Question

If `7/3 L` of a gas at room temperature exerts a pressure of `21 kPa` on its container, what pressure will the gas exert if the container's volume changes to `15/7 L`?

Answer 1

The pressure is `=22.9kPa`

Explanation:

Use Boyle's Law

`P_1V_1=P_2V_2`

`P_1=21kPa`

`V_1=7/3l`

`V_2=15/7l`

`P_2=P_1V_1/V_2=21*7/3/(15/7)=(21*7*7)/(3*15)=22.9kPa`