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ThermodynamicsNEET Physics · Class 11 · NCERT Chapter 11

32 NEET previous-year questions on Thermodynamics, each with the correct answer and a step-by-step solution. Filter by topic and expand any question to see how to solve it.

PYQ frequency · topic × year

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18
19
20
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24
First law
1
1
1
1
1
Isothermal
1
1
1
1
Adiabatic
1
1
1
1
1
Cp / Cv / γ
1
1
1
1
PV work
1
1
1
Cyclic
1
1
1
Carnot engine
1
1
1
1
Refrigerator COP
1
1
1
Second law
1

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All (32)
First law (5)
Isothermal (4)
Adiabatic (5)
Cp / Cv / γ (4)
PV work (3)
Cyclic (3)
Carnot engine (4)
Refrigerator COP (3)
Second law (1)

A

100 J

B

200 J

C

300 J

D

400 J

Solution

First law: .

A

B

C

D

Solution

For an ideal gas, depends only on . Isothermal constant . Then .

A

B

C

D

Solution

Adiabatic: . Work done on gas . So . Compression heats the gas.

A

Newton's third law

B

Conservation of momentum

C

Conservation of energy (including heat)

D

Conservation of mass

Solution

The first law extends conservation of energy to include heat as a form of energy transfer.

A

20 J

B

80 J

C

100 J

D

180 J

Solution

.

A

B

C

D

Solution

Isothermal: with . Integrating gives .

A

750 J

B

1000 J

C

1730 J

D

2500 J

Solution

Isothermal: . Closest to 1730 J.

A

B

C

D

const

Solution

Isothermal for an ideal gas. Then , neither zero unless V is unchanged.

A

Positive

B

Zero

C

Negative

D

Equal to

Solution

Compression: , so and . The work is negative; the gas has work done on it.

A

const

B

const

C

const

D

const

Solution

Quasi-static adiabatic: const. Equivalently const.

A

600 K

B

900 K

C

1200 K

D

1500 K

Solution

const .

A

Less steep

B

Same

C

γ times steeper

D

Half as steep

Solution

Slope of adiabat / slope of isotherm = . Adiabat is always steeper.

A

Increases

B

Decreases

C

Remains the same

D

First increases, then decreases

Solution

Adiabatic compression: , work is done ON the gas. So and temperature rises. Bicycle pump, diesel engine.

A

B

C

D

Solution

.

Now , so . Therefore .

A

B

C

D

Solution

Mayer's relation: .

A

7/5

B

5/3

C

4/3

D

1

Solution

Monoatomic: , , , .

A

B

C

D

Solution

Diatomic at room T: (3 translational + 2 rotational). .

A

B

C

D

Solution

. For : . For : .

A

200 J

B

400 J

C

600 J

D

1000 J

Solution

.

A

Slope of the curve

B

Area under the curve

C

Length of the curve

D

Final pressure × volume

Solution

= area under the curve on a PV plot.

A

Depends only on the end states

B

Depends on the path taken

C

Is always zero

D

Equals the change in internal energy

Solution

Work is a path function; different paths between the same two states give different W.

A

Zero

B

Maximum

C

Equal to net heat absorbed

D

Equal to net work done

Solution

State function: depends only on initial and final states. In a cycle these are the same, so .

A

Zero

B

C

D

Cannot be determined

Solution

Cycle: , so . Clockwise cycle enclosed area = positive work = . So .

A

A heat engine

B

A refrigerator

C

A free expansion

D

An adiabatic process

Solution

Counter-clockwise net work is done ON the gas refrigerator (or heat pump) cycle.

A

B

C

D

Solution

.

A

400 K

B

500 K

C

600 K

D

750 K

Solution

.

A

200 K

B

250 K

C

300 K

D

400 K

Solution

Carnot: .

A

A heat engine with

B

A heat engine with

C

A Carnot engine

D

A refrigerator

Solution

Kelvin Planck: no engine can convert all absorbed heat into work. So is forbidden, since some heat must be rejected to a cold reservoir.

A

B

C

D

Solution

.

A

50 J

B

150 J

C

250 J

D

300 J

Solution

.

A

0.33

B

2

C

3

D

4

Solution

.

A

Energy is conserved

B

Heat cannot flow spontaneously from a colder to a hotter body

C

Entropy of the universe is constant

D

Internal energy is a state function

Solution

Clausius statement: heat does not flow spontaneously from cold to hot. To reverse the natural direction, work must be supplied, which is exactly what a refrigerator does.

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