ‘X is (Molar mass of oxygen is 32 g mol^-1)
(a) 8
(b) 16
(c) 20
(d)12
Answer:
(a) 8

Question 17.
The deviation of real gases from ideal behavior is measured in terms of
(a) expansivity factor
(b) molar mass
(c) pressure
(d) compressibility factor
Answer:
(d) compressibility factor

Question 18.
The gases which deviate from ideal behavior at
(a) low temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and low pressure
(d) high temperature and high pressure
Answer:
(b) high temperature and low pressure

Question 19.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called temperature
(a) inversion
(b) ideal
(c) Boyle
(d) reversible
Answer:
(c) Boyle

Question 20.
Ideal gas equation for ‘n’ moles is
(a) \(\frac{P}{R}=\frac{n T}{V}\)

(b) PV = \(\frac{n R}{T}\)

(c) \(\frac{V}{T}=\frac{n P}{R}\)

(d) \(\frac{P}{T}=\frac{n R}{V}\)
Answer:
(d) \(\frac{P}{T}=\frac{n R}{V}\)

Question 21.
The pressure correction introduced by Van der Waals is,
P_ideal =
(a) P + \(\frac{a V^{2}}{n^{2}}\)

(b) P + \(\frac{a n}{V^{2}}\)

(c) P + \(\frac{a n^{2}}{V^{2}}\)

(d) P + \(\frac{a V}{n^{2}}\)
Answer:
(c) P + \(\frac{a n^{2}}{V^{2}}\)

Question 22.
The volume correction introduced by Van der Waals is, V_ideal =
(a) V – nb
(b) V + nb
(c) \(\frac{V}{n b}\)
(d) b – nV
Answer:
(b) V + nb

Question 23.
Van der waals equation for one mole of a gas is
(a) (P + \(\frac{a}{V}\))(V-b) = RT
(b) (P – \(\frac{a}{V^{2}}\))(V + -b) = RT
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT
(d) (P + \(\frac{a}{V}\))(V + b) = RT
Answer:
(c) (P + \(\frac{a}{V^{2}}\))(V – b) = RT

Question 24.
The unit of Van der Waals constant V is
(a) atm lit mol^-1
(b) atm lit² mol^-2
(c) atm lit^-2 mol²
(d) atm lit^-1 mol²
Answer:
(b) atm lit² mol^-2

Question 25.
The unit of Van der waals constant ‘b’ is
(a) lit mol^-1
(b) lit mol
(c) atm lit mol^-1
(d) atm lif^-1 mol^-2
Answer:
(a) lit mol^-1

Question 26.
The temperature above which a gas cannot be liquefied even at high pressure is called is ________ temperature.
(a) ideal
(b) inversion
(c) critical
(d) real
Answer:
(c) critical

Question 27.
The gas used in pressure-volume isotherm study of Andrew’s experiment is
(a) N₂
(b) H₂S
(c) NH₃
(d) CO₂
Answer:
(d) CO₂

Question 28.
Which of the following gas has highest critical temperature?
(a) NH₃
(b) CO₂
(c) N₂
(d) CH₄
Answer:
(a) NH₃

Question 29.
The relationship between critical volume and Vander waals constant is
(a) V_c =\(\frac{a}{R b}\)
(b) V_c = 3b
(c) V_c = \(\frac{8 a}{27 R b}\)
(d) V_c = 8b
Answer:
(b) V_c = 3b

Question 30.
The temperature below which a gas obeys Joule-Thomson effect is called ______ temperature.
(a) critical
(b) Inversion
(c) ideal
(d) real
Answer:
(b) Inversion

Question 31.
In the equation PV = nRT, which one cannot be numerically equal to R?
(a) 8.31 × 10⁷ ergs K^-1 mol^-1
(b) 8.31 × 10⁷ dynes cm K^-1 mol^-1
(c) 8.317J K^-1 mol^-1
(d) 8.317L atm K^-1 mol^-1
Answer:
(d) 8.317L atm K^-1 mol^-1

