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(C) The change in internal energy $\Delta U$ is a state function,meaning it depends only on the initial and final states.For an isothermal process,the

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$30$

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(C) The change in internal energy $\Delta U$ is a state function,meaning it depends only on the initial and final states.For an isothermal process,the temperature remains constant,so the change in internal energy $\Delta U_{	ext{iso}} = 0$.Given the total change in internal energy $\Delta U_{	ext{total}} = -30 J$,and since the path consists of two isothermal processes and one adiabatic process,we have $\Delta U_{	ext{total}} = \Delta U_{	ext{iso1}} + \Delta U_{	ext{adiabatic}} + \Delta U_{	ext{iso2}}$.Substituting the values: $-30 J = 0 + \Delta U_{	ext{adiabatic}} + 0$,which gives $\Delta U_{	ext{adiabatic}} = -30 J$.For an adiabatic process,the first law of thermodynamics states $\Delta Q = \Delta U + W$. Since $\Delta Q = 0$,we have $W_{	ext{adiabatic}} = -\Delta U_{	ext{adiabatic}}$.Therefore,$W_{	ext{adiabatic}} = -(-30 J) = 30 J$.

Column-$I$	Column-$II$
$(A)$ $PV$ vs $V$ (Isothermal expansion)	$(P)$ $W$ > 0
$(B)$ $P$ vs $T$ (Isochoric heating)	$(Q)$ $W$ < 0
$(C)$ $P$ vs $V$ (Isobaric expansion)	$(R)$ $\Delta Q$ > 0
$(D)$ $V$ vs $T$ (Isobaric compression)	$(S)$ $\Delta U$ > 0
	$(T)$ $\Delta U$ < 0

$A$. Isobaric	$I$. $\Delta Q = \Delta W$
$B$. Isochoric	$II$. $\Delta Q = \Delta U$
$C$. Adiabatic	$III$. $\Delta Q = 0$
$D$. Isothermal	$IV$. $\Delta Q = \Delta U + P \Delta V$

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