Obj-2 Specific Heat Capacities of Gases HC Verma Solutions Vol-2 Class-12 Ch-27

Obj-2 Specific Heat Capacities of Gases HC Verma Solutions Vol-2 Class-12 Ch-27. Step by Step Solutions of Objective -2 (MCQ-2) Questions of Chapter-27 Specific Heat Capacities of Gases (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Obj-2 Specific Heat Capacities of Gases HC Verma Solutions Vol-2 Class-12 Ch-27

Board ISC and other board
Publications Bharti Bhawan Publishers
Chapter-27 Specific Heat Capacities of Gases
Class 12
Vol  2nd
writer HC Verma
Book Name Concept of Physics
Topics Solution of Objective-2 (MCQ-2) Questions
Page-Number 77

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Specific Heat Capacities of Gases Objective-2 (MCQ-2) Questions

HC Verma Solutions of Ch-27  Vol-2 Concept of Physics for Class-12
(Page-77)

Question-1 :-

A gas kept in a container of finite conductivity is suddenly compressed. The process
(a) must be very nearly adiabatic
(b) must be very nearly isothermal
(c) may be very nearly adiabatic
(d) may be very nearly isothermal

Answer-1 :-

The options (c) and (d) are correct

Explanation:

Due to sudden compression, the gas did not get sufficient time for heat exchange. So, no heat exchange occurred. Therefore, the process may be adiabatic. For any process to be isothermal, its temperature should remain constant, i.e. pressure and volume should change simultaneously while their product (temperature) should be constant.

Question-2 :-

Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.

(a) Q = 0

(b) W = 0

(c) Q ≠ W

(d) Q = W

Answer-2 :-

The option (d) Q = W is correct
Explanation:

In an isothermal process, temperature of the system stays constant, i.e. there’s no change in internal energy. Thus, U = 0, where U denotes the change in internal energy of the system. According to the first law of thermodynamics, heat supplied to the system is equal to the sum of change in internal energy and work done by the system, such that Q = U + W. As U = 0, Q = W.

Question-3 :-

Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process.
(a) Q = 0
(b) W = 0
(c) Q = W
(d) Q ≠ W

Answer-3 :-

The options (a) and (d) are correct

Explanation:

In an adiabatic process, no heat is supplied to the system; so, Q = 0. According to the first law of thermodynamics, heat given to any system is equal to the sum of the change in internal energy and the work done on the system. So, Q = W+U and as Q = 0, W = -U and  Q ≠ W.

Answer-6 :-

The options (a) and (b) are correct

Explanation:

The temperature of one mole of a gas kept in a container of fixed volume is increased by 1 degree Celsius if 3 calories, i.e. 12.54 J of heat is added to it. So, its molar heat capacity, C​v = 12.54 J   JK-1 mol-1, as molar heat capacity at fixed volume is the heat supplied to a mole of gas to increase its temperature by a degree. For a monatomic gas,

​C​v ≃ 3/2R =1.5 × 8.314 =12.54 JK-1mol-1. Among the given gases, only helium and argon are inert and, hence, monoatomic. Therefore, the gas may be helium or argon.

Question-7 :-

Four cylinders contain equal number of moles of argon, hydrogen, nitrogen and carbon dioxide at the same temperature. The energy is minimum in

(a) argon

(b) hydrogen

(c) nitrogen

(d) carbon dioxide

Answer-7 :-

The option (a) argon is correct

Explanation:

The energy of a gas is measured as CvT. All the four cylinders are at the same temperature but the gases in them have different values of  Cv, such that it is least for the monatomic gas and keeps on increasing as we go from monatomic to tri-atomic. Among the above gases, argon is monatomic, hydrogen and nitrogen are diatomic and carbon dioxide is tri-atomic. Therefore, the energy is minimum in argon.

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