Match the List-I with List-II: List-I (I | vedclass

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(D) For a reversible isothermal expansion, the work done is given by $w = -nRT \ln \frac{V_f}{V_i}$. Thus, $(A) \rightarrow (II)$.$(B)$ For free expan

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$(A)-(II), (B)-(I), (C)-(III), (D)-(IV)$

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(D) For a reversible isothermal expansion, the work done is given by $w = -nRT \ln \frac{V_f}{V_i}$. Thus, $(A) \rightarrow (II)$.$(B)$ For free expansion, the external pressure $p_{ex} = 0$, so $w = -p_{ex} \Delta V = 0$. Thus, $(B) \rightarrow (I)$.$(C)$ For irreversible expansion, the work done is $w = -p_{ex}(V_f - V_i)$. Thus, $(C) \rightarrow (III)$.$(D)$ For irreversible compression, the work done is $w = -p_{ex}(V_f - V_i) = -p_{ex}(-(V_i - V_f)) = -p_{ex}(V_i - V_f)$ is not standard, but based on the provided options, the work done is $w = -p_{ex}(V_f - V_i)$. However, looking at the options, $(D)$ corresponds to $(IV)$ as $w = -p_{ex}(V_i - V_f)$ is the expression for compression where $V_i > V_f$. Thus, $(D) \rightarrow (IV)$.

List-$I$ (Isothermal process for ideal gas system)	List-$II$ (Work done)
$(A)$ Reversible expansion	$(I)$ $w = 0$
$(B)$ Free expansion	$(II)$ $w = -nRT \ln \frac{V_f}{V_i}$
$(C)$ Irreversible expansion	$(III)$ $w = -p_{ex}(V_f - V_i)$
$(D)$ Irreversible compression	$(IV)$ $w = -p_{ex}(V_i - V_f)$

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