Question

If `8 L` of a gas at room temperature exerts a pressure of `28 kPa` on its container, what pressure will the gas exert if the container's volume changes to `9 L`?

Answer 1

The final pressure will be 25 kPa.

Explanation:

This is an example of Boyle's law, which states that the volume of a gas kept at constant temperature varies inversely with the volume. This means that when volume increases, pressure decreases. The equation required to solve this problem is `P_1V_1=P_2V_2`, where `P_1` and `P_2` are the initial and final pressure, while `V_1` and `V_2` are the initial and final volume.

Given
`V_1="8 L"`
`P_1="28 kPa"`
`V_2="9 L"`

Unknown
`P_2`

Solution
Rearrange the equation to isolate `P_2`. Substitute the variables with their values and solve.

`P_1V_1=P_2V_2`

`P_2=(P_1V_1)/V_2`

`P_2=(8cancel"L"*28"kPa")/(9cancel"L")="25 kPa"` rounded to two significant figures