Question

If `19 L` of a gas at room temperature exerts a pressure of `2 kPa` on its container, what pressure will the gas exert if the container's volume changes to `12 L`?

Answer 1

`P_2=3,17 kPa`

Explanation:

`P_1=2kPa`
`V_1=19L`
`V_2=12L`
`P_2=?`

`"formula :"`

`P_1*V_1=P_2*V_2" Boyle's law"`

`2*19=P_2*12`

`P_2=(2*19)/12`

`P_2=3,17 kPa`