JEE Main 2026 Apr 8 Shift 2, Physics Q11: Thermodynamic Processes

JEE Main2026Apr 8, Shift 2Physics

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
$- 2 P V (33 - 1)$
B
$\frac{4}{3} P V$
C
$- \frac{4}{3} P V$
D
$\frac{3}{4} P V$

Solution

For a monoatomic ideal gas, $γ = \frac{5}{3}$ . Adiabatic: $P V^{γ} =$ const.

$P_{2} = P (\frac{V}{27 V})^{5/3} = \frac{P}{2 7 ^{5/3}} = \frac{P}{3 ^{5}} = \frac{P}{243}$ .

$P_{2} V_{2} = \frac{P}{243} (27 V) = \frac{27}{243} P V = \frac{P V}{9}$ .

Change in internal energy in an adiabatic process: $Δ U = - W = \frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1}$ .

$Δ U = \frac{\frac{P V}{9} - P V}{\frac{5}{3} - 1} = \frac{- \frac{8 P V}{9}}{\frac{2}{3}} = - \frac{8 P V}{9} \cdot \frac{3}{2} = - \frac{4 P V}{3}$ .

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JEE Main 2026 Apr 8 Shift 2, Physics Q11: Thermodynamic Processes

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