# A gas occupies a volume of 1309*mL at "STP". At constant temperature, what volume will it occupy at a pressure of 941*mm*Hg?

Jul 24, 2017

This is a question that has been asked by someone who has never used a mercury manometer. I get a final volume of ${V}_{2} \cong 1000 \cdot m L$.

#### Explanation:

$\text{1 atmosphere}$ of pressure will support a column of mercury that is $760 \cdot m m \cdot H g$. It is very UNWISE to put a mercury column under a pressure of GREATER than $1 \cdot a t m$. Why? Because, you run the risk of breaking the vessel, or overshooting the manometer's height, and getting mercury all over the lab, where it will inhabit every nook and cranny.

So we first convert ${P}_{1}$ and ${P}_{2}$ to kosher units......

$\text{STP}$ specifies $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$

And the quoted ${P}_{2} \equiv \frac{941 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 1.24 \cdot a t m .$

And old $\text{Boyle's Law}$ holds that for a given quantity of gas.......

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.

And thus V_2=(P_1V_1)/P_2=(1*atm*1309*mL)/(1.24*atm)=??*mL.

Remember what I said in regard to the use of $m m \cdot H g$ as a pressure measurement. You should NEVER quote a pressure $> 760 \cdot m m \cdot H g$.