Question

If `7/6 L` of a gas at room temperature exerts a pressure of `6 kPa` on its container, what pressure will the gas exert if the container's volume changes to `2/3 L`?

Answer 1

10.5 k Pa

Explanation:

As volume reduction process doesn't allow any change in mass or temperature of the gas,we can apply Boyle's law i.e `PV = constant`
Hence we can write

`P1 V1 = P2 V2` (`P2` is the pressure exerted by the gas on the container after volume reduction)
Or, `(7/6)6` = `P2 *(2/3)`
Or, `P2` = `21/2 k Pa` or `10.5 k Pa`