Match List-I with List-II. Given V_1 and V_2 | vedclass

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(C) . For a reversible isothermal expansion, the work done is $w = -nRT \ln(V_2/V_1)$. Since $\Delta U = 0$ for an isothermal process, $q = -w = nRT \

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) . For a reversible isothermal expansion, the work done is $w = -nRT \ln(V_2/V_1)$. Since $\Delta U = 0$ for an isothermal process, $q = -w = nRT \ln(V_2/V_1)$. Thus, $A-II$.
$B$. In free expansion, the gas expands against zero external pressure $(P_{ext} = 0)$, so $w = 0$. For an isothermal process, $\Delta U = 0$, hence $q = 0$. Thus, $B-I$.
$C$. For an irreversible compression, the work done is $w = -P_{ext}(V_2 - V_1) = P_{ext}(V_1 - V_2)$. Thus, $C-III$.
$D$. For a cyclic reversible process, the entropy change of the system is zero $(\oint dS = 0)$. Since $dS = dq_{rev}/T$, it follows that $\oint \frac{dq_{rev}}{T} = 0$. Thus, $D-IV$.
The correct matching is $A-II, B-I, C-III, D-IV$.

List-$I$ (Isothermal process)	List-$II$ (Expression)
$A$. Reversible expansion	$I$. $q = 0$
$B$. Free expansion	$II$. $q = nRT \ln \frac{V_2}{V_1}$
$C$. Irreversible Compression	$III$. $w = -P_{ext}(V_1 - V_2)$
$D$. Cyclic reversible	$IV$. $\frac{q_{rev}}{T} = 0$

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