Question

If `19 L` of a gas at room temperature exerts a pressure of `4 kPa` on its container, what pressure will the gas exert if the container's volume changes to `7 L`?

Answer 1

`11*10^3 Pa`

Explanation:

When the containers volume changes to `7 L` from `19 L`
some work must be done hence from Boyle's law We can say that

`P_1V_1=P_2V_2`

where `PV="costant"`

so by substituting the values we get

`19*4*10^3=7*P_2`

`P_2=10.9*10^3~=11*10^3`