A gas occupies a volume of `250 mL` at `0.98` bar of Hg and `25^(@)C`. What additionl pressure is required to reduce the volume of the gas to `200 mL` at the same temperature ?

Answer

`0.245 "bar"`

Explanation

`P_(1) = 0.98` bar , `P_(2) = ?`
`V_(1) = 250 mL , V_(2) = 200 mL`
Since temperature is constant, Boyles' Law is applicable
`P_(1)V_(1) = P_(2)V_(2)` or `P_(2) = (P_(1)V_(1))/(V_(2)) = ((0.98 "bar") xx (250 mL))/((200mL)) = 1.225 "bar"`
Additional pressure to be applied `= (1.225 - 0.98) = 0.245 "bar"`.