# If 14/3 L of a gas at room temperature exerts a pressure of 45 kPa on its container, what pressure will the gas exert if the container's volume changes to 13/5 L?

Feb 15, 2016

$\frac{1050}{13} \approx 80.77 k P a$

#### Explanation:

Assuming the temperature remains constant, this is an application of Boyle's Law, which states that $P \cdot V$ (pressure times Volume) remains constant at constant temperature.
So, if

${V}_{1} = \frac{14}{3}$ and ${P}_{1} = 45$

and ${V}_{2} = \frac{13}{5}$ then since

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

$\frac{14}{3} \cdot 45 = {P}_{2} \cdot \frac{13}{5}$

$\frac{14 \cdot 45 \cdot 5}{3 \cdot 13} = {P}_{2}$

${P}_{2} = \frac{1050}{13} \approx 80.77 k P a$