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

An open vessel at 27^(@)C is heated untill (3)/(5) parts of the air in

1.	An open vessel at 27^(@)C is heated untill (3)/(5) parts of the air in it has been expelled. Assuming that thevolume of the vessel remains constant, find the temperature to which the vessel has been heated.
Answer» Solution :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"`