# If 13/4 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 5/6 L?

Nov 14, 2016

The pressure is $= 62.4 k P a$

#### Explanation:

We apply Boyle's law.

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

${P}_{1} = 16 k P a$

${V}_{1} = \frac{13}{4} l$

${V}_{2} = \frac{5}{6} l$

So ${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{16 \cdot \frac{13}{4}}{\frac{5}{6}} = 4 \cdot 13 \cdot \frac{6}{5} = \frac{312}{5} = 62.4 k P a$