Question

If `9/4 L` of a gas at room temperature exerts a pressure of `16 kPa` on its container, what pressure will the gas exert if the container's volume changes to `9/7 L`?

Answer 1

The pressure is `=28kPa`

Explanation:

We apply Boyle's Law

`P_1V_1=P_2V_2`

The initial pressure is `P_1=16kPa`

The initial volume is `V_1=9/4L`

The final volume is `V_2=9/7`

The final pressure is

`P_2=V_1/V_2*P_1=(9/4)/(9/7)*16`

`=7/4*16=28kPa`