# If 3 L of a gas at room temperature exerts a pressure of 25 kPa on its container, what pressure will the gas exert if the container's volume changes to 5 L?

Jul 3, 2017

The new pressure is $= 15 k P a$

#### Explanation:

We apply Boyle's Law

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

The initial pressure is ${P}_{1} = 25 k P a$

The initial volume is ${V}_{1} = 3 L$

The final volume is ${V}_{2} = 5 L$

The final pressure is

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

$= \frac{3}{5} \cdot 25 = 15 k P a$

Jul 3, 2017

The final volume will be $\text{15 kPa}$.

#### Explanation:

This question involves Boyle's law, which states that the volume of a given amount of gas held at constant temperature, varies inversely with the applied pressure. This means that if the pressure increases, the volume decreases, and vice-versa. The equation for this law is:

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

Known

${P}_{1} = \text{25 kPa}$

${V}_{1} = \text{3 L}$

${V}_{2} = \text{5 L}$

Unknown

${P}_{2}$

Solution

Rearrange the equation to isolate ${P}_{2}$. Substitute your data into the equation and solve.

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

P_2=(25"kPa"xxcolor(red)cancel(color(black)("3L")))/(5color(red)cancel(color(black)("L")))="15 kPa"