#### Answer

a) Isochoric process
b) $T_{1}$ = 336$^{\circ}$C and $T_{2}$ = 1556$^{\circ}$C

#### Work Step by Step

a) From the figure, it can be seen that volume remains constant. Therefore it can be concluded that this is an Isochoric process (Vertical line).
b) What we have:
number of moles (n) = 0.0040 mol
From the figure, we can see that the initial volume $V_{1}$ = $200cm_{3}$ and the initial pressure $p_{1}$ = 1 atm.
Before we can use the ideal gas law to find initial Temperature ($T_{1}$), we must convert the above to the relevant SI units:
n = 0.0040 mol
$V_{1}$ = 200$cm_{3}$ = $200\times10^{-6}m^{3}$
$p_{1}$ = 1 atm = $1.013 \times10^{5}Pa$
R is a constant (8.31 J/mol K)
Now we can use the ideal gas law:
$p_{1}$$V_{1}$ = nR$T_{1}$
So $T_{1}$ = $\frac{p_{1}V_{1}}{nR}$ = $\frac{(1.013\times10^{5})(200\times10^{-6})}{0.0040 \times 8.31}$ = 609.5K.
Convert Kelvins (K) to Celsius by subtracting 273:
T$^{\circ}C$ = 609.5K - 273 = 336$^{\circ}$C.
To calculate the final temperature, we can use
$\frac{p_{1}V_{1}}{T_{1}}$ = $\frac{p_{2}V_{2}}{T_{2}}$.
Since this is an isochoric process, $V_{1}$ = $V_{2}$.
The final temperature $T_{2}$ will be
$T_{2}$ = $\frac{p_{2}V_{2}T_{1}}{p_{1}V_{1}}$ = $\frac{3(V_{1})(609.5)}{(1)(V_{1})}$ = 1829 K = 1556$^{\circ}$C.