Question 32.
A sample of a given mass of gas at constant temperature occupies a volume of 95 cm³ under a pressure of 10.13 × 10⁴ Nm^-2. At the same temperature, its volume at pressure of 10.13 × 10⁴ Vm^-2 is
(a) 190 cm³
(b) 93 cm³
(c) 46.5 cm³
(d) 4.75 cm³
Answer:
(b) 93 cm³

Question 33.
The number of moles of H₂ in 0.224 L of hydrogen gas at STP (273 K, 1 atm) assuming ideal gas behavior is
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(c) 0.01

Question 34.
Use of hot air balloons in sports and meteorological observations is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Charle’s law
Answer:
(d) Charle’s law

Question 35.
To what temperature must a neon gas sample be heated to double its pressure, if the initial volume of gas at 75°C is decreased by 15.0% by cooling the gas?
(a) 319°C
(b) 592°C
(c) 128°C
(d) 60°C
Answer:
(a) 319°C

Question 36.
7.0 g of a gas at 300K and 1 atm occupies a volume of 4.1 litre. What is the molecular mass of the gas?
(a) 42
(b) 38.24
(c) 14.5
(d) 46.5
Answer:
(a) 42

Question 37.
If most probable velocity is represented by a and fraction possessing it by / then with increase in temperature which one of the following is correct?
(a) α increases, f decreases
(b) α decreases, f increases
(c) Both α and f decrease
(d) Both α and f increase
Answer:
(a) α increases, f decreases

Question 38.
The rate of diffusion of gases A and B of molecular weight 36 and 64 are in the ratio
(a) 9 : 16
(b) 4 : 3
(c) 3 : 4
(d) 16 : 9
Answer:
(b) 4 : 3

Question 39.
To which of the following gaseous mixtures Dalton’s law is not applicable?
(a) Ne + He + SO₂
(b) NH₃ + HCl + HBr
(c) O₂ + N₂ +CO₂
(d) N₂ + H₂ + O₂
Answer:
(b) NH₃ + HCl + HBr

Question 40.
Which of the following is true about gaseous state?
(a) Thermal energy = molecular interaction
(b) Thermal energy >> molecular interaction
(c) Thermal energy << molecular interaction
(d) molecular forces >> those in liquids
Answer:
(b) Thermal energy >> molecular interaction

Question 41.
50 mL of gas A effuses through a pinhole in 146 seconds. The same volume of CO₂ under identical condition effuses in 115 seconds. The molar mass of A is
(a) 44
(b) 35.5
(c) 71
(d) None of these
Answer:
(c) 71

Question 42.
Gas deviates from ideal gas nature because molecules
(a) are colourless
(b) attract each other
(c) contain covalent bond
(d) show Brownian movement
Answer:
(b) attract each other

Question 43.
Which of the following gases is expected to have largest value of van der Waals constant ‘a’?
(a) He
(b) H₂
(c) NH₃
(d) O₂
Answer:
(c) NH₃

Question 44.
In van der Waals equation of state for a real gas, the term that accounts for intermolecular forces is
(a) V_m – b
(b) P + \(\frac{a}{V m^{2}}\)
(c) RT
(d) \(\frac{1}{R T}\)
Answer:
(b) P + \(\frac{a}{V m^{2}}\)

Question 45.
What are the most favourable conditions to liquefy a gas?
(a) High temperature and low pressure
(b) Low temperature and high pressure
(c) High temperature and high pressure
(d) Low temperature and low pressure
Answer:
(b) Low temperature and high pressure

Question 46.
Which of the following has a non-linear relationship?
(a) P vs V
(b) P vs \(\frac{1}{V}\)
(c) both (a) and (b)
(d) none of these
Answer:
(a) P vs V

Question 47.
Why is that the gases show ideal behavior when the volume occupied is large?
(a) So that the volume of the molecules can be neglected in comparison to it.
(b) So that the pressure is very high
(c) So that the Boyle temperature of the gas is constant
(d) all of these
Answer:
(a) So that the volume of the molecules can be neglected in comparison to it.

