# If 9/4 L of a gas at room temperature exerts a pressure of 16 kPa on its container, what pressure will the gas exert if the container's volume changes to 9/7 L?

May 28, 2017

The pressure is $= 28 k P a$

#### Explanation:

We apply Boyle's Law

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

The initial pressure is ${P}_{1} = 16 k P a$

The initial volume is ${V}_{1} = \frac{9}{4} L$

The final volume is ${V}_{2} = \frac{9}{7}$

The final pressure is

${P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1} = \frac{\frac{9}{4}}{\frac{9}{7}} \cdot 16$

$= \frac{7}{4} \cdot 16 = 28 k P a$