Question

If `3 L` of a gas at room temperature exerts a pressure of `7 kPa` on its container, what pressure will the gas exert if it the container's volume changes to `1 L`?

Answer 1

From the ideal gas law we have that

`P_1*V_1=P_2*V_2=>P_2=(P_1*V_1)/(V_2)`

Hence

`P_2=(3*7*10^4)/1=21*10^4=21kPa`