Question 48.
At a given temperature, pressure of a gas obeying Van der Waals equation is
(a) less than that of an ideal gas
(b) more than that of an ideal gas
(c) more or less depending on the nature of gas
(d) equal to that of an ideal gas
Answer:
(a) less than that of an ideal gas

Question 49.
NH₃ gas is liquefied more easily than N₂. Hence
(a) Van der Waals constant a and b of NH₃ > that of N₂
(b) van der Waals constants a and b of NH₃ < that of N₂
(c) a(NH₃) > a(N₂) but b(NH3) < b(N₂)
(d) a(NH₃) < a(N₂) but b(NH₃) > b(N₂)
Answer:
(c) a(NH₃) > a(N₂) but b(NH3) < b(N₂)

Question 50.
Maximum deviation from ideal gas is expected from
(a) CH_4(g)
(b) NH_3(g)
(c) H_2(g)
(d) N_2(g)
Answer:
(b) NH_3(g)

II. Very short question and answers (2 Marks):

Question 1.
Define Pressure? Give its unit?
Answer:
Pressure is defined as force divided by the area to which the force is applied. The SI unit of pressure is pascal which is defined as 1 Newton per square meter (Nm^-2).
Pressure = \(\frac{\text { Force }\left(\mathrm{N}(\text { or }) \mathrm{Kg} \mathrm{ms}^{-2}\right)}{\text {Area }\left(\mathrm{m}^{2}\right)}\)

Question 2.
State Gay Lussac ‘s law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P ∝ T or \(\frac{P}{T}\) = constant K
If P₁ and P₂ are the pressures at temperatures T₁ and T₂, respectively, then from Gay Lussac’s law
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 3.
State Dalton’s law of partial pressure.
Answer:
The total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton s law of partial pressures.
For a mixture containing three gases 1, 2, and 3 with partial pressures p₁, p_2, and p₃ in a container with volume V, the total pressure Ptotal will be give by
P_total = p₁ + p₂ + p₃

Question 4.
What is Compressibility factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT. This is termed as compressibility factor. Mathematically,
Z = \(\frac{P V}{n R T}\)

Question 5.
Define Critical temperature (T_c) of a gas?
Answer:
Critical temperature (T_c) of a gas is defined as the temperature above which it cannot be liquefied even at high pressure.

Question 6.
How does cooling is produced in Adiabatic Process?
Answer:
In Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10^-4 K i.e., as low as 0 K can be achieved.

Question 7.
How do you understand PV relationship?
Answer:
The PV relationship can be understood as follows. The pressure is due to the force of the gas particles on the walls of the container. If a given amount of gas is compressed to half of its volume, the density is doubled and the number of particles hitting the unit area of the container will be doubled. Hence, the pressure would increase two fold.

Question 8.
Write a note on Consequence of Boyle’s law.
Answer:
The pressure-density relationship can be derived from the Boyle’s law as shown below.
P₁V₁ = P₂V₂
P₁ \(\frac{m}{d_{1}}\) = P₂ \(\frac{m}{d_{2}}\)
where “m” is the mass, d₁ and d₂ are the densities of gases at pressure P₁ and P₂.
\(\frac{P_{1}}{d_{1}}=\frac{P_{2}}{d_{2}}\)
In other words, the density of a gas is directly proportional to pressure.

Question 9.
A gas cylinder can withstand a pressure of 15 aim. The pressure of cylinder is measured 12 atm at 27°C. Upto which temperature limit the cylinder will not burst?
Answer:
Cylinder will burst at that temperature when it attains the pressure of 15 atm
P₁ = 12 atm
T = 27° C = (27 + 273)K = 300K
P₂ = 15 atm; T = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

T₂ = \(\frac{15 \times 300}{12}\) = 375 K
= (375 – 273)°C = 102°C

Question 10.
Write a note on application of Dalton’s law.
Answer:
In a reaction involving the collection of gas by downward displacement of water, the pressure of dry vapor collected can be calculated using Dalton’s law.
P_{dry gas collected} = P_total – P_watervapour
P_watervapour has generally referred as aqueous tension and its values are available for air at various temperatures.

