Question

If `11/6 L` of a gas at room temperature exerts a pressure of `9 kPa` on its container, what pressure will the gas exert if the container's volume changes to `5/4 L`?

Answer 1

The pressure is `=13.2kPa`

Explanation:

We apply Boyle's Law

`P_1V_1=P_2V_2`

The initial pressure is `P_1=9kPa`

The initial volume is `V_1=11/6L`

The final volume is `V_2=5/4L`

The final pressure is

`P_2=V_1/V_2*P_1`

`=(11/6)/(5/4)*9`

`=13.2kPa`