We could use the combined law of gases:
[tex]\frac{P_1\cdot V_1}{T_1}=\frac{P_2\cdot V_2}{T_2}[/tex]Replacing the values of the problem, we got that:
[tex]\begin{gathered} P_1=4.11kPa \\ V_1=5.22L \\ T_1=405.4K \\ P_2=8.22kPa \\ T_2=124.8K \\ V_2=?_{} \end{gathered}[/tex]Now, we could solve the equation given for V2 with all these given values:
[tex]\begin{gathered} \frac{P_1\cdot V_1}{T_1}=\frac{P_2\cdot V_2}{T_2}\to V_2=\frac{P_1\cdot V_1\cdot T_2}{T_1\cdot P_2}_{} \\ \\ V_2=\frac{4.11kPa\cdot5.22L\cdot124.8K}{405.4K\cdot8.22kPa} \\ \\ V_2=0.80L^{}_{} \end{gathered}[/tex]Therefore, the new volume of the gas is 0.80L.
