# 500.0 mL of a gas is collected at 745.0 mm Hg. What will the volume be at standard pressure?

Apr 6, 2016

This is an application of Boyle's Law.

#### Explanation:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$. This form is useful in that we do not have to convert volume and pressure to standard forms. We can even use the non-standard pressure measurement of $m m$ $H g$.

$1$ $a t m$ $=$ $760$ $m m$ $H g$. So the volume of your gas should slightly reduce:

${V}_{2} = \frac{{P}_{1} \times {V}_{1}}{{P}_{2}}$ $=$ $\frac{745 \cdot m m \cdot H g \times 500.0 \cdot m L}{760 \cdot m m \cdot H g}$ $=$ ?? $m L$.

Note that here (reasonably) a constant temperature is assumed.