Question 11.
State Charles law.
Answer:
For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature.
\(\frac{V}{T}\) = constant (at constant pressure)

Question 12.
What happens when a balloon is moved from an ice cold water bath to a boiling water bath?
Answer:
If a balloon is moved from an ice cold water bath to a boiling water bath, the temperature of the gas increases. As a result, the gas molecules inside the balloon move faster and gas expands. Hence, the volume increases.

Question 13.
Write notes on coefficient of expansion(α).
Answer:
The relative increase in volume per °C (α) is equal to \(\frac{V}{V_{0} T}\)
Therefore, \(\frac{V}{V_{0} T}\)
TV = V₀(αT + 1)
Charles found that the coefficient of expansion is approximately equal to 1/273. It means that at constant temperature for a given mass, for each degree rise in temperature, all gases expand by 1/273 of their volume at 0°C.

Question 14.
State Gay-Lussac law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P ∝ T (at constant volume)

Question 15.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 16.
Distinguish between diffusion and effusion.
Answer:
The property of gas that involves the movement of the gas molecules through another gases is called diffusion. Effusion is another process in which a gas escapes from a container through a very small hole.

Question 17.
State Graham’s law of diffusion.
Answer:
The rate of diffusion or effusion is inversely proportional to the square root of molar mass. This statement is called Graham’s law of diffusion/effusion.
Mathematically, rate of diffusion ∝ \(\frac{1}{M}\)

Question 18.
What is Boyle temperature?
Answer:
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.

III. Short Question and Answers (3 Marks):

Question 1.
Write notes on Boyle’s point.
Answer:
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point. The Boyle point varies with the nature of the gas. Above the Boyle point, for real gases, Z > 1, i.e., the real gases show positive deviation. Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with the increase in pressure.

Question 2.
What are the different methods of liquefaction of gases?
Answer:
There are different methods used for liquefaction of gases:

1. In Linde’s method, Joule-Thomson effect is used to get liquid air or any other gas.
2. In Claude’s process, the gas is allowed to perform mechanical work in addition to Joule-Thomson effect so that more cooling is produced.
3. In Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10^-4 K i.e., as low as 0 K can be achieved.

Question 3.
48 litre of dry N₂ is passed through 36g of H₂O at 27°C and this results In a loss of 1.20 g of water. Find the vapour pressure of water?
Answer:
Water loss is observed because of escape of water molecules with N₂ gas. These water vapour occupy the volume of N₂ gas i.e., 48 litres.
Using,
PV = \(\frac{m}{M}\)RT;
V = 48L
m = 1.2g;
M = 18u;
R = 0.082 L atm K^-1mol^-1
T = 27 + 273 = 300 K
P = \(\frac{1.2}{18} \times \frac{0.0821 \times 300}{48}\) = 0.034 atm

Question 4.
A flask of capacity one litre is heated from 25°C to 35°C. What volume of air will escape from the flask?
Answer:
Applying, \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V₁ = 1L;
T₁ = 25 + 273 = 298 K
V₂ = ?
T₂ = 35 + 273 = 308 K
V₂ = \(\frac{1}{298}\) × 308 = 1.033 L
Capacity of Flask = 1 L
So, volume of air escaped = 1.033 – 1 = 0.033 L = 33 mL

Question 5.
Discuss the graphical representation of Boyle’s law.
Answer:
Boyle’s law is applicable to all gases regardless of their chemical identity (provided the pressure is low). Therefore, for a given mass of a gas under two different sets of conditions at constant temperature we can write,

Question 6.
A certain gas takes three times as long to effuse out as helium. Find its molecular mass.
Answer:

