Laws of Thermodynamics Obj-2 HC Verma Solutions Ch-26 Class-12 Vol-2 Concept of Physics for Class-12. Step by Step Solutions of Objective -2 (MCQ-2) Questions of Chapter-26 Laws of Thermodynamics (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-12 Physics.

## Laws of Thermodynamics Obj-2 (MCQ-2) HC Verma Solutions Ch-26 Vol-2 Concept of Physics for Class-12

 Board ISC and other board Publications Bharti Bhawan Publishers Chapter-26 Laws of Thermodynamics Class 12 Vol 2nd writer HC Verma Book Name Concept of Physics Topics Solution of Objective-2 (MCQ-2) Questions Page-Number 61, 62

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Exercise

Laws of Thermodynamics Obj-2 (MCQ-2)

### HC Verma Solutions of Ch-26  Vol-2 Concept of Physics for Class-12

(Page-61)

#### Question-1 :-

The pressure p and volume V of an ideal gas both increase in a process.

(a) Such a process is not possible.

(b) The work done by the system is positive.

(c) The temperature of the system must increase.

(d) Heat supplied to the gas is equal to the change in internal energy.

The options (b) and (c) are correct

Explanation:

Work done by the system depends on the change in volume during the process (ΔW = pΔV), where p be the pressure and ΔV be the change in volume of the ideal gas​. Since in this process the volume is continuously increasing, the work done by the system will be positive.

According to the ideal gas equation, pV=nRT.

Since both p and V are increasing, temperature (T ) must also increase.

#### Question-2 :-

In a process on a system, the initial pressure and volume are equal to the final pressure and volume.

(a) The initial temperature must be equal to the final temperature.

(b) The initial internal energy must be equal to the final internal energy.

(c) The net heat given to the system in the process must be zero.

(d) The net work done by the system in the process must be zero.

The options (a) and (b) are correct

Explanation:

(a) Let initial pressure, volume and temperature be P1, V1 and T1 and final pressure, volume and temperature be P2, V2 and T2. Then, (b) Internal energy is given by ΔU = nCvΔT.

Since ΔT = 0, ΔU = 0.

(c) This may not be true because the system may be isothermal and in between, heat may have been given to the system. Also, the system may have done mechanical expansion work and returned back to its original state.

(d) Not necessary because an isothermal system may have done work absorbing heat from outside and coming back to the same state losing heat.

### Laws of Thermodynamics Obj-2 (MCQ-2)

HC Verma Solutions of Ch-26  Vol-2 Concept of Physics for Class-12

(Page-62)

#### Question-3 :-

A system can be taken from the initial state p1, V1 to the final state p2, V2 by two different methods. Let ∆Q and ∆W represent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods?

(a) ∆Q

(b) ∆W

(c) ∆Q + ∆W

(d) ∆Q − ∆W

The option (d) ∆Q − ∆W is correct

Explanation:

A system is taken from an initial state to the final state by two different methods. So, work done and heat supplied in both the cases will be different as they depend on the path followed. On the other hand, internal energy of the system (U) is a state function, i.e. it only depends on the final and initial state of the process. They are the same in the above two methods.

Using the first law of thermodynamics, we get

∆Q = ∆U – ∆W

⇒ ∆U = ∆Q – ∆W

Here, ∆U is the change in internal energy, ∆Q is the heat given to the system and ∆W is the work done by the system.

#### Question-4 :-

Refer to figure. Let ∆U1 and ∆U2 be the change in internal energy in processes A and B respectively, ∆Q be the net heat given to the system in process A + B and ∆W be the net work done by the system in the process A + B. (a) ∆U1 + ∆U2 = 0

(b) ∆U1 − ∆U2 = 0

(c) ∆Q − ∆W = 0

(d) ∆Q + ∆W = 0

The options (a) and (c) are correct

Explanation:

The process that takes place through A and returns back to the same state through B is cyclic. Being a state function, net change in internal energy, ∆U will be zero, i.e.

∆U1 (Change in internal energy in process A) = ∆​U2 (Change in internal energy in process B)

⇒ ΔU = ∆U1 + ∆U2 = 0

Here, ∆U is the total change in internal energy in the cyclic process.

Using the first law of thermodynamics, we get

ΔQ − ΔW = ΔU

Here, ∆Q is the net heat given to the system in process A + B and ∆W is the net work done by the system in the process A + B.

Thus,

∆Q – ∆W = 0

#### Question-5 :-

The internal energy of an ideal gas decreases by the same amount as the work done by the system.

(a) The process must be adiabatic.

b) The process must be isothermal.

(c) The process must be isobaric.

(d) The temperature must decrease.

The options (a) and (d) are correct

Explanation:

Using the first law of thermodynamics, we get

ΔQ = ΔW + ΔU

⇒ ΔQ = 0. . . . . . . . . [∵ ΔW = − ΔU ]

Thus, no heat is exchanged in the process, i.e. the process is adiabatic and since the internal energy is decreasing, the temperature must decrease because the gas is an ideal gas.

On the other hand, volume and pressure of the gas are varying, leading to positive work done. So, the process cannot be isochoric and isobaric.

—: End of Laws of Thermodynamics Obj-2 (mcq-2) HC Verma Solutions Vol-2 Chapter-26 :–