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(C) Since internal energy is a state function,the change in internal energy depends only on the initial and final states. Therefore,$\Delta U_{ABC} =

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$460$

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(C) Since internal energy is a state function,the change in internal energy depends only on the initial and final states. Therefore,$\Delta U_{ABC} = \Delta U_{AC}$.For process $AB$ (isochoric,$\Delta V = 0$):$\Delta W_{AB} = 0$$\Delta Q_{AB} = \Delta U_{AB} = 400 \, J$.For process $BC$ (isobaric,$P = 6 	imes 10^4 \, Pa$):$\Delta W_{BC} = P \Delta V = 6 	imes 10^4 	imes (4 	imes 10^{-3} - 2 	imes 10^{-3}) = 6 	imes 10^4 	imes 2 	imes 10^{-3} = 120 \, J$.$\Delta Q_{BC} = \Delta U_{BC} + \Delta W_{BC} \implies 100 = \Delta U_{BC} + 120 \implies \Delta U_{BC} = -20 \, J$.Total change in internal energy for path $ABC$ is:$\Delta U_{AC} = \Delta U_{AB} + \Delta U_{BC} = 400 - 20 = 380 \, J$.For process $AC$,the work done $\Delta W_{AC}$ is the area under the line $AC$ in the $PV$ diagram,which is a trapezoid:$\Delta W_{AC} = 	ext{Area} = \frac{1}{2} 	imes (P_A + P_C) 	imes (V_C - V_A) = \frac{1}{2} 	imes (2 	imes 10^4 + 6 	imes 10^4) 	imes (4 	imes 10^{-3} - 2 	imes 10^{-3}) = \frac{1}{2} 	imes 8 	imes 10^4 	imes 2 	imes 10^{-3} = 80 \, J$.Using the first law of thermodynamics for process $AC$:$\Delta Q_{AC} = \Delta U_{AC} + \Delta W_{AC} = 380 + 80 = 460 \, J$.

Column $I$	Column $II$
$(A)$ An insulated container has two chambers separated by a valve. Chamber $I$ contains an ideal gas and Chamber $II$ has a vacuum. The valve is opened.	$(p)$ The temperature of the gas decreases
$(B)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-2}$	$(q)$ The temperature of the gas increases or remains constant
$(C)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-4/3}$	$(r)$ The gas loses heat
$(D)$ An ideal monoatomic gas expands such that its pressure $P$ and volume $V$ follow the behavior shown in the graph	$(s)$ The gas gains heat

Column $I$	Column $II$
$(p)$ isobaric	$(x)$ $\frac{PV(1 - 2^{1 - \gamma})}{\gamma - 1}$
$(q)$ isothermal	$(y)$ $PV$
$(r)$ adiabatic	$(z)$ $PV \ln 2$

The figure below shows two paths that may be taken by a gas to go from a state $A$ to a state $C.$ In process $AB,$ $400 \, J$ of heat is added to the system and in process $BC,$ $100 \, J$ of heat is added to the system. The heat absorbed by the system in the process $AC$ will be ...... $J$

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