An ideal gas exists in a state with pressure | vedclass

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Question

(A) For a cyclic process, the change in internal energy $\Delta U = 0$. Therefore, the total heat exchanged $Q_T$ is equal to the total work done $W$

Vedclass · Accepted Answer

$P_0 V_0(2 \ln 2 - 0.75)$

Vedclass · Answer

(A) For a cyclic process, the change in internal energy $\Delta U = 0$. Therefore, the total heat exchanged $Q_T$ is equal to the total work done $W$ by the system, i.e., $Q_T = W$.$1$. Isothermal expansion from $(P_0, V_0)$ to $(P_0/4, 4V_0)$:The work done $W_1 = nRT \ln(V_f/V_i) = P_0 V_0 \ln(4V_0/V_0) = P_0 V_0 \ln 4 = 2 P_0 V_0 \ln 2$.$2$. Isobaric compression from $(P_0/4, 4V_0)$ to $(P_0/4, V_0)$:The work done $W_2 = P \Delta V = (P_0/4)(V_0 - 4V_0) = (P_0/4)(-3V_0) = -0.75 P_0 V_0$.$3$. Isochoric heating from $(P_0/4, V_0)$ to $(P_0, V_0)$:The work done $W_3 = 0$ since volume is constant.Total work done $W = W_1 + W_2 + W_3 = 2 P_0 V_0 \ln 2 - 0.75 P_0 V_0 = P_0 V_0(2 \ln 2 - 0.75)$.Since $Q_T = W$, the total heat exchanged is $P_0 V_0(2 \ln 2 - 0.75)$.

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