An open vessel at `27^(@)C` is heated until `3//5` of the air in it is expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated.

Answer

750

Explanation

As the vessel is open , pressure and volume remain constant . Thus if `n_(1)` moles are present at `T_(1) and n_(2)` moles are present at `T_(2)` , we can write `PV = n_(1)RT_(1) ` and also `PV = n_(2)RT_(2)`
Hence , `n_(1)RT_(1)=n_(2)RT_(2)`
or `n_(1)T_(1)=n_(2)T_(2)`
or `(n_(1))/(n_(2)) = (T_(2))/(T_(1)) " "...(i)`
Suppose the no. of moles of air originally present = n
After heating no . of moles of air expelled = `3/5 n `
` :. ` No. of moles left after heating = `n - 3/5 n = 2/5 n `
Thus `n_(1) = n , T_(1) = 300 K , n_(2) = 2/5 n , T_(2) = ?`
Substituting in eqn . (i) , we get
`n/(2/5n) = (T_(2))/300 or 5/2 = (T_(2))/300 or T_(2) = 750 K `
Alternatively , suppose the volume of the vessel = V , i.e., Volume of air initially at `27^(@) C = V `
Volume of air expelled = `3/5 V `
` :.` Volume of air left at `27^(@)C = 2/5 V `
However , on heating to `T^(@)K ` , it would beome = V
As pressure remains constant , (vessel being open ) ,
`(V_(1))/(T_(1))=(V_(2))/(T_(2)),i.e., (2//5V)/(300K) = V/(T_(2))`
or `T_(2) = 750 K `