One mole of a mono-atomic ideal gas is taken through a cycle ABCDA as shown in the P-V diagram, then (a) for process A to B, heat is lost and internal energy increases. (b) for process B to C, heat is lost and internal energy increases. (c) for process C to D, heat is gained and internal energy increases. (d) for process D to A, heat is gained and internal energy increases.

Answer: (c). Use T ∝ PV: PV at A = 9pV, B = 3pV, C = pV, D = 9pV. A to B: T falls, so Δ U 0 and Q = nCPΔ T > 0: heat is gained. D to A: T is the same at both ends (isothermal compression), Δ U = 0; (d) is wrong. So the answer is (c).

One mole of a mono-atomic ideal gas is taken through a cycle ABCDA as… | MHT CET 2026 (12 May, Shift 1)

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P-V Diagrams & Cyclic Processes

One mole of a mono-atomic ideal gas is taken through a cycle ABCDA as shown in the P-V diagram, then

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Answer: (c)

Use $T \propto PV$ : $PV$ at A $= 9pV$ , B $= 3pV$ , C $= pV$ , D $= 9pV$ .
A to B: $T$ falls, so $\Delta U < 0$ ; (a) is wrong.
B to C: $T$ falls, so $\Delta U < 0$ ; (b) is wrong.
C to D: isobaric expansion, $T$ rises from $pV/R$ to $9pV/R$ , so $\Delta U > 0$ and $Q = nC_P\Delta T > 0$ : heat is gained.
D to A: $T$ is the same at both ends (isothermal compression), $\Delta U = 0$ ; (d) is wrong.
So the answer is (c).

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P-V Diagrams & Cyclic Processes question from MHT CET 2026 (12 May, Shift 1)

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