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(D) For an expansion from $V_0$ to $V$ $(V > V_0)$:$1$. Work done $(W = \int P dV)$ is equal to the area under the $P-V$ curve. Since the isothermal c

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None of the above.

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(D) For an expansion from $V_0$ to $V$ $(V > V_0)$:$1$. Work done $(W = \int P dV)$ is equal to the area under the $P-V$ curve. Since the isothermal curve is above the adiabatic curve for expansion,the work done by the gas is greater in the isothermal process.$2$. For the isothermal process,the final temperature $T_f = T_0$.$3$. For the adiabatic process,$T_0 V_0^{\gamma-1} = T_f V^{\gamma-1}$. Since $V > V_0$ and $\gamma > 1$,$T_f = T_0 (V_0/V)^{\gamma-1} $4$. For the isothermal process,$P_f = P_0 (V_0/V)$.$5$. For the adiabatic process,$P_f = P_0 (V_0/V)^{\gamma}$. Since $\gamma > 1$ and $(V_0/V) Since the isothermal process results in higher final pressure and temperature,and higher work done by the gas,none of the statements $A, B, C$ are correct.

Two gases have the same initial pressure,volume,and temperature. They expand to the same final volume,one adiabatically and the other isothermally. Which of the following statements is correct regarding the final state?

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