Question

If `9/7 L` of a gas at room temperature exerts a pressure of `12 kPa` on its container, what pressure will the gas exert if the container's volume changes to `8/5 L`?

Answer 1

The new pressure is `=9.64kPa`

Explanation:

Apply Boyle's Law

`P_1V_1=P_2V_2` at constant temperature

The initial pressure is `P_1=12kPa`

The initial volume is `V_1=9/7L`

The final volume is `V_2=8/5L`

The final pressure is

`P_2=V_1/V_2P_1=(9/7)/(8/5)*12=9.64kPa`