# At constant temperature, a 3.8*L volume of gas at 765*mm*Hg, was expanded to give a pressure of 0.500*atm. What is the new volume?

May 4, 2017

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ under conditions of constant temperature.

#### Explanation:

We need to know that $1 \cdot a t m$ will support a column of mercury that is $760 \cdot m m$ high, and thus a mercury column may used as a highly visual measurement of pressure:

$1 \cdot a t m \equiv 760 \cdot m m \cdot H g$ (note that given the schemozzle that occurs when you spill mercury in the lab, you really should not put mercury under a pressure of MORE than 1 atmosphere). Here.......

${P}_{1} = \frac{765 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 1.01 \cdot a t m .$

We solve for ${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2 = \frac{1.01 \cdot a t m \times 3.8 \cdot L}{0.500 \cdot a t m} = 7.60 \cdot L$