An ideal monoatomic gas with pressure P, volu | vedclass

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(D) For an isothermal process, the temperature remains constant, so $PV = 	ext{constant}$.Initial state: $(P, V)$. Final state: $(P_i, 2V)$.$PV = P_i

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$2^{-2/3}$

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(D) For an isothermal process, the temperature remains constant, so $PV = 	ext{constant}$.Initial state: $(P, V)$. Final state: $(P_i, 2V)$.$PV = P_i(2V) \implies P_i = \frac{P}{2} \implies P = 2P_i \quad ...(i)$For an adiabatic process, $PV^{\gamma} = 	ext{constant}$.Initial state: $(P, V)$. Final state: $(P_a, 2V)$.$PV^{\gamma} = P_a(2V)^{\gamma} \implies P_a = P \left(\frac{V}{2V}ight)^{\gamma} = P(2)^{-\gamma}$For a monoatomic gas, the adiabatic index $\gamma = \frac{5}{3}$.Substituting $P = 2P_i$ and $\gamma = \frac{5}{3}$ into the adiabatic equation:$P_a = (2P_i)(2)^{-5/3} = P_i \cdot 2^1 \cdot 2^{-5/3} = P_i \cdot 2^{1 - 5/3} = P_i \cdot 2^{-2/3}$Therefore, the ratio $\frac{P_a}{P_i} = 2^{-2/3}$.

An ideal monoatomic gas with pressure $P$, volume $V$ and temperature $T$ is expanded isothermally to a volume $2V$ and a final pressure $P_i$. If the same gas is expanded adiabatically to a volume $2V$, the final pressure is $P_a$. The ratio $\frac{P_a}{P_i}$ is

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