# A gas occupies 100.0 mL at a pressure of 780 mm Hg. What is the volume of the gas when its pressure is increased to 880 mm Hg?

Apr 23, 2017

This question has been set by someone who has never used a mercury manometer............we gets ${V}_{2} = 87.6 \cdot m L$

#### Explanation:

One atmosphere of pressure will support a column of mercury that is $760 \cdot m m$ high. A manometer is thus a highly visual measure of the strength of a vacuum, or as a measure of atmospheric pressure.

If you put the column under a pressure greater than $1 \cdot a t m$ you are likely to get mercury all over the laboratory, where it will inhabit every crack and every cranny. This is a MAJOR cleanup job, which contract cleaners would not touch.

And so P_1="780 mm Hg"/("760 mm Hg"*atm^-1)=1.026*atm.

And so P_2="880 mm Hg"/("760 mm Hg"*atm^-1)=1.158*atm.

And now we use old $\text{Boyle's Law}$, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2.} \ldots .$, and solve for

${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2 = \frac{1.026 \cdot a t m \times 100.0 \cdot m L}{1.158 \cdot a t m}$

$= 87.6 \cdot m L$

The volume is slightly decreased, as we would expect, why?