# A 2.5*L volume of gas at 800*mm*Hg pressure, is expanded with constant temperature to a 5.0*L volume. What is the resultant pressure of the gas?

May 8, 2017

Well ${P}_{1} {V}_{1} = {P}_{2} {V}_{2.} \ldots \ldots \ldots .$ given constant $T$, which is old $\text{Boyle's Law} \ldots \ldots \ldots \ldots \ldots . .$
And so ${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{800 \cdot m m \cdot H g \times 0.250 \cdot L}{0.500 \cdot L} =$
$400 \cdot m m \cdot H g$.
Just to ADD that you DO NOT measure a pressure in $m m \cdot H g$ OVER $1 \cdot a t m$ ($1 \cdot a t m \equiv 760 \cdot m m \cdot H g$). This is an inappropriate question posed by someone who has NEVER used a mercury manometer, and certainly NEVER cleaned up a mercury spill.