Question

If `8 L` of a gas at room temperature exerts a pressure of `32 kPa` on its container, what pressure will the gas exert if the container's volume changes to `5 L`?

Answer 1

The pressure is `=51.2kPa`

Explanation:

We apply Boyle's Law

`P_1V_1=P_2V_2`

`P_1=32kPa`

`V_1=8 L`

`V_2=5L`

`P_2=(P_1V_1)/V_2=(32*8)/5=51.2kPa`