An open vessel contains air at `27^(@)C`. At what temperature should the vessel be heated so that 1/4h of air escapes from the vessel? Assume that volume of the vessel remains same on heating.

Explanation

Initial temperature of air`=(273+27)K=300K`. Let the amount of air in the vessel at `27^(@)C=x` mol. Suppose, the vessel is heated to a temperature of `T_(1)K` so that 1/4 th of air escapes from the vessel.
So, the amount of air in the vessel at `T_(1)K=x-(x)/(4)=(3)/(4)x`
mol
since, the vessel is open, the pressure of air in the vessel either at 300K or at `T_(1)K` is equal to atmospheric pressure, i.e., 1 atm. again, volume of the vessel does not changes because of heating.
Therefore, at temperature 300K,
`x xx 0.0821xx300=(3)/(4)x xx0.0821xxT_(1) or T_(1)=400K`
`therefore t=(400-273)^(@)C=127^(@)C`.