Answer
a) $V_{2}$ = $V_{1}$
b) $T_{2}$ =$\frac{T_{1}}{3}$
Work Step by Step
a) This is an isochoric process, which means that volume remains constant. Thus $V_{2}$ = $V_{1}$
b) For this question, we can use the ideal gas law:
$\frac{p_{1}V_{1}}{T_{1}}$ = $\frac{p_{2}V_{2}}{T_{2}}$,
where:
$p_{1}$ is the initial pressure
$p_{2}$ is the final pressure
$T_{1}$ is the initial temperature
$T_{2}$ is the final temperature
$V_{1}$ is the initial volume
$V_{2}$ is the final volume.
So now we solve for $T_{2}$:
$T_{2}$ = $\frac{p_{2}V_{2}T_{1}}{p_{1}V_{1}}$ = $\frac{\frac{1}{3}p_{1}V_{1}T_{1}}{p_{1}V_{1}}$ = $\frac{1}{3}T_{1}$ = $\frac{T_{1}}{3}$