A sample of a gas occupies `100 dm^(3)` at `1` bar pressure and at `27^(o)C`. If the volume of the gas is reduced to `5 dm^(3)` at the same temperature, what additional pressure must be be applied?

Explanation

From the given data : `P_(1) = 1 "bar"` , `P_(2) = ?`
`V_(1) = 100 dm^(3)` , `V_(2) = 2 dm^(3)`
Since temperature is constant, Boyle's Law can be applied
`P_(1)V_(1) = P_(2)P_(2)` or `P_(2) = (P_(1)V_(1))/(V_(2))`
`P_(2) = ((1 "bar") xx (10 dm^(3)))/((5 dm^(3))) = 20` bar
`:.` Additional pressure applied `= 20 - 1 = 19` bar