# A gas occupies 12.3 at a pressure of 40.0 KPa. What is the volume when the pressure is increased to 60.0 KPa?

May 11, 2017

The volume is $= 8.2 u$, whatever the chosen units were to begin with...

#### Explanation:

We apply Boyle's Law at constant temperature and constant number of $\text{mol}$s:

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

The initial pressure is ${P}_{1} = 40 k P a$

The initial volume is ${V}_{1} = 12.3$

The final pressure is $60 k P a$

The final volume is

${V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1}$

$= \frac{40}{60} \cdot 12.3 = 8.2 u$