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(C) For an ideal gas,$PV = nRT$,so $V = (nR/P)T$. Work done $W = \int P dV$. From $PV = nRT$,$P dV + V dP = nR dT$. For processes $AB$ and $CD$,$T$ is

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$2$

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(C) For an ideal gas,$PV = nRT$,so $V = (nR/P)T$. Work done $W = \int P dV$. From $PV = nRT$,$P dV + V dP = nR dT$. For processes $AB$ and $CD$,$T$ is constant,so $dT = 0$,which implies $P dV = -V dP$. Thus,$W = \int_{P_i}^{P_f} P dV = -\int_{P_i}^{P_f} V dP = -\int_{P_i}^{P_f} (nRT/P) dP = -nRT \ln(P_f/P_i) = nRT \ln(P_i/P_f)$. Let the lines passing through the origin be $T = m_1 P$ and $T = m_2 P$. For process $AB$,$T = T_1$,$P_A = T_1/m_1$,$P_B = T_1/m_2$. $W_{AB} = nRT_1 \ln(P_A/P_B) = nRT_1 \ln((T_1/m_1) / (T_1/m_2)) = nRT_1 \ln(m_2/m_1)$. Similarly,for process $CD$,$W_{CD} = nRT_2 \ln(m_2/m_1)$. Given $W_{AB} = 2 W_{CD}$,we have $nRT_1 \ln(m_2/m_1) = 2 nRT_2 \ln(m_2/m_1)$. Therefore,$T_1 = 2 T_2$,which gives $T_1/T_2 = 2$.

On a $TP$ diagram,two moles of an ideal gas perform processes $AB$ and $CD$. If the work done by the gas in the process $AB$ is two times the work done in the process $CD$,then what is the value of $T_1/T_2$?

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