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(C) For a thermally insulated container, the process is adiabatic.For an adiabatic process, the relation between pressure $P$ and volume $V$ is given

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) For a thermally insulated container, the process is adiabatic.For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$, where $\gamma$ is the adiabatic index.For a monoatomic gas, $\gamma = \frac{5}{3}$.Given the initial volume is $V_1$ and final volume $V_2 = \frac{V_1}{8}$.Using the relation $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$, we get:$\frac{P_2}{P_1} = (\frac{V_1}{V_2})^{\gamma} = (\frac{V_1}{V_1/8})^{\gamma} = 8^{\gamma}$.Substituting $\gamma = \frac{5}{3}$, we get:$\frac{P_2}{P_1} = 8^{5/3} = (2^3)^{5/3} = 2^5 = 32$.Thus, the ratio of final pressure to initial pressure is $32:1$.

$A$ monoatomic gas is stored in a thermally insulated container and the gas is suddenly compressed to $(\frac{1}{8})^{\text{th}}$ of its initial volume. The ratio of final pressure to initial pressure is (in $:1$)

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