A thermodynamic system undergoes a two-step process. In the first step $A o B$ (isochoric), $400 \

Question

A thermodynamic system undergoes a two-step process. In the first step $A 	o B$ (isochoric), $400 \mathrm{~J}$ of heat is added. In the second step $B 	o C$ (isobaric), $500 \mathrm{~J}$ of heat is added. Given the states are $A(2 	imes 10^4 \mathrm{~Pa}, 1 	imes 10^{-3} \mathrm{~m}^3)$, $B(6 	imes 10^4 \mathrm{~Pa}, 1 	imes 10^{-3} \mathrm{~m}^3)$, and $C(6 	imes 10^4 \mathrm{~Pa}, 4 	imes 10^{-3} \mathrm{~m}^3)$, the change in internal energy of the system for the overall process $A 	o C$ is

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(b) : Total heat added $\Delta Q = Q_{AB} + Q_{BC} = 400 + 500 = 900 \mathrm{~J}$. Work done in isochoric process $A 	o B$ is $W_{AB} = 0$. Work done in isobaric process $B 	o C$ is $W_{BC} = P_B(V_C - V_B) = 6 	imes 10^4 (4 	imes 10^{-3} - 1 	imes 10^{-3}) = 6 	imes 10^4 	imes 3 	imes 10^{-3} = 180 \mathrm{~J}$. Total work $\Delta W = 180 \mathrm{~J}$. By the first law of thermodynamics, $\Delta U = \Delta Q - \Delta W = 900 - 180 = 720 \mathrm{~J}$.