Question

If `7/5 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 `13/12 L`?

Answer 1

The new pressure is `=20.7kPa`

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=7/5L`

The final volume is `V_2=13/12L`

The final pressure is

`P_2=V_1/V_2*P_1`

`=(7/5)/(13/12)*16=20.7kPa`