# If 3/7 L of a gas at room temperature exerts a pressure of 9 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/9 L?

Oct 16, 2016

${P}_{2} = \frac{243}{35} k P a$

#### Explanation:

Let ${P}_{1} = 9 \textcolor{w h i t e}{.} k P a$
Let ${V}_{1} = \frac{3}{7} \textcolor{w h i t e}{.} L$
Let ${V}_{2} = \frac{9}{5} \textcolor{w h i t e}{.} L$

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

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

${P}_{2} = \frac{243}{35} k P a$

Oct 16, 2016

$\frac{243}{35}$kPa

#### Explanation:

Boyle's Law for gases states that pressure and volume vary inversely when temperature and mass are held constant. This can be expressed with the formula:

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

${P}_{1} = 9$kPa

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

${P}_{2} =$?

${V}_{2} = \frac{5}{9} L$

$9 \cdot \frac{3}{7} = {P}_{2} \cdot \frac{5}{9}$

$\frac{9}{5} \cdot 9 \cdot \frac{3}{7} = {P}_{2} \cdot \frac{\cancel{5}}{\cancel{9}} \cdot \frac{\cancel{9}}{\cancel{5}} \textcolor{w h i t e}{a a a}$Multiply both sides by 9/5

$\frac{243}{35} = {P}_{2}$

${P}_{2} = \frac{243}{35}$kPa