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$(C)$ Free expansion of an ideal gas into a vacuum is an adiabatic process where $W = 0$ and $Q = 0$, so $\Delta U = 0$. For an ideal gas, $\Delta U =

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$(A) \rightarrow p, (B) \rightarrow p \& s, (C) \rightarrow p \& s, (D) \rightarrow q \& p$

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$(C)$ Free expansion of an ideal gas into a vacuum is an adiabatic process where $W = 0$ and $Q = 0$, so $\Delta U = 0$. For an ideal gas, $\Delta U = nC_v\Delta T = 0$, implying $\Delta T = 0$. Thus, the temperature remains constant. This matches $(q)$.$(B)$ Polytropic process $PV^x = 	ext{constant}$ with $x = 2$. For a monoatomic gas, $\gamma = 5/3$. Since $x > \gamma$, the molar heat capacity $C = C_v + R/(1-x) = 3R/2 + R/(1-2) = 3R/2 - R = R/2 > 0$. Also, $T \propto PV \propto V^{-2} \cdot V = V^{-1}$, so as $V$ increases, $T$ decreases. Since $C > 0$ and $\Delta T $(C)$ Polytropic process with $x = 4/3$. Since $x  0$, meaning the gas gains heat. Matches $(p)$ and $(s)$.$(D)$ The graph shows $V$ increasing, if $P$ decreases as $V$ increases, the gas does work. If the process is such that $T$ increases, it gains heat. Based on standard interpretation of such problems, $(D)$ matches $(q)$ and $(s)$.

Column $I$	Column $II$
$(A)$ An insulated container has two chambers separated by a valve. Chamber $I$ contains an ideal gas and Chamber $II$ has a vacuum. The valve is opened.	$(p)$ The temperature of the gas decreases
$(B)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-2}$	$(q)$ The temperature of the gas increases or remains constant
$(C)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-4/3}$	$(r)$ The gas loses heat
$(D)$ An ideal monoatomic gas expands such that its pressure $P$ and volume $V$ follow the behavior shown in the graph	$(s)$ The gas gains heat

Column $I$	Column $II$
$(p)$ isobaric	$(x)$ $\frac{PV(1 - 2^{1 - \gamma})}{\gamma - 1}$
$(q)$ isothermal	$(y)$ $PV$
$(r)$ adiabatic	$(z)$ $PV \ln 2$

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