The temperature of a given mass of air was changed from `15^(@)C` to `-15^(@)C`. If the pressure remains unchanged and the initial volume was `100 mL`, what should be the final volume?

Explanation

Given
`T_(1)=15^(@)C` or `15+273.15 K=288.15 K`
`V_(1)=100 mL`
`T_(2)= -15^(@)C=273.15-15=258.15 K`
`V_(2)=?`
We know `(V_(1))/(T_(1))=(V_(2))/(T_(2))`
` :. V_(2)=(V_(1))/(T_(1))xxT_(2)=(100xx258.15)/(288.15)=89.58 mL`