Chemistry: Molecular Approach (4th Edition)

$4.4\times10^{2}$ mmHg
Given that $V_{1}$ = 5.6 L, $p_{1}$ = 735 mmHg and $V_{2}$ = 9.4 L According to Boyle's law, $p_{1}V_{1}=p_{2}V_{2}$. Then, $p_{2} = \frac{p_{1}V_{1}}{V_{2}} = \frac{735mmHg\times5.6L}{9.4L}$ = $4.4\times10^{2}$ mmHg