One mole of an ideal gas undergoes two differ | vedclass

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(B) For cycle $I$:Work done $W_I = W_a + W_b + W_c + W_d$$W_a = P \Delta V = (4P_0)(2V_0 - V_0) = 4P_0V_0$$W_b = nRT \ln(V_f/V_i) = P_i V_i \ln(V_f/V_

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$2$

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(B) For cycle $I$:Work done $W_I = W_a + W_b + W_c + W_d$$W_a = P \Delta V = (4P_0)(2V_0 - V_0) = 4P_0V_0$$W_b = nRT \ln(V_f/V_i) = P_i V_i \ln(V_f/V_i) = (4P_0)(2V_0) \ln(4V_0/2V_0) = 8P_0V_0 \ln 2$$W_c = P \Delta V = (2P_0)(V_0 - 4V_0) = -6P_0V_0$$W_d = 0$ (isochoric)$W_I = 4P_0V_0 + 8P_0V_0 \ln 2 - 6P_0V_0 = 8P_0V_0 \ln 2 - 2P_0V_0 = 2P_0V_0(4 \ln 2 - 1)$For cycle $II$:Work done $W_{II} = W_{a'} + W_{b'} + W_{c'} + W_{d'}$$W_{a'} = P_i V_i \ln(V_f/V_i) = (4P_0)(V_0) \ln(2V_0/V_0) = 4P_0V_0 \ln 2$$W_{b'} = 0$ (isochoric)$W_{c'} = P \Delta V = (P_0)(V_0 - 2V_0) = -P_0V_0$$W_{d'} = 0$ (isochoric)$W_{II} = 4P_0V_0 \ln 2 - P_0V_0 = P_0V_0(4 \ln 2 - 1)$Ratio $W_I / W_{II} = [2P_0V_0(4 \ln 2 - 1)] / [P_0V_0(4 \ln 2 - 1)] = 2$

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

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