A thermodynamical system is changed from state$\left( {{p_1}{V_1}} \right){\rm{to}}\left( {{p_2},{V_2}} \right)$ by two different processes, the quantity which will remain same will be

${\rm{\Delta }}Q - {\rm{\Delta }}W$

${\rm{\Delta }}Q + {\rm{\Delta }}W$

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NEET • PHYSICS • THERMODYNAMICS

Thermodynamics practice · Easy

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Showing 1-30 of 40 questions

1 Easy

1 of 40

Understanding +4 | -1

When heat is added to a system, which of the following is not possible?

View answer and explanation

Correct option: C

Neither internal energy increases nor work is done by the system

Option C (Neither internal energy increases nor work is done by the system) is correct because if heat is added to a system, at least one of the internal energy or work done by the system must increase according to the first law of thermodynamics.

2 Easy

2 of 40

Application +4 | -1

If sink and source temperature of a refrigerator are

4^{\circ} \mathrm{C}

and

15^{\circ} \mathrm{C}

respectively. Then efficiency of refrigerator is

View answer and explanation

Correct option: B

0.0382

Option A (0.0382) is correct because the efficiency of a refrigerator is calculated using the formula

\eta = 1 - \frac{{T_2}}{{T_1}}

, where

T_1

and

T_2

are the source and sink temperatures in Kelvin.

3 Easy

3 of 40

Understanding +4 | -1

Temperature is a measure of coldness or hotness of an object. This definition is based on

View answer and explanation

Correct option: A

Zeroth law of thermodynamics

Option A (Zeroth law of thermodynamics) is correct because the zeroth law defines thermal equilibrium and provides the basis for temperature measurement.

4 Easy

4 of 40

Understanding +4 | -1

The internal energy of a system remains constant
when it undergoes

(A) a cycle process

(B) an isothermal process

(C) an adiabatic process

(D) any process in which the heat given out by the
system is equal to the work done on the system

Choose the correct statement

View answer and explanation

Correct option: D

A, B, D

The correct answer is D (A, B, D) because internal energy remains constant in a cyclic process, an isothermal process, and any process where heat given out equals work done on the system.

5 Easy

5 of 40

Application +4 | -1

Find the amount of workdone to increase the temperature of one mole of an ideal gas by

30^{\circ} \mathrm{C}

if it is expanding under the condition

V \propto T^{2 / 3},(R=1.99 \mathrm{cal} / \mathrm{mol}-\mathrm{K})

View answer and explanation

Correct option: A

167

Option A (167) is correct because the work done is calculated using the given relation between volume and temperature and the ideal gas constant.

6 Easy

6 of 40

Understanding +4 | -1

Assertion: Internal energy of an ideal gas does not depend upon volume of the gas.
Reason: Internal energy of ideal gas depends on temperature of gas.

View answer and explanation

Correct option: B

If both assertion and reason are true but reason is not correct explanation of the assertion

Option B (If both assertion and reason are true but reason is not correct explanation of the assertion) is correct because while both statements are true, the reason doesn't directly explain why the assertion is true. The assertion is true because ideal gases have no intermolecular forces.

7 Easy

7 of 40

Application +4 | -1

For a refrigerator, heat absorbed from source is

800 \mathrm{~J}

and heat supplied to sink is

500 \mathrm{~J}

then find coefficient of performance.

View answer and explanation

Correct option: C

\frac{5}{3}

Option C (

\frac{5}{3}

) is correct because the coefficient of performance (COP) for a refrigerator is the ratio of heat absorbed from the cold reservoir to the work done.

8 Easy

8 of 40

Analyzing +4 | -1

Assertion: The ratio

\frac{C_{{V}}}{{C}_{{P}}}

for a monatomic gas is less than for a diatomic gas.
Reason: The molecules of a monatomic gas have more degrees of freedom than those of a diatomic gas.

View answer and explanation

Correct option: C

If Assertion is true but Reason is false

Option C (Assertion is true but Reason is false) is correct because the ratio of specific heats is indeed lower for monatomic gases, but this is because monatomic gases have *fewer* degrees of freedom.

9 Easy

9 of 40

Application +4 | -1

0.08 \mathrm{~kg}

air is heated at constant volume through

5^{\circ} \mathrm{C}.

The specific heat of air at constant volume is

0.17 \mathrm{~kcal} / \mathrm{kg}^{\circ} \mathrm{C}

and

\mathrm{J}=4.18 \text{ joule}/ \mathrm{cal}.

