Match the thermodynamic processes given under | vedclass

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(A) . Freezing of water at $273 \ K$ and $1 \ atm$ is a reversible phase change at equilibrium,so $\Delta G = 0$. Since entropy decreases during freez

Vedclass · Accepted Answer

$A$ $\rightarrow (R, T); B$ $\rightarrow (P, Q, S); C$ $\rightarrow (P, Q, S); D$ $\rightarrow (P, Q, S, T)$

Vedclass · Answer

(A) . Freezing of water at $273 \ K$ and $1 \ atm$ is a reversible phase change at equilibrium,so $\Delta G = 0$. Since entropy decreases during freezing,$\Delta S_{sys} < 0$. Thus,$A ightarrow (R, T)$.
$B$. Expansion of an ideal gas into a vacuum is free expansion $(w=0)$. Under isolated conditions,$q=0$,so $\Delta U = q + w = 0$. Thus,$B ightarrow (P, Q, S)$.
$C$. Mixing of ideal gases in an isolated container implies $q=0$ and $w=0$ (no external work),so $\Delta U = 0$. Thus,$C ightarrow (P, Q, S)$.
$D$. Reversible heating and cooling back to the initial state is a cyclic process. For any cyclic process,the change in state functions is zero,so $\Delta U = 0$. Since it is reversible,$\Delta S_{total} = 0$,but $\Delta S_{sys}$ is not necessarily zero. However,for a cycle,$\Delta U = 0$. In this specific reversible cycle,$q 
eq 0$ and $w 
eq 0$ over the path,but $\Delta U = 0$. The options provided suggest $D ightarrow (P, Q, S, T)$ is not strictly correct as $q 
eq 0$ and $w 
eq 0$. Re-evaluating the options,$A ightarrow (R, T)$ and $B ightarrow (P, Q, S)$ match option $A$.

Column $I$	Column $II$
$A$. Freezing of water at $273 \ K$ and $1 \ atm$	$P$. $q=0$
$B$. Expansion of $1 \ mol$ of an ideal gas into a vacuum under isolated conditions	$Q$. $w=0$
$C$. Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container	$R$. $\Delta S_{sys} < 0$
$D$. Reversible heating of $H_{2(g)}$ at $1 \ atm$ from $300 \ K$ to $600 \ K$,followed by reversible cooling to $300 \ K$ at $1 \ atm$	$S$. $\Delta U=0$
	$T$. $\Delta G=0$

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