# A gas occupies 3.30 liters and is at 850 mmHg. What would the pressure be in atm, if the new volume were increased to 5.63 liters?

Oct 3, 2016

This is a clear application of $\text{Boyle's Law}$ $P \propto \frac{1}{V}$

The new pressure is approx. $0.7 \cdot a t m$.

#### Explanation:

This is a clear application of $\text{Boyle's Law}$ $P \propto \frac{1}{V}$

And thus, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

We note that $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$

${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2$ $=$ $\left(\frac{850 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}\right) \times \frac{3.30 \cdot L}{5.63 \cdot L}$

$=$ ??atm

This is not a good question. It has not been proposed by a chemist. You do NOT use a mercury manometer to measure pressure above one atmosphere. If you do so, you'll get mercury all over the lab, and I'm not going to clean it up, and no one else is either.