An ideal gas having pressure P, volume V and temperature T undergoes a thermodynamic process in which dW = 0 and dQ < 0. Then, for the gas (a) V will increase (b) T will increase (c) P first decreases and then increases (d) T will decrease

Answer: (d). dW = 0 ⇒ volume constant (isochoric). First law: dU = dQ - dW = dQ < 0, so internal energy falls. For an ideal gas U ∝ T, so T decreases (and P ∝ T also decreases). So the answer is (d) T will decrease.

An ideal gas having pressure P, volume V and temperature T undergoes… | MHT CET 2026 (20 May, Shift 1)

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Heat, Work & First Law of Thermodynamics

An ideal gas having pressure P, volume V and temperature T undergoes a thermodynamic process in which $dW = 0$ and $dQ < 0$ . Then, for the gas

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Answer: (d)

$dW = 0 \Rightarrow$ volume constant (isochoric).
First law: $dU = dQ - dW = dQ < 0$ , so internal energy falls.
For an ideal gas $U \propto T$ , so $T$ decreases (and $P \propto T$ also decreases).
So the answer is (d) T will decrease.

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Heat, Work & First Law of Thermodynamics question from MHT CET 2026 (20 May, Shift 1)

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