Question 7.
The vapour pressure of water at 80°C is 355.5 mm of Hg. A 100 mL vessel contains water saturated with O₂ at 80°C, the total pressure being 760 mm of Hg. The contents of the vessel were pumped into a 50 mL vessel at the same temperature. What is the partial pressure of O₂?
Answer:
P_total = P_H2O + P_O2
P_O2 = 760 – 355.5 = 404.5
When the contents were pumped into 50 mL vessel
P₁ = 404.5
V₁ = 100 mL
P₂ =?
V₂ = 50 mL
P₁V₁ = P₂V₂
P₂ = \(\frac{404.5 \times 100}{50}\) = 809 mm
Thus, PO₂ in 50 mL vessel = 809 mm

Question 8.
Write notes on compressibility factor for real gases.
Answer:
The compressibility factor Z for real gases can be rewritten,
Z = PV_real / nRT …………..(1)
V_ideal = \(\frac{n R T}{P}\) ……………(2)
substituting (2) in (1)
where V_real is the molar volume of the real gas and V_ideal is the molar volume of it when it behaves ideally.

Question 9.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V, and n and their relationships were governed by the gas laws studied so far.
Boyle’s law V ∝ P¹
Charles law V ∝ T
Avogadro’s law V ∝ n
We can combine these equations into the following general equation that describes the physical behaviour of all gases.
V ∝ \(\frac{n T}{P}\)

V = \(\frac{n R T}{P}\)
where, R is the proportionality constant called universal gas constant.
The above equation can be rearranged to give the ideal gas equation
PV = nRT

Question 10.
State and explain Dalton’s law of partial pressure.
Answer:
John Dalton stated that “the total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton’s law of partial pressures, i.e., for a mixture containing three gases 1, 2, and 3 with partial pressures p₁, p_2, and p₃ in a container with volume V, the total pressure P_total will be given by
P = p₁ + p₂ + p₃

IV. Long Question and Answers (5 Marks):

Question 1.
Derive Van der waals equation of state.
Answer:
J.D.Van der Waals made the first mathematical analysis of real gases. His treatment provides us an interpretation of real gas behaviour at the molecular level. He modified the ideal gas equation PV = nRT by introducing two correction factors, namely, pressure correction and volume correction.

Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.

Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.
P’ ∝ p²
p = \(\frac{n}{v}\)

where n is the number of moles of gas and V is the volume of the container
⇒ P’ = \(\frac{n^{2}}{V^{2}}\)
⇒ P’ = a\(\frac{n^{2}}{V^{2}}\)

where a is the proportionality constant and depends on the nature of gas.
Therefore, ideal P = P + a\(\frac{n^{2}}{V^{2}}\)

Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V’ to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\)π(2r)³
= \(\left|\frac{-8}{(3 \pi)}\right|\) = 8V_m

where V_m is a volume of a single molecule
Excluded volume for single-molecule = \(\frac{8 V_{m}}{2}\) = 4V_m

Excluded volume for n molecule = n(4V_m) = nb
Where b is van der Waals constant which is equal to 4 V_m
=> V’ = nb
V_ideal = V nb

Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

Question 2.
Derive the relationship between Van der Waals constants and critical constants.
Answer:
The van der Waals equation for n moles is
(P + \(\frac{a n^{2}}{V}\))(V – nb) = nRT
For 1 mole
(p + \(\frac{a}{V^{2}}\))(V – b) = RT
From the equation we can derive the values of critical constants P_c, V_c, and T_c in terms of a and b, the van der Waals constants, On expanding the above equation
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0

Multiply above equation by
\(\frac{V_{2}}{P}\)(PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT) = 0

V³ + \(\frac{a V}{P}\) + -bV² – \(\frac{a b}{p}\) – \(\frac{R T V^{2}}{P}\) = 0

when the above equation is rearranged in powers of V

V³ \(\frac{R T}{P}\) + bV² \(\frac{a}{P}\) V – \(\frac{a b}{P}\) = 0.

The above equation is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point, all these three solutions of V_c are equal to the critical volume. The pressure and temperature becomes P_c and T_c respectively
i.e., V = V_c
V – V_c = 0
V – (V_c)³ = 0
V³ – 3V_cV² + 3VV_c² – V_c³ = 0.
As equation identical with equation above, we can equate the cocfficients of V_c, V and constant terms.

