Laws of Thermodynamics Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-26

Laws of Thermodynamics Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-26 Concept of Physics for Class-12. Step by Step Solutions of Objective -1 (MCQ-1) Questions of Chapter-26 Calorimetry (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Laws of Thermodynamics Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-26

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-1 (MCQ-1) Questions
Page-Number 61

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Laws of Thermodynamics Objective-1 (MCQ-1) Questions

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

(Page-61)

Question-1 :-

The first law of thermodynamics is a statement of

(a) conservation of heat

(b) conservation of work

(c) conservation of momentum

(d) conservation of energy

Answer-1 :-

The option (d) conservation of energy is correct
Explanation:

Heat is a form of energy. Since the first law of thermodynamics deals with the conservation of heat, it actually refers to the conservation of energy in the broader sense.

The first law of thermodynamics is just the restatement of the law of conservation of energy. We observe that the energy supplied to a system will contribute to change in its internal energy and the amount of work done by the system on its surroundings.

Question-2 :-

If heat is supplied to an ideal gas in an isothermal process,

(a) the internal energy of the gas will increase

(b) the gas will do positive work

(c) the gas will do negative work

(d) the said process is not possible

Answer-2 :-

The option (b) the gas will do positive work is correct
Explanation:

Internal energy of the gas,

U = ∫ Cv dT

Here, Cv is the specific heat at constant volume and dT is the change in temperature.

If  the process is isothermal, i.e. dT = 0, dU = 0.

Using the first law of thermodynamics, we get

ΔQ = ΔU + ΔW

But ΔU = 0

⇒ ΔQ = +ΔW

Here, ΔW is the positive work done by the gas.

Question-3 :-

Figure shows two processes A and B on a system. Let ∆Q1 and ∆Q2 be the heat given to the system in processes A and B respectively. Then

Figure shows two processes A and B on a system. Let ∆Q1 and ∆Q2 be the heat given to the system in processes A and B respectively. Then .

(a) ∆Q1 > ∆Q2

(b) ∆Q1 = ∆Q2

(c) ∆Q1 < ∆Q2

(d) ∆Q1 ≤ ∆Q2

Answer-3 :-

The option (a) ∆Q1 > ∆Q2 is correct
Explanation:

Both the processes A and B have common initial and final points. So, change in internal energy, ∆U is same in both the cases. Internal energy is a state function that does not depend on the path followed.

In the P-V diagram, the area under the curve represents the work done on the system, ∆W. Since area under curve A > area under curve B, ∆W​1> ∆W2.

Now,

∆Q1 = ∆U + ∆W1

∆Q2 = ∆U + ∆W2

But ∆W1 > ∆W2

⇒ ∆Q1 > ∆Q2

Here, ∆Q1 and ∆Q2 denote the heat given to the system in processes A and B, respectively.

Question-4 :-

Refer to figure. Let ∆U1 and ∆U2 be the changes in internal energy of the system in the process A and B. Then

Refer to figure. Let ∆U1 and ∆U2 be the changes in internal energy of the system in the process A and B. Then. Then .

(a) ∆U1 > ∆U2

(b) ∆U1 = ∆U2

(c) ∆U1 < ∆U2

(d) ∆U1 ≠ ∆U2

Answer-4 :-

The option (b) ∆U1 = ∆U2 is correct
Explanation:

The internal energy of the system is a state function, i.e. it only depends on the initial and final point of the process and doesn’t depend on the path followed. Both processes A and B have common initial and final points. Therefore, change in internal energy in process A is equal to the change in internal energy in process​ B. Thus,

∆U1 = ∆U2 = 0

Question-5 :-

Consider the process on a system shown in figure. During the process, the work done by the system

Consider the process on a system shown in figure. During the process, the work done by the system .

(a) continuously increases

(b) continuously decreases

(c) first increases then decreases

(d) first decreases then increases

Answer-5 :-   (Laws of Thermodynamics Obj-1 HC Verma )

The option (a) continuously increases is correct
Explanation:

Work done by a system, W= ∫ PdV

Here,

P = Pressure on the system

dV = change in volume.

Since dV is positive, i.e. the volume is continuously increasing, work done by the system also continuously increases.

Question-6 :-

Consider the following two statements.

(A) If heat is added to a system, its temperature must increase.

(B) If positive work is done by a system in a thermodynamic process, its volume must increase.

(a) Both A and B are correct

(b) A is correct but B is wrong

(c) B is correct but A is wrong

(d) Both A and B are wrong

Answer-6 :-

The option (c) B is correct but A is wrong is correct
Explanation:

If heat is added to a system in an isothermal process, then there’ll be no change in the temperature.

Work done by system, ΔW = P Δ V

⇒ ΔW = Positive ⇒ ΔV = Positive

Here,

P = Pressure

ΔV = change in volume

Question-7 :-

An ideal gas goes from the state i to the state f as shown in figure. The work done by the gas during the process

An ideal gas goes from the state i to the state f as shown in figure. The work done by the gas during the process

(a) is positive

(b) is negative

(c) is zero

(d) cannot be obtained from this information

Answer-7 :-

The option (c) is zero is correct
Explanation:

Work done by the gas during the process,

ΔW = P Δ V

Here,

P = Pressure

ΔV = change in volume

Since the process described in the figure is isochoric, P = kT. As volume remains constant (ΔV = 0), ΔW = 0.

Question-8 :-

Consider two processes on a system as shown in figure.

The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ∆W1 and ∆W2 be the work done by the system in the processes A and B respectively.

The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ∆W1 and ∆W2 be the work done by the system in the processes A and B respectively.

(a) ∆W1 > ∆W2

(b) ∆W1 = ∆W2

(c) ∆W1 < ∆W2

(d) Nothing can be said about the relation between ∆W1 and ∆W2

Answer-8 :-  (Laws of Thermodynamics Obj-1 HC Verma )

The option (c) ∆W1 < ∆W2 is correct
Explanation:

Work done by the system, ∆W = P ∆ V

here,

P = Pressure in the process

∆V = Change in volume during the process

Let Vi and Vf​  be the volumes in the initial states and final states for processes A and B, respectively. Then,

Let Vi and Vf​ be the volumes in the initial states and final states for processes A and B, respectively. Then,

Question-9 :-

A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder.

(a) increases

(b) decreases

(c) remains constant

(d) increases or decreases depending on the nature of the gas

Answer-9 :-

The option (b) decreases is correct
Explanation:

As the piston of a metallic cylinder containing gas is moved to compress the gas, the volume in which the gas is contained reduces, leading to increase in pressure and temperature. When the time elapses, the heat generated radiates through the metallic cylinder as metals are good conductors of heat. Consequently, the pressure of the gas in the cylinder decreases because of decrease in the temperature.

 

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