# If 5/7 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 8/9 L?

Mar 7, 2017

The pressure is $= 7.23 k P a$

#### Explanation:

We apply Boyle's Law

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

${P}_{1} = 9 k P a$

${V}_{1} = \frac{5}{7} L$

${V}_{2} = \frac{8}{9} L$

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

$= \frac{\frac{5}{7}}{\frac{8}{9}} \cdot 9$

$= \frac{5}{7} \cdot \frac{9}{8} \cdot 9$

$= \frac{405}{56}$

$= 7.23 k P a$