Question

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`?

Answer 1

The new pressure is `=15kPa`

Explanation:

We apply Boyle's Law

`P_1V_1=P_2V_2`

The initial pressure is `P_1=25kPa`

The initial volume is `V_1=3L`

The final volume is `V_2=5L`

The final pressure is

`P_2=V_1/V_2*P_1`

`=3/5*25=15kPa`

Answer 2

The final volume will be `"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_1V_1=P_2V_2`

Known

`P_1="25 kPa"`

`V_1="3 L"`

`V_2="5 L"`

Unknown

`P_2`

Solution

Rearrange the equation to isolate `P_2`. Substitute your data into the equation and solve.

`P_2=(P_1V_1)/(V_2)`

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