The process CD is shown in the diagram. As th | vedclass

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(B) The temperature of an ideal gas is given by $PV = nRT$,which implies $T \propto PV$. At point $C$,$P_C = 3p_0$ and $V_C = v_0$,so $T_C \propto (3p

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Temperature first increases and then decreases

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(B) The temperature of an ideal gas is given by $PV = nRT$,which implies $T \propto PV$. At point $C$,$P_C = 3p_0$ and $V_C = v_0$,so $T_C \propto (3p_0)(v_0) = 3p_0v_0$. At point $D$,$P_D = p_0$ and $V_D = 3v_0$,so $T_D \propto (p_0)(3v_0) = 3p_0v_0$. Since $T_C = T_D$,the temperature at the start and end points is the same. For a linear process $P = mV + c$,the product $PV = V(mV + c) = mV^2 + cV$ is a quadratic function of $V$. The maximum value of this product occurs at the midpoint of the process. Thus,the temperature first increases and then decreases as the system moves from $C$ to $D$.

The process $CD$ is shown in the diagram. As the system is taken from $C$ to $D$,what happens to the temperature of the system?

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