# If 16 L of a gas at room temperature exerts a pressure of 7 kPa on its container, what pressure will the gas exert if the container's volume changes to 2 L?

Nov 17, 2016

The pressure is $= 56 k P a$

#### Explanation:

Use Boyle's Law

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

${P}_{1} = 7 k P a$

${V}_{1} = 16 l$

${V}_{2} = 2 l$

So, ${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = 7 \cdot \frac{16}{2} = 56 k P a$