Given P_A = 3 times 10^4 , Pa,P_B = 8 times 1 | vedclass

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(A) From the $P-V$ diagram,process $AB$ is an isochoric process (constant volume),so the work done $W_{AB} = 0 \, J$.Process $BC$ is an isobaric proce

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$560$

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(A) From the $P-V$ diagram,process $AB$ is an isochoric process (constant volume),so the work done $W_{AB} = 0 \, J$.Process $BC$ is an isobaric process (constant pressure) at $P_B = 8 	imes 10^4 \, Pa$. The volume changes from $V_A$ to $V_D$ (since $V_B = V_A$ and $V_C = V_D$).Work done in process $BC$ is $W_{BC} = P_B(V_C - V_B) = 8 	imes 10^4 	imes (5 	imes 10^{-3} - 2 	imes 10^{-3}) = 8 	imes 10^4 	imes 3 	imes 10^{-3} = 240 \, J$.Total work done in path $ABC$ is $W_{AC} = W_{AB} + W_{BC} = 0 + 240 = 240 \, J$.Total heat absorbed in path $ABC$ is $\Delta Q_{AC} = \Delta Q_{AB} + \Delta Q_{BC} = 600 + 200 = 800 \, J$.Using the first law of thermodynamics,$\Delta Q_{AC} = \Delta U_{AC} + W_{AC}$.Substituting the values,$800 = \Delta U_{AC} + 240$.Therefore,the change in internal energy is $\Delta U_{AC} = 800 - 240 = 560 \, J$.

Column-$I$	Column-$II$
$(a)$ Heat engine	$(i)$ $\eta = \frac{Q_2}{Q_1 - Q_2}$
$(b)$ Heat pump	$(ii)$ $\eta = \frac{Q_1 - Q_2}{Q_1}$
	$(iii)$ $\eta = \frac{T_1 - T_2}{T_1}$

Given $P_A = 3 \times 10^4 \, Pa$,$P_B = 8 \times 10^4 \, Pa$,$V_A = 2 \times 10^{-3} \, m^3$,and $V_D = 5 \times 10^{-3} \, m^3$. An ideal gas absorbs $600 \, J$ of heat in the process $AB$ and $200 \, J$ of heat in the process $BC$. Find the change in internal energy between $A$ and $C$ in $J$.

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