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(C) From the ideal gas equation,$PV = nRT$,so $T \propto PV$.At point $A$,$P_A = P_0$ and $V_A = V_0$,so $T_A \propto P_0 V_0$.At point $B$,$P_B = 2P_

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$\frac{2}{7}$

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(C) From the ideal gas equation,$PV = nRT$,so $T \propto PV$.At point $A$,$P_A = P_0$ and $V_A = V_0$,so $T_A \propto P_0 V_0$.At point $B$,$P_B = 2P_0$. Let the volume at $B$ be $V_B$. Then $T_B \propto (2P_0) V_B$.Thus,$\frac{T_A}{T_B} = \frac{P_0 V_0}{2P_0 V_B} = \frac{V_0}{2V_B}$.From the geometry of the $PV$ diagram,the slope of line $AB$ is $	an 60^{\circ} = \frac{2P_0 - P_0}{V_B - V_0} = \frac{P_0}{V_B - V_0}$.So,$V_B - V_0 = \frac{P_0}{	an 60^{\circ}} = \frac{P_0}{\sqrt{3}}$.The slope of line $BC$ is $-	an 30^{\circ} = \frac{P_0 - 2P_0}{4V_0 - V_B} = \frac{-P_0}{4V_0 - V_B}$.So,$4V_0 - V_B = \frac{P_0}{	an 30^{\circ}} = P_0 \sqrt{3}$.Dividing the two equations: $\frac{V_B - V_0}{4V_0 - V_B} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.$3V_B - 3V_0 = 4V_0 - V_B \implies 4V_B = 7V_0 \implies V_B = \frac{7}{4}V_0$.Substituting $V_B$ into the ratio: $\frac{T_A}{T_B} = \frac{V_0}{2(7/4)V_0} = \frac{V_0}{(7/2)V_0} = \frac{2}{7}$.

Four moles of an ideal monoatomic gas undergoes a cyclic process $ABCA$ as shown in the figure. Find the ratio of the temperature of $A$ and $B$.

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