Answer
(a) $V_2 = V_1$
(b) $T_2 = \frac{T_1}{3}$
Work Step by Step
(a) Since this is an isochoric process, the volume does not change. Therefore, $V_2 = V_1$.
(b) We can find an expression for the initial temperature as:
$P_1V_1 = nRT_1$
$T_1 = \frac{P_1V_1}{nR}$
We can find the new pressure as:
$P_2V_2 = nRT_2$
$T_2 = \frac{P_2V_2}{nR}$
$T_2 = \frac{(P_1/3)V_1}{nR}$
$T_2 = \frac{1}{3}~\frac{P_1V_1}{nR}$
$T_2 = \frac{1}{3}~T_1$
