A3.00Ltank contains air at 3.00 atmand at $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$. The tank is sealed and cooled until the pressure is1.00atm. (a) What is the temperature then in degree celsius? Assume that the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the volume when the pressure again becomes3.00 atm?

The relation between pressure, volume, temperature and number of moles is expressed by the ideal gas equationPV=nRT.

In the first case, the volume and the number of moles is kept constant. Rearranging the equation,$P=\frac{nR}{V}T$, which means, $P=acons\tan t*T$or $P\propto T$. So, as there are two cases before cooling() and after cooling($P_1,T_1$), then it becomes, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$.

Thus, the relation for final temperature is $T_2=\frac{P_2{T}_1}{P_1}$.

In the second case, the pressure $P_1$and volume $V_2$are changed, to $P_3$and $V_3$, with temperature and number of moles kept constant. Then, ideal gas equation is arranged as $V=\frac{nRT}{P}$which implies,$V\propto \frac{1}{P}$or $PV=acons\tan t$. For the two cases, $P_2{V}_2=P_3{V}_3$. Thus, the final volume $V_3=\frac{P_2{V}_2}{P_3}$.

In degree Celsius,$T_2=97.67K-273=-175.{33}^{o}C.$.

Hence, the final temperature when the tank is kept at a constant volume is $-175.{33}^{o}C.$

(b)The final volume is found from $V_3=\frac{P_2{V}_2}{P_3}$.

Substituting the values of $P_2=1atm,V_2=3L\left(provided\right),P_3=3atm$ in $V_3=\frac{P_2{V}_2}{P_3}$,

localid="1668075026754" $$\begin{gathered} V_3=\frac{1*3}{3} \\ {V}_3=1 \end{gathered}$$

Hence, the final volume after final compression to 3atm is 1L.