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(D) Work done is equal to the area under the $P-V$ graph. As shown in the graph,the isothermal curve lies above the adiabatic curve for the same expan

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All of the above.

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(D) Work done is equal to the area under the $P-V$ graph. As shown in the graph,the isothermal curve lies above the adiabatic curve for the same expansion,thus the work done by the gas is greater for the isothermal process.For the isothermal process: $T_{2} = T_{0}$.For the adiabatic process: $T_{0}(V_{0})^{\gamma-1} = T(V)^{\gamma-1}$.Since $V > V_{0}$,it follows that $T For the isothermal process: $P_{0}V_{0} = P_{i}V \implies P_{i} = P_{0} \left(\frac{V_{0}}{V}\right)$.For the adiabatic process: $P_{0}(V_{0})^{\gamma} = P_{a}(V)^{\gamma} \implies P_{a} = P_{0} \left(\frac{V_{0}}{V}\right)^{\gamma}$.Since $\gamma > 1$ and $\frac{V_{0}}{V}  \left(\frac{V_{0}}{V}\right)^{\gamma}$,therefore $P_{i} > P_{a}$.Since all statements are correct,the correct option is $D$.

Column $I$	Column $II$
$A$. Ratio of $\frac{\Delta Q}{\Delta U}$ in an isobaric process	$1$. $\frac{T_1}{T_1-T_2}$
$B$. Ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process	$2$. $\frac{T_2}{T_1-T_2}$
$C$. Coefficient of performance of a refrigerator	$3$. $\frac{\gamma}{\gamma-1}$
$D$. Coefficient of performance of a heat pump	$4$. $\gamma$

Two gases have the same initial pressure,volume,and temperature. They expand to the same final volume,one adiabatically and the other isothermally.

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