## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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