# If 7/3 L of a gas at room temperature exerts a pressure of 21 kPa on its container, what pressure will the gas exert if the container's volume changes to 15/7 L?

Nov 29, 2016

The pressure is $= 22.9 k P a$

#### Explanation:

Use Boyle's Law

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

${P}_{1} = 21 k P a$

${V}_{1} = \frac{7}{3} l$

${V}_{2} = \frac{15}{7} l$

${P}_{2} = {P}_{1} {V}_{1} / {V}_{2} = 21 \cdot \frac{7}{3} / \left(\frac{15}{7}\right) = \frac{21 \cdot 7 \cdot 7}{3 \cdot 15} = 22.9 k P a$