What is the increase in volume, when the temperature of `600 mL` of air increases from `27^(@)C` to `47^(@)C` under constant pressure?

Explanation

Charles' law is applicable as the pressure and amount remains constant.
`(V_(1))/(T_(1))=(V_(2))/(T_(2))` or `V_(1)=(T_(1))/(T_(2))xxV_(2)`
`V_(1)=(320 K)/(300 K)xx600 mL=640 mL`
Increase in volume of air is
`640 mL- 600 mL=40 mL`