For glass `gamma_(g) = 3 alpha_(g) = 3 xx 9 xx 10^(-6)`
` = 27 xx 10^(-6) .^(@)C^(-1)`
Volume of air in the flask, ` V =1 litre = 1000 cm^(3)`
Let `V_(m)` be the volume of mercury in the flask and `V_(a)` be the volume of air in the flask.
Volume of gas flask = volume of mercury + volume of air
`V =V_(m) + V+(a)`
As `V_(a) ` remains constant, when temperature is raised by `Delta T` , so
` Delta V = Delta V_(m)`
or, `V gamma_(g) Delta T = V_(m) gamma_(m) DeltaT`
` or V_(m) = (gamma_(g)V)/(gamma_m) = ((27 xx 10^(-6)) xx 1000)/(1.8 xx 10^(-4)) = 150 cm^(3)`.