Q61 / 75 · JEE Main 2026 (08 April Shift 2) · Physics
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 $V_{final}=27V$ is ________ J.
- A$-2PV(3\sqrt{3}-1)$
- B$\dfrac{4}{3}PV$
- Correct answerC$-\dfrac{4}{3}PV$
- D$\dfrac{3}{4}PV$
Solution
For a monoatomic ideal gas, the ratio of specific heats is $\gamma = \dfrac{5}{3}$.
In an adiabatic process, $PV^{\gamma} = \text{constant}$.
$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$
$P V^{5/3} = P_2 (27V)^{5/3}$
$P_2 = P \left(\dfrac{1}{27}\right)^{5/3} = \dfrac{P}{243}$
The change in internal energy is given by:
$\Delta U = \dfrac{P_2 V_2 - P_1 V_1}{\gamma - 1}$
Substituting the values:
$\Delta U = \dfrac{\left(\dfrac{P}{243}\right)(27V) - PV}{\dfrac{5}{3} - 1}$
$\Delta U = \dfrac{\dfrac{PV}{9} - PV}{\dfrac{2}{3}}$
$\Delta U = \dfrac{-\dfrac{8}{9}PV}{\dfrac{2}{3}} = -\dfrac{4}{3}PV$
Answer: $-\dfrac{4}{3}PV$
In an adiabatic process, $PV^{\gamma} = \text{constant}$.
$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$
$P V^{5/3} = P_2 (27V)^{5/3}$
$P_2 = P \left(\dfrac{1}{27}\right)^{5/3} = \dfrac{P}{243}$
The change in internal energy is given by:
$\Delta U = \dfrac{P_2 V_2 - P_1 V_1}{\gamma - 1}$
Substituting the values:
$\Delta U = \dfrac{\left(\dfrac{P}{243}\right)(27V) - PV}{\dfrac{5}{3} - 1}$
$\Delta U = \dfrac{\dfrac{PV}{9} - PV}{\dfrac{2}{3}}$
$\Delta U = \dfrac{-\dfrac{8}{9}PV}{\dfrac{2}{3}} = -\dfrac{4}{3}PV$
Answer: $-\dfrac{4}{3}PV$
Hence, the correct answer is option: 'C'.