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(C) For an ideal gas,the entropy change of the system is given by $\Delta S_{sys} = nR \ln(\frac{V_2}{V_1})$.$\Delta S_{sys} = 1 	imes 8.3 	imes \ln

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(C) For an ideal gas,the entropy change of the system is given by $\Delta S_{sys} = nR \ln(\frac{V_2}{V_1})$.$\Delta S_{sys} = 1 	imes 8.3 	imes \ln(\frac{8}{2}) = 8.3 	imes \ln(4) = 8.3 	imes 2 	imes \ln(2) = 8.3 	imes 2 	imes 0.693 = 11.5038$ $J$ $K^{-1}$ $mol^{-1}$.The entropy change of the surroundings is $\Delta S_{surr} = \frac{-q_{rev}}{T} = \frac{-q_{sys}}{T}$. Since the process is against constant external pressure,$q_{sys} = -w = P_{ext}(V_2 - V_1)$.$\Delta S_{surr} = \frac{-P_{ext}(V_2 - V_1)}{T} = \frac{-1 	ext{ } atm 	imes (8 - 2) 	ext{ } L}{300 	ext{ } K} = \frac{-6 	ext{ } L 	ext{ } atm}{300 	ext{ } K}$.Using $1$ $L$ $atm = 100$ $J$,$\Delta S_{surr} = \frac{-6 	imes 100}{300} = -2$ $J$ $K^{-1}$ $mol^{-1}$.$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = 11.5 - 2 = 9.5$ $J$ $K^{-1}$ $mol^{-1}$.

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	$IV$. $\frac{q_{rev}}{T} = 0$

Expansion of $1$ $mol$ of an ideal gas takes place from $2$ $L$ to $8$ $L$ at $300$ $K$ against a constant external pressure of $1$ $atm$. Calculate $\Delta S_{total}$ in $J$ $K^{-1}$ $mol^{-1}$.
(Given: $R = 8.3$ $J$ $K^{-1}$ $mol^{-1}$,$1$ $L$ $atm = 100$ $J$,$\ln 2 = 0.693$) (in $.5$)

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Expansion of $1$ $mol$ of an ideal gas takes place from $2$ $L$ to $8$ $L$ at $300$ $K$ against a constant external pressure of $1$ $atm$. Calculate $\Delta S_{total}$ in $J$ $K^{-1}$ $mol^{-1}$.(Given: $R = 8.3$ $J$ $K^{-1}$ $mol^{-1}$,$1$ $L$ $atm = 100$ $J$,$\ln 2 = 0.693$) (in $.5$)