-3V_cV² = –\(\frac{R T_{C}}{P}\) + bV²

3V_c = \(\frac{R T_{C}}{P_{C}}\) + b ……….(1)

3V_c² = \(\frac{a}{P_{C}}\) …………..(2)

V_c³ = \(\frac{a b}{P_{C}}\) …………..(3)

Divide equation (3) by equation (2)

\(\frac{V_{C}}{3}\) = b
i.e., V_c = 3b ……………(4)

when equation (4) is substituted in (2)

The critical constants can be calculated using the values of vander walls constant of a gas and vice versa.

a = 3V_c² P_c and b = \(\frac{V_{C}}{3}\)

Question 3.
Explain Andrew’s isotherm for Carbon dioxide.
Answer:
Thomas Andrew gave the first complete data on pressure-volume-temperature of a substance in the gaseous and liquid states. He plotted isotherms of carbon dioxide at different temperatures which is shown in Figure. From the plots we can infer the following.

At low temperature isotherms, for example, at 130°C as the pressure increases, the volume decreases along AB and is a gas until the point B is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist and the pressure remains constant. At C, the gas is completely converted into liquid. If the pressure is higher than at C, only the liquid is compressed so, there is no significant change in the volume. The successive isotherms shows similar trend with the shorter flat region. i.e., The volume range in which the liquid and gas coexist becomes shorter.

At the temperature of 31.1°C the length of the shorter portion is reduced to zero at point P. In other words, the CO₂ gas is liquefied completely at this point. This temperature is known as the liquefaction temperature or critical temperature of CO₃. At this point the pressure is 73 atm. Above this temperature, CO₃ remains as a gas at all pressure values. It is then proved that many real gases behave in a similar manner to carbon dioxide.

Question 4.
Explain Boyle’s law experiment:
Answer:
Robert Boyle performed a series of experiments to j study the relation between the pressure and volume of gases. The schematic diagram of the apparatus j used Boyle is shown in figure.

Mercury was added through the open end of the apparatus such that the mercury level on both ends are equal as shown in the figure (a). Add more amount of mercury until the volume of the trapped air is reduced to half of its original volume as shown in figure (b). The pressure exerted on the gas by the addition of excess mercury is given by the difference in mercury levels of the tube.

Initially the pressure exerted by the gas is equal to 1 atm as the difference in height of the mercury levels is zero. When the volume is reduced to half, the difference in mercury levels increases to 760 mm. Now the pressure exerted by the gas is equal to 2 atm. It led him to conclude that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
Mathematically, the Boyle’s law can be written as
V ∝ \(\frac{1}{P}\) ……….(1)
(T and n are fixed, T-temperature, n- number of moles)
V = k × \(\frac{1}{P}\) ……….(2)
k – proportionality constant When we rearrange equation (2)
PV = k at constant temperature and mass.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 12 Mineral Nutrition Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition

11th Bio Botany Guide Mineral Nutrition Text Book Back Questions and Answers

Part -I

Question 1.
Identify correct match.
1. Die back disease of citrus -(i) Mo
2. Whip tail disease – (ii) Zn
3. Brown heart of turnip -(iii) Cu
4. Little leaf -(iv) B

Answer:
b) 1 (iii) 2 (i) 3 (iv) 4 (ii)

Question 2.
If a plant is provided with all mineral nutrients but, Mn concentration is increased, what will be the deficiency?
(a) Mn prevent the uptake of Fe, Mg but not Ca
(b) Mn increase the uptake of Fe, Mg and Ca
(c) Only increase the uptake of Ca
(d) Prevent the uptake Fe, Mg, and Ca
Answer:
(a) Mn prevent the uptake of Fe, Mg but not Ca

Question 3.
The element which is not remobilized?
a) Phosphorous
b) Potassium
c) Calcium
d) Sulphur
Answer:
c) Calcium

