A thermodynamic process is shown in the figur | vedclass

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

Vedclass · Accepted Answer

$560$

Vedclass · Answer

(A) From the given $P-V$ graph:Process $AB$ is an isochoric process (volume is constant),so work done $W_{AB} = 0$.Process $BC$ is an isobaric process (pressure is constant at $P_B = 8 	imes 10^4 \, Pa$),so work done $W_{BC} = P_B(V_C - V_B)$.Since $V_C = V_D = 5 	imes 10^{-3} \, m^3$ and $V_B = V_A = 2 	imes 10^{-3} \, m^3$,we have:$W_{BC} = 8 	imes 10^4 	imes (5 - 2) 	imes 10^{-3} = 8 	imes 10^4 	imes 3 	imes 10^{-3} = 240 \, J$.Total work done in process $AC$ (path $A ightarrow B ightarrow C$) is $W_{AC} = W_{AB} + W_{BC} = 0 + 240 = 240 \, J$.Total heat added in process $AC$ 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,$\Delta U_{AC} = 800 - 240 = 560 \, J$.

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