The change in its internal energy is approximately

View answer and explanation

Correct option: A

284 J

The change in internal energy is calculated using the formula

\Delta U = mc_v\Delta T

. Substituting the given values, we get

\Delta U = 0.08 \times 0.17 \times 5 \times 4186.8 \approx 284 J

10 Easy

10 of 40

Application +4 | -1

One mole of an ideal gas is compressed adiabatically from

6 \, \mathrm{Ltr}

2 \, \mathrm{Ltr}

. The initial pressure of the gas is

10^5 \mathrm{~N} / \mathrm{m}^2

. Then final pressure of gas is if molar specific heat of gas at constant volume

=\frac{3}{2} R .\left[(3)^{5 / 3}=6.19\right]

View answer and explanation

Correct option: A

6.19 \times 10^5 \mathrm{~N} / \mathrm{m}^2

Option A (

6.19 \times 10^5 \mathrm{~N} / \mathrm{m}^2

) is correct because the final pressure is calculated using the adiabatic process equation,

P_1V_1^\gamma = P_2V_2^\gamma

, where

\gamma = C_p/C_v

11 Easy

11 of 40

Understanding +4 | -1

Assertion: In adiabatic process work is independent of path.
Reason: In adiabatic process work done is equal to negative of change in internal energy.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion

Option A is correct because in an adiabatic process, no heat is exchanged with the surroundings (Q=0), so the work done is equal to the negative change in internal energy (W = -ΔU), making it path-independent.

12 Easy

12 of 40

Understanding +4 | -1

A thermodynamical system is changed from state

\left( {{p_1}{V_1}} \right)\mathrm{to}\left( {{p_2},{V_2}} \right)

by two different processes, the quantity which will remain same will be

View answer and explanation

Correct option: D

\mathrm{\Delta }Q - \mathrm{\Delta }W

Option D (

\mathrm{\Delta }Q - \mathrm{\Delta }W

) is correct because the change in internal energy, which is a state function, is equal to

\mathrm{\Delta }Q - \mathrm{\Delta }W

according to the first law of thermodynamics.

13 Easy

13 of 40

Analyzing +4 | -1

A monoatomic ideal gas expanded adiabatically to

\eta

times of its initial volume. If the ratio of final rate of collisions of molecules with unit area of container's walls to the initial rate is

\frac{1}{\eta^x}

. Find the value of

x

View answer and explanation

Correct option: A

1.33

Option A (1.33) is correct because it is derived from the relationship between collision rate, volume, and adiabatic expansion.

14 Easy

14 of 40

Application +4 | -1

In a thermodynamic process, pressure of fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If initial internal energy of the gas was 40 J, what will be final internal energy ?

View answer and explanation

Correct option: B

28 J

Option B (28 J) is correct because the final internal energy is calculated using the first law of thermodynamics, accounting for heat released and work done on the gas.

15 Easy

15 of 40

Understanding +4 | -1

Statement-1: When an ideal gas is taken from a given thermodynamic state

A

to another given thermodynamic state

B

by any polytrophic process, the change in the interval energy of the system will be the same in all processes.
Statement-2: Internal energy of the gas depends only upon its absolute temperature.

View answer and explanation

Correct option: A

Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1

Option A is correct because the change in internal energy of an ideal gas depends only on its initial and final temperatures, and the internal energy itself is a function of absolute temperature only. Therefore, the change in internal energy will be the same regardless of the path taken.

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16 Easy

16 of 40

Application +4 | -1

An iron ball of mass

0.2 \mathrm{~kg}

is heated to

100^{\circ} \mathrm{C}

and put into a block at ice at

0^{\circ} \mathrm{C} \,25 \mathrm{~g}

of ice melts. If the latent heat of fusion of ice is

80 \mathrm{~cal} \mathrm{~g}^{-1}

, then the specific heat of iron in

\mathrm{cal} \mathrm{~g}^{-1} \,{}^\circ\mathrm{C}

View answer and explanation

Correct option: B

0.1

Option B (0.1) is correct because the specific heat of iron is calculated by equating the heat lost by the iron ball to the heat gained by the ice during melting.

17 Easy

17 of 40

Application +4 | -1

A girl weighing 42 kg eats bananas whose energy is 980 calories. If this energy is used to go to height

h

, the value of

h

View answer and explanation

Correct option: A

10 m

Option A (10 m) is correct because the potential energy gained by the girl equals the energy from the bananas, so

mgh = energy

, and solving for h gives 10m.

18 Easy

18 of 40

Understanding +4 | -1

The increase in internal energy of a system is equal to the work done on the system. Which process does the system undergo?

View answer and explanation

Correct option: B

Adiabatic

Option B is correct because if the increase in internal energy of a system is equal to the work done on the system (ΔU = W), and from the first law of thermodynamics ΔU = Q - W, then Q = 0, which defines an adiabatic process.

