Question

If `7/5` `L` of a gas at room temperature exerts a pressure of `3` `kPa` on its container, what pressure will the gas exert if the container's volume changes to `5/3` `L`?

Answer 1

Start with the Combined Gas Law: `(P_1V_1)/T_1=(P_2V_2)/T_2`

Temperature is constant, so `(P_1V_1)=(P_2V_2) to P_2=(P_1V_1)/V_2 = (3*(7/5))/(5/3) = (21/5)/(5/3)=36/25` `kPa`

Explanation:

We could memorise Boyle's, Charles' and Gay-Lussac's Laws, which relate various combinations of pressure, volume and temperature, but I find it simpler to just memorise the Combined Gas Law and then eliminate the thing that is kept constant.

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

In this case, the temperature is kept constant, so we can ignore it:

`(P_1V_1)=(P_2V_2)`

Rearranging:

`P_2=(P_1V_1)/V_2`