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.

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))`
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)/(5)n)=(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 become=V
As pressure remains constant, (vessel being open), `(V_(1))/(T_(1))=(V_(2))/(T_(2)),i.e., (2//5" V")/(300" K")=(V)/(T_(2))" or " T_(2)=750" K"`