Initial pressure and volume of a…

Single correct+4 / -1Answer matches NTA’s final key

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.

$-2PV\left(3\sqrt3 - 1\right)$
$\frac43 PV$
$-\frac43 PV$
$\frac34 PV$

Solution

For a monoatomic gas $\gamma = \frac53$ . Along the adiabat

$P_f = P\left(\frac{V}{27V}\right)^{5/3} = \frac{P}{27^{5/3}} = \frac{P}{243}, \qquad P_fV_f = \frac{27PV}{243} = \frac{PV}{9}.$

$\Delta U = nC_V\Delta T = \frac{P_fV_f - PV}{\gamma - 1} = \frac32\left(\frac{PV}{9} - PV\right) = -\frac43 PV.$

The gas does work at the cost of its internal energy, so $\Delta U$ is negative.

Answer: $-\frac43 PV$ .

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Thermodynamics

Solution

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