An ideal gas in a thermally insulated vessel | vedclass

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(A) Since the vessel is thermally insulated,the heat exchange $q = 0$.Since the gas expands against zero external pressure $(P_{ex} = 0)$,the work don

Vedclass · Accepted Answer

$A, B, C$

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(A) Since the vessel is thermally insulated,the heat exchange $q = 0$.Since the gas expands against zero external pressure $(P_{ex} = 0)$,the work done $w = -P_{ex} \Delta V = 0$.According to the first law of thermodynamics,$\Delta U = q + w = 0 + 0 = 0$.For an ideal gas,internal energy $U$ is a function of temperature only,so $\Delta U = 0$ implies $\Delta T = 0$,which means $T_2 = T_1$.Using the ideal gas equation $PV = nRT$,since $n, R,$ and $T$ are constant,$P_1 V_1 = P_2 V_2$.The process is an adiabatic irreversible expansion (free expansion),so the relation $P_2 V_2^\gamma = P_1 V_1^\gamma$ is not applicable.Therefore,statements $(A), (B),$ and $(C)$ are correct.

Column $I$	Column $II$
$A$. $CO_{2(s)} \rightarrow CO_{2(g)}$	$p$. phase transition
$B$. $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$	$q$. allotropic change
$C$. $2H_{(g)} \rightarrow H_{2(g)}$	$r$. $\Delta H$ is positive
$D$. $P_{(\text{white, solid})} \rightarrow P_{(\text{red, solid})}$	$s$. $\Delta S$ is positive
	$t$. $\Delta S$ is negative

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