Question 4.
Match the correct combination.

a) A-1 B-3 C-4 D-2
b) A-2 B-1 C-3 D-4
c) A-4 B-3 C-1 D-2
d) A-4 B-2 C-1 D-3
Answer:
c) A-4 B-2 C-1 D-3

Question 5.
Identify the correct statement:
(i) Sulphur is essential for amino acids Cystine and Methionine
(ii) Low level of N, K, S and Mo affect the cell division
(iii) Non – leguminous plant Alnus which contain bacterium Frankia
(iv) Denitrification carried out by nitrosomonas and nitrobacter.

(a) (i), (ii) are correct
(b) (i), (ii), (iii) are correct
(c) I only correct
(d) all are correct
Answer:
(b) (i), (ii), (iii) are correct

Question 6.
Nitrogen is present in the atmosphere in huge amounts but higher plants fail to utilize it. Why?
Answer:
1. Plants absorb minerals from the soil along with water with the help of Roots. Minerals are absorbed as salts.

2. Nitrogen is present in large quantities in the atmosphere in a gaseous form, the gaseous nitrogen must be fixed in the form of Nitrate salts in the soil to facilitate absorption by plant roots.

3. Nitrogen fixation can occur 2 ways by

• Non – Biological means (Industrial process or by lighting)
• Biological means (Bacteria / Cyanobacteria Fungi)
• Therefore higher plants con not utilize the atmospheric Nitrogen.

Question 7.
Why is that in certain plants, deficiency symptoms appear first in younger parts of the plants while in others, they do so in mature organs?
Answer:
When deficiency symptoms appear first, we can notice the differences in old and younger leaves. It is mainly due to mobility’ of minerals. Based on this, they are classified into
1. Actively mobile minerals and
2. Relatively immobile minerals

a) Actively mobile minerals: Nitrogen, Phosphorus, Potassium, Magnesium, Chlorine, Sodium, Zinc and Molybdenum. Deficiency symptoms first appear on old and senescent leaves due to active movement of minerals to younger leaves, than the older leaves.

b) Relatively immobile minerals: Calcium, Sulphur, Iron, Boron and Copper. Here, deficiency symptoms first appear on young leaves due to the immobile nature of minerals.

Question 8.
Plant A in a nutrient medium shows whiptail disease plant B in a nutrient medium shows a little leaf disease. Identify mineral deficiency of plant A and B?
Answer:
Mineral deficiency of plant A and B:

1. Plant A is deficient in the mineral molybdenum (Mo).
2. Plant B is deficient in the mineral zinc (Zn).

Question 9.
Write the role of nitrogenase enzyme in nitrogen fixation?
Answer:
Nitrogen fixation is the first step in Nitrogen cycle, during which gaseous nitrogen from the atmosphere is fixed. It required nitrogenase enzyme complex nitrogenase is active only in anaerobic condition. To create this anaerobic condition, a pigment known as leghaemoglobin is synthesized in the nodules which acts as oxygen scavenger and removes oxygen.

Question 10.
Explain the insectivorous mode of nutrition in angiosperms?
Answer:
Plants which are growing in nitrogen deficient areas develop insectivorous habit to resolve nitrogen deficiency.

1. Nepenthes (Pitcher plant): Pitcher is a modified leaft and contains digestive enzymes. Rim of the pitcher is provided with nectar glands and acts as an attractive lid. When insect is trapped, proteolytic enzymes will digest the insect.
2. Drosera (Sundew): It consists of long club shaped tentacles which secrete sticky digestive fluid which looks like a sundew.
3. Utricularia (Bladder wort): Submerged plant in which leaf is modified into a bladder to collect insect in water.
4. Dionaea (Venus fly trap): Leaf of this plant modified into a colourful trap. Two folds of lamina consist of sensitive trigger hairs and when insects touch the hairs it will close.

