500 mL of nitrogen at `27^(@)C` is cooled to `-5^(@)C` at the same pressure. The new volume becomes

Explanation

Initial conditions
`V_(1)=500ml`
`T_(1)=273+27=300K` Final conditions `V_(2)=?`
`T_(2)=273+(-5)= 268K`
By applying Charle’s law, `(V_(1))/(T_(1))=(V_(2))/(T_(2))`
`(500)/(300)=(V_(2))/(268)`
`V_(2)= (500)/(300)xx268 = 446.66 ml`