JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Heat & Thermodynamics

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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.

(A) $- 2 PV (33 - 1)$
(B) $\frac{4}{3} PV$
(C) $- \frac{4}{3} PV$ ✓ Correct
(D) $\frac{3}{4} PV$

Answer: (C)

Solution

For a monoatomic gas

γ = \frac{5}{3}

. The adiabatic relation

P V^{γ} = P_{2} (27 V)^{γ}

gives

P_{2} = \frac{P}{2 7 ^{5/3}} = \frac{P}{3 ^{5}} = \frac{P}{243}

. The change in internal energy is

Δ U = \frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1} = \frac{\frac{P}{243} ( 27 V ) - P V}{2/3} = \frac{P V ( \frac{1}{9} - 1 )}{2/3} = \frac{- \frac{8}{9} P V}{2/3} = - \frac{4}{3} P V

. The change in internal energy is

- \frac{4}{3} P V

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option

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