Insectivorous Plants

1. Nepenthes (Pitcher Plant)

2. Drosera (Sundew)

3. Dlonaca (Venus Fly tray)

Part – II

11th Bio Botany Guide Mineral Nutrition Additional Important Questions and Answers

I. Choose the Correct Answers

Question 1.
Plants naturally obtain nutrients from:
(a) atmosphere
(b) water
(c) soil
(d) all of these
Answer:
(d) all of these

Question 2.
The minerals placed under the list of unclassified minerals are
a) Carbon. Hydrogen, & Oxygen
b) Sodium. Silicon. Cobalt and selenium
c) Copper, Iron, Cadmium, and selenium
d) Magnesium, Sulphur, & Manganese
Answer:
b) Sodium, Silicon, Cobalt, and Selenium

Question 3.
Who coined the term ‘Hydroponics’:
(a) Julius Von Sachs
(b) William Frederick Goerick
(c) Liebig
(d) Wood word
Answer:
(b) William Frederick Goerick

Question 4.
Skeletal elements are
a) Carbon, Hydrogen, and Oxygen
b) Nitrogen, Phosphorus, and Calcium
c) Potassium, Magnesium, and Sulphur
d) Nitrogen, Sulphur and Phosphorus
Answer:
a) Carbon, Hydrogen, and Oxygen

Question 5.
Actively mobile minerals are:
(a) nitrogen and phosphorus
(b) iron and manganese
(c) sodium and cobalt
(d) silicon and selenium
Answer:
(a) nitrogen and phosphorus

Question 6.
Which chelating agent found in soil are produced by bacteria?
a) Siderophores
b) EDTA
c) Auxin
d) Gibberellin
Answer:
a) Siderophores

Question 7.
Molybdenum is essential for the reaction of:
(a) hydrolase enzyme
(b) nitrogenase enzyme
(c) carboxylase enzyme
(d) dehydrogenase enzyme
Answer:
(b) nitrogenase enzyme

Question 8.
Minerals that play important role for activation of enzymes involved in Respiration are
a) Molybdenum and Boron
b) Boron and Silicon
c) Calcium and Magnesium
d) Magnesium and Manganese
Answer:
d) Magnesium and Manganese

Question 9.
Essential component of aminoacids like Cystine, Cysteine and Melhionine is
a) Potassium
b) Magnesium
c) Sulphur
d) Calcium
Answer:
c) Sulphur

Question 10.
Which of the element is involved in the synthesis of DNA and RNA:
(a) calcium
(b) magnesium
(c) sulphuric
(d) potassium
Answer:
(b) magnesium

Question 11.
Delay in flowering is due to the deficiency of
a) N, S, Mo
b) Ca, Mg, Mn
c) C, H, O
d) N,P,K
Answer:
a) N,S,Mo

Question 12.
Kheria disease of Rice and Internal cork of Apple are caused by the deficiency of
a) Calcium and Maganese
b) Zinc and Boron
c) Copper and Manganese
d) Boron and Nickel
Answer:
b) Zinc and Boron

Question 13.
Indicate the correct statements:
(i) Iron is the essential element for the synthesis of chlorophyll and carotenoid
(ii) Iron is the activator of carboxylene enzyme
(iii) Iton is the component of cytochrome
(iv) lvon is the component of plastocyanin

(a) (i) and (ii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(d) (i) and (iii)

Question 14.
The enzyme that is a constituent of urease and dehydrogenase are
a) Molybdenum
b) Boron
c) Nickel
d) Zinc
Answer:
c) Nickel

Question 15.
A membrane bound bacterium formed inside the nodule is called
a) Bacteriod
b) Plasmid
c) Nucleoid
d) Noduloid
Answer:
a) Bacteriod

Question 16.
The increased concentration of manganese in plants will prevent the uptake of:
(a) calcium and potassium
(b) sodium and potassium
(c) boron and silicon
(d) iron and magnesium
Answer:
(d) iron and magnesium

Question 17.
Plants need one of the following minerals for ATP and meristematic tissue formation
a) K, N
b) N, Cu
c) N, Ca
d) P, N
Answer:
d) P, N