Question

If `8/3` `L` of a gas at room temperature exerts a pressure of `30` `kPa` on its container, what pressure will the gas exert if the container's volume changes to `7/4` `L`?

Answer 1

The final pressure is `45.7` `kPa`.

Explanation:

The combined gas law is:

`(P_1V_1)/T_1=(P_2V_2)/T_2`

In this case, the temperature doesn't change, so it can be removed:

`P_1V_1=P_2V_2`

Rearranging:

`P_2=(P_1V_1)/V_2=(30*(8/3))/(7/4)=45.7` `kPa`