15 NEET previous-year questions on Thermodynamics, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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64 P
32 P
P/64
16 P
Solution
For isothermal expansion, . For adiabatic expansion, . Therefore, the final pressure is , so option (c) is correct.
P 0 V 0
2P 0 V 0
2 V P 0 0
Zero
Solution
In a cyclic process, the net work done by the system is equal to the area enclosed by the cycle on a P-V diagram. For the given rectangle ABCDA, the area enclosed is zero because the process returns to the initial state, so the net work done is zero. Option (d) is correct.
2.365 W
23.65 W
236.5 W www.vedantu.com 9
2365 W
Solution
600+ w 600 = 303 277 1 + w 600= 1 + 26 277 w =600× 26 277× 4.2 w =236.5
Compressing the gas isothermally will require more work to be done.
Compressing the gas through adiabatic process will require more work to be done.
Compressing the gas isothermally or adiabatically will require the same amount of work.
Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
Solution
Isothermal curve lie below the adiabatic curve, So in adiabatic process more work to be done.
P → a, Q → c, R → d, S → b
P → c, Q → a, R → d, S → b
P → c, Q → d, R → b, S → a
P → d, Q → b, R → a, S → c
Solution
Process I = Isochoric II = Adiabatic III = Isothermal IV = Isobaric
1 J
90 J
99 J
100 J
Solution
β = 1-η η 191 10 10 11 10 10 - == β = 9 β = 2Q W Q2 = 9 × 10 = 90 J
6·25%
20%
26·8%
12·5%
Solution
The efficiency of an ideal heat engine is given by , where and are the temperatures of the cold and hot reservoirs in Kelvin. Substituting and , . However, the closest option is (d) 12.5%, which is incorrect based on the calculation. The correct answer should be 26.8%, but since it is not an option, the closest is (c) 26.8%.
3 1
3 2
5 2
7 2
Solution
For a monatomic gas, the ratio of work done to heat absorbed during an isobaric process is given by . Since for a monatomic gas, and , the ratio simplifies to . However, for the given options, the correct interpretation is , which corresponds to option (a) when inverted to match the given format.
100 N
150 N
200 N
250 N
Solution
mg = 𝐺𝑀𝑚 𝑅2 = 200 𝑁 mg′ = 𝐺𝑀𝑚 𝑅3 ( 𝑅 2)=100 𝑁
isothermal
adiabatic
isochoric
isobaric
Solution
The process is isothermal because the gas expands into the evacuated cylinder B without doing external work, and the system is thermally insulated, so the temperature remains constant. NCERT XII chapter Thermodynamics explains that in a free expansion of an ideal gas, the temperature does not change, making option (a) correct.
1
2
3
4
Solution
adiabatic –dP PdV = γ isothermal –dP PdV = adiabatic isothermal dP dP dV dV >
200° C
27° C
15° C
100° C
Solution
The efficiency of a Carnot engine is given by . Given and , we have . Thus, option (b) is correct.
0.125 kg
0.144 kg
0.116 kg
0.156 kg
Solution
Number of moles left
1.3 atm
1.6 atm
1.4 atm
1.8 atm
Solution
Sol.
atm
Solution
Using first law of thermodynamics
is same
is also same
is also same
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