JEE MAIN - Physics (2025 - 8th April Evening Shift - No. 1)
A monoatomic gas having $ \gamma = \frac{5}{3} $ is stored in a thermally insulated container and the gas is suddenly compressed to $ \left( \frac{1}{8} \right)^{\text{th}} $ of its initial volume. The ratio of final pressure and initial pressure is:
($\gamma$ is the ratio of specific heats of the gas at constant pressure and at constant volume)
Explanation
To find the ratio of final pressure ($ P_f $) to initial pressure ($ P_i $), we use the adiabatic condition:
$ P_i V_i^\gamma = P_f V_f^\gamma $
We aim to express the pressure ratio:
$ \frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^\gamma $
Given that the final volume $ V_f $ is $ \frac{1}{8} $ of the initial volume $ V_i $, this simplifies to:
$ \left(\frac{V_i}{V_f}\right) = 8 $
Thus, the pressure ratio becomes:
$ \frac{P_f}{P_i} = 8^\frac{5}{3} $
Calculating $ 8^\frac{5}{3} $ gives:
$ \frac{P_f}{P_i} = 32 $
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