# Convert 77.0 L at 18.0 mm of Hg to its new volume at standard pressure.

Dec 15, 2014

The answer is $1.8 L$.

Standard Temperature and Pressure stipulates a pressure of $100.0 k P a$. In order to solve this problem, we'll use Boyle's law, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, which states that the pressure and volume of a gas are proportional to one another.

Let's calculate the volume by converting $m m H g$ to $k P a$ first.

$18 m m H g \cdot \frac{1.0 a t m}{760 m m H g} \cdot \frac{101.325 k P a}{1.0 a t m} = 2.4 k P a$

So, ${V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1} = \frac{2.4 k P a}{100.0 k P a} \cdot 77 L = 1.8 L$

Now calculate the volume by converting $k P a$ to $m m H g$.

$100.0 k P a \cdot \frac{1 a t m}{101.325 k P a} \cdot \frac{760 m m H g}{1 a t m} = 750 m m H g$

${V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1} = \frac{18 m m H g}{750 m m H g} \cdot 77 L = 1.8 L$