19 Easy

19 of 40

Understanding +4 | -1

In a cyclic process, work done by the system is

View answer and explanation

Correct option: B

Equal to heat given to the system

The correct answer is B (Equal to heat given to the system) because in a cyclic process, the change in internal energy is zero, so the work done equals the heat added.

20 Easy

20 of 40

Application +4 | -1

At constant volume the specific heat of a gas is

\frac{{3R}}{2},

then the value of '

\gamma

' will be

View answer and explanation

Correct option: C

\frac{5}{3}

Option C (5/3) is correct because γ = Cp/Cv, and Cp can be found using the relation Cp = Cv + R.

21 Easy

21 of 40

Application +4 | -1

Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are

10^5 \mathrm{~N} / \mathrm{m}^2

and 6 litre respectively. The final volume of the gas is 2 litre. Molar specific heat of the gas at constant volume is

\frac{3 R}{2}

View answer and explanation

Correct option: A

957

The correct answer is 957. The work done during an adiabatic compression is calculated using the formula

W = \frac{P_2V_2 - P_1V_1}{1-\gamma}

, where

\gamma = 1 + \frac{2}{f}

and f is the degrees of freedom.

22 Easy

22 of 40

Application +4 | -1

An ideal gas has a molar heat capacity

\mathrm{C}=\frac{9 \mathrm{R}}{2}

in the process

\mathrm{PT}=

constant. Find the degree of freedom of the gas molecule.

(\mathrm{R}=

ideal gas constant)

View answer and explanation

Correct option: A

Option A (5) is correct because the molar heat capacity is related to the degrees of freedom by the equation C = Cv + R/2, and Cv = (f/2)R.

23 Easy

23 of 40

Understanding +4 | -1

A gas has

n

degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be

View answer and explanation

Correct option: A

\frac{n}{n+2}

Option A (

\frac{n}{n+2}

) is correct because the ratio of specific heat at constant volume (

C_v

) to specific heat at constant pressure (

C_p

) is given by

\frac{C_v}{C_p} = \frac{1}{\gamma} = \frac{n/2}{n/2 + 1} = \frac{n}{n+2}

, where n is the degrees of freedom.

24 Easy

24 of 40

Understanding +4 | -1

The work done in an isothermal expansion of a gas depends upon

View answer and explanation

Correct option: C

Both temperature and expansion ratio

Option C is correct because the work done in an isothermal expansion of a gas depends on both the temperature and the expansion ratio (ratio of final to initial volume).

25 Easy

25 of 40

Understanding +4 | -1

Statement-1: A quasi-static process is so called because it is a sudden and large change of the system.
Statement-2: An adiabatic process is no quasi-static because it is a sudden and large change of the system.

View answer and explanation

Correct option: B

Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1

Option D is correct. Statement-1 is false because a quasi-static process is a slow, gradual change. Statement-2 is true because an adiabatic process can be a sudden change, and therefore not quasi-static.

26 Easy

26 of 40

Application +4 | -1

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is:

View answer and explanation

Correct option: B

\frac{9}{4} R

Option B (

\frac{9}{4} R

) is correct because the molar specific heat of the mixture is calculated using the weighted average of the individual molar specific heats.

27 Easy

27 of 40

Understanding +4 | -1

Heat added to a system is equal to

View answer and explanation

Correct option: D

Sum of above all the three factors

Option D (Sum of above all the three factors) is correct because the heat added to a system can increase its internal kinetic energy, increase its internal potential energy, and/or be used to do work by the system, as described by the first law of thermodynamics.

28 Easy

28 of 40

Application +4 | -1

A monoatomic gas at pressure

P

and volume

V

is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be :

View answer and explanation

Correct option: C

32 P

Option C (

32 P

) is correct because for a monoatomic gas undergoing adiabatic compression, the final pressure is 32 times the initial pressure when the volume is reduced to one-eighth.

29 Easy

29 of 40

Understanding +4 | -1

{C_P}

and

{C_V}

denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then

{C_P} - {C_V} =

View answer and explanation

Correct option: A

\frac{R}{{28}}

Option A (

\frac{R}{{28}}

) is correct because it represents the difference between specific heats of nitrogen per unit mass, considering its molar mass.

30 Easy

30 of 40

Understanding +4 | -1

Assertion: The internal energy of a real gas is function of both, temperature and volume.
Reason: At constant pressure, some heat is spent in expansion of the gas.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of the assertion

Option A (If both assertion and reason are true and reason is the correct explanation of the assertion) is correct because the internal energy of a real gas depends on both temperature and volume due to intermolecular forces, and at constant pressure, some heat is used for expansion, indicating volume dependence.

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Key Takeaways

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