A gas occupies `0.6 dm^(3)` under a pressure of `0.92` bar. Find under what pressure, the volume of gas will be reduced by 20 precent of its original volume, temperature remaining constant.

Answer

`1.15 "bar"`

Explanation

`P_(1) = 0.92 "bar" , P_(2) = ?`
`V_(1) = 0.6 m^(3) , V_(2) = 0.6- (0.6 xx 20)/(100) = 0.6- 0.12 = 0.48 dm^(3)`.
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)) = ((0.92 "bar") xx (0.6 dm^(3)))/((0.48 dm^(3))) = 1.15 "bar"`