Q126 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
An ideal gas having pressure $P$, volume $V$ and temperature $T$ is expanded isothermally, to a volume $3V$ and final pressure $P_I$. The same gas is expanded adiabatically to a volume $3V$, the final pressure being $P_A$. The ratio $\dfrac{P_A}{P_I}$ is $\left(\dfrac{C_P}{C_V} = \gamma\right)$
- A$3\gamma$
- B$3^\gamma$
- Correct answerC$3^{(1-\gamma)}$
- D$3^{(\gamma-1)}$
Solution
For the isothermal expansion, the temperature remains constant. Using Boyle's law:
$P_1 V_1 = P_2 V_2$
$P V = P_I (3V)$
$P_I = \dfrac{P}{3}$
For the adiabatic expansion, the relation between pressure and volume is:
$P_1 V_1^\gamma = P_2 V_2^\gamma$
$P V^\gamma = P_A (3V)^\gamma$
$P_A = \dfrac{P}{3^\gamma}$
Taking the ratio of the final pressures:
$\dfrac{P_A}{P_I} = \dfrac{\dfrac{P}{3^\gamma}}{\dfrac{P}{3}} = \dfrac{3}{3^\gamma} = 3^{1-\gamma}$
Answer: $3^{(1-\gamma)}$
$P_1 V_1 = P_2 V_2$
$P V = P_I (3V)$
$P_I = \dfrac{P}{3}$
For the adiabatic expansion, the relation between pressure and volume is:
$P_1 V_1^\gamma = P_2 V_2^\gamma$
$P V^\gamma = P_A (3V)^\gamma$
$P_A = \dfrac{P}{3^\gamma}$
Taking the ratio of the final pressures:
$\dfrac{P_A}{P_I} = \dfrac{\dfrac{P}{3^\gamma}}{\dfrac{P}{3}} = \dfrac{3}{3^\gamma} = 3^{1-\gamma}$
Answer: $3^{(1-\gamma)}$
Hence, the correct answer is option: 'C'.