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.
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100 J
200 J
300 J
400 J
Solution
First law: .
Solution
For an ideal gas, depends only on . Isothermal constant . Then .
Solution
Adiabatic: . Work done on gas . So . Compression heats the gas.
Newton's third law
Conservation of momentum
Conservation of energy (including heat)
Conservation of mass
Solution
The first law extends conservation of energy to include heat as a form of energy transfer.
20 J
80 J
100 J
180 J
Solution
.
Solution
Isothermal: with . Integrating gives .
750 J
1000 J
1730 J
2500 J
Solution
Isothermal: . Closest to 1730 J.
const
Solution
Isothermal for an ideal gas. Then , neither zero unless V is unchanged.
Positive
Zero
Negative
Equal to
Solution
Compression: , so and . The work is negative; the gas has work done on it.
const
const
const
const
Solution
Quasi-static adiabatic: const. Equivalently const.
600 K
900 K
1200 K
1500 K
Solution
const .
Less steep
Same
γ times steeper
Half as steep
Solution
Slope of adiabat / slope of isotherm = . Adiabat is always steeper.
Increases
Decreases
Remains the same
First increases, then decreases
Solution
Adiabatic compression: , work is done ON the gas. So and temperature rises. Bicycle pump, diesel engine.
Solution
.
Now , so . Therefore .
Solution
Mayer's relation: .
7/5
5/3
4/3
1
Solution
Monoatomic: , , , .
Solution
Diatomic at room T: (3 translational + 2 rotational). .
Solution
. For : . For : .
200 J
400 J
600 J
1000 J
Solution
.
Slope of the curve
Area under the curve
Length of the curve
Final pressure × volume
Solution
= area under the curve on a PV plot.
Depends only on the end states
Depends on the path taken
Is always zero
Equals the change in internal energy
Solution
Work is a path function; different paths between the same two states give different W.
Zero
Maximum
Equal to net heat absorbed
Equal to net work done
Solution
State function: depends only on initial and final states. In a cycle these are the same, so .
Zero
Cannot be determined
Solution
Cycle: , so . Clockwise cycle enclosed area = positive work = . So .
A heat engine
A refrigerator
A free expansion
An adiabatic process
Solution
Counter-clockwise net work is done ON the gas refrigerator (or heat pump) cycle.
Solution
.
400 K
500 K
600 K
750 K
Solution
.
200 K
250 K
300 K
400 K
Solution
Carnot: .
A heat engine with
A heat engine with
A Carnot engine
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.
Solution
.
50 J
150 J
250 J
300 J
Solution
.
0.33
2
3
4
Solution
.
Energy is conserved
Heat cannot flow spontaneously from a colder to a hotter body
Entropy of the universe is constant
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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