Physics · Thermodynamics

JEE Main 2026 — 8 April, Evening Shift — Question 64

Initial pressure and volume of a monoatomic ideal gas are P and V. The change in internal energy of this gas in adiabatic expansion to volume $V_{final}$ = 27 V is _________ J.

Option A:
$-2PV(3\sqrt{3} - 1)$
Option B:
$\frac{4}{3} PV$
Option C:
$-\frac{4}{3} PV$
Correct
Option D:
$\frac{3}{4} PV$

Answer: C

Step-by-step solution

& P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \\ & \gamma = \frac{5}{3} \\ & PV^{5/3} = P_2 \times (27V)^{5/3} \\ & P_2 = \frac{P}{(27)^{5/3}} = \frac{P}{3^5} \\ \\ & \Delta U \Rightarrow nC_v \Delta T = \frac{nR\Delta T}{(\gamma - 1)} \\ & = \frac{P_2 V_2 - P_1 V_1}{(\gamma - 1)} \\ & \Rightarrow \frac{\frac{P}{3^5} \times 3^3 \cdot V - PV}{\frac{5}{3} - 1} \Rightarrow \frac{-PV \left\{ 1 - \frac{1}{9} \right\}}{\frac{2}{3}} \\ & = -PV \times \frac{8}{9} \times \frac{3}{2} \\ & \Rightarrow -\frac{4}{3} PV \end{aligned}$$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Different Thermodynamic Processes