To solve this question, we need to use the combined gas equation which is given as
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ p=\text{pressure} \\ v=\text{volume} \\ t=\text{temperature} \end{gathered}[/tex]Now we should write down the given data
[tex]\begin{gathered} v_1=28.4l \\ p_1=724\operatorname{mm}hg \\ T_1=303K \\ v_2=15.5L \\ T_2=379K \\ p_2=\text{ ?} \end{gathered}[/tex]So, we would use the previous equation i wrote down.
[tex]\begin{gathered} \frac{p_1v_1}{t_1}=\frac{p_2v_2}{t_2} \\ \frac{724\times28.4}{303}=\frac{p_2\times15.5}{379} \\ 67.8600=\frac{p_2\times15.5}{379} \\ 15.5p_2=67.86\times379 \\ 15.5p_2=25718.94 \\ \text{divide both sides by the coefficent of p2} \\ \frac{15.5p_2}{15.5}=\frac{25718.94}{15.5} \\ p_2=1659.28\operatorname{mm}hg \end{gathered}[/tex]From the calculations above, the final pressure of the gas is 1659.28mmHg
