Question

If `7/4 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 `13/12 L`?

Answer 1

The pressure is `=25.85kPa`

Explanation:

We apply Boyle's Law

`P_1V_1=P_2V_2`

`P_1=7/4L`

`V_1=16kPa`

`V_2=13/12L`

`P_2=(P_1V_1)/V_2`

`=(7/4*16)/(13/12)=28/13*12=25.85kPa`