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Thermodynamics is one of the most dependable chapters in JEE Main. Four processes and one first law — that is very nearly the whole thing. What the paper really tests is whether you know which quantity is zero in which process, because once you know that, the first law hands you the answer in a single line. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 2 questions every year, which is 8 marks.
- The first law applied to the four standard processes is asked every year.
- Work done in each process and the molar specific heats fill the rest.
What the syllabus says
In the syllabus (Unit 8, Thermodynamics): Thermal equilibrium and the concept of temperature, the zeroth law of thermodynamics, heat, work and internal energy, the first law of thermodynamics, isothermal and adiabatic processes, and the second law of thermodynamics with reversible and irreversible processes.
Taken out in 2024: the Carnot engine and its efficiency.
Never in the JEE Main syllabus: heat engines and refrigerators. They are in NCERT, so they still turn up in old question banks.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| First law | ΔQ = ΔU + ΔW |
| Change in internal energy | ΔU = nCvΔT, for every process |
| Isothermal process | ΔU = 0, W = nRT ln(V2/V1) |
| Isobaric process | W = PΔV = nRΔT |
| Isochoric process | W = 0, ΔQ = ΔU |
| Adiabatic process | ΔQ = 0, PVγ = constant |
| Adiabatic, temperature and volume | TVγ−1 = constant |
| Adiabatic work | W = (P1V1 − P2V2)/(γ − 1) |
| Cyclic process | ΔU = 0 over one complete cycle |
| Second law of thermodynamics | heat does not flow on its own from a colder body to a hotter one; processes are reversible or irreversible |
Note the second line carefully. ΔU = nCvΔT holds in every process, not only at constant volume. The Cv is there because internal energy depends on temperature alone, and it does not care how the gas got there.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. In an isothermal process carried out on an ideal gas, the change in internal energy is
- (a) zero
- (b) nCvΔT
- (c) nRT
- (d) equal to the heat supplied
Q2. The work done by an ideal gas expanding isothermally from V1 to V2 is
- (a) nRT ln(V2/V1)
- (b) nRΔT
- (c) PΔV
- (d) zero
Q3. In an adiabatic process, the quantity that stays zero is the
- (a) work done
- (b) heat exchanged with the surroundings
- (c) change in internal energy
- (d) change in temperature
Q4. An ideal gas is expanded isothermally. For this process
- (a) the internal energy rises
- (b) the internal energy falls
- (c) the heat taken in equals the work done by the gas
- (d) no heat is exchanged
Q5. The work done by a gas in an isochoric process is
- (a) PΔV
- (b) zero
- (c) nRΔT
- (d) nCvΔT
Q6. For an adiabatic process, the correct relation between temperature and volume is
- (a) TVγ = constant
- (b) TVγ−1 = constant
- (c) TγV = constant
- (d) T/V = constant
Q7. In any natural process, the entropy of the universe
- (a) decreases
- (b) increases
- (c) stays exactly constant
- (d) becomes zero
Q8. A gas expands from 1 m3 to 3 m3 against a constant pressure of 105 Pa. The work done by it is
- (a) 105 J
- (b) 2 × 105 J
- (c) 3 × 105 J
- (d) zero
Q9. Over one complete cycle of a cyclic process, the change in internal energy is
- (a) zero
- (b) positive
- (c) negative
- (d) equal to the work done
Q10. An ideal gas is taken round a closed cycle. Over one full cycle
- (a) the change in internal energy is zero, so the heat absorbed equals the work done
- (b) the work done is zero
- (c) the heat absorbed is zero
- (d) the internal energy rises
Answer key, with a hint for each
| Q1 | (a) | Internal energy depends only on temperature, and in an isothermal process the temperature does not change. |
| Q2 | (a) | The pressure changes throughout, so you must integrate PdV with P = nRT/V, which gives the logarithm. |
| Q3 | (b) | Adiabatic means no heat enters or leaves. The temperature certainly does change. |
| Q4 | (c) | For an ideal gas the internal energy depends only on temperature, so ΔU = 0 in an isothermal change. The first law ΔU = Q − W then gives Q = W — every joule of heat taken in leaves again as work. |
| Q5 | (b) | Isochoric means constant volume, and with ΔV = 0 there is no work at all. |
| Q6 | (b) | Put P = nRT/V into PVγ = constant. |
| Q7 | (b) | This is the second law. Entropy stays constant only in an ideal reversible process, which never happens exactly. |
| Q8 | (b) | W = PΔV = 105 × 2. |
| Q9 | (a) | The gas returns to its starting state, so every state function including U comes back to its original value. |
| Q10 | (a) | Internal energy is a state function. The gas returns to the same state, so ΔU = 0 and the first law gives Q = W over the cycle. |
Now mark yourself honestly
JEE Main marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 25 Physics questions are out of 100.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in January.
The full chapter, on a clock
Ten questions tell you a little. Twenty-five tell you the truth.
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