# If 11/6 L of a gas at room temperature exerts a pressure of 9 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/4 L?

Jul 18, 2017

The pressure is $= 13.2 k P a$

#### Explanation:

We apply Boyle's Law

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

The initial pressure is ${P}_{1} = 9 k P a$

The initial volume is ${V}_{1} = \frac{11}{6} L$

The final volume is ${V}_{2} = \frac{5}{4} L$

The final pressure is

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

$= \frac{\frac{11}{6}}{\frac{5}{4}} \cdot 9$

$= 13.2 k P a$