A $$2\,L$$  vessel is filled with air at $$50{\,^ \circ }C$$  and a pressure of $$3\,atm.$$  The temperature is now raised to $$200{\,^ \circ }C.$$   A valve is now opened so that the pressure inside drops to one atm. What will be the fraction of the total number of moles inside escaped on opening the valve ? ( Assume no change in the volume of the container. )

Solution :
$${\text{Given that,}}V = 2\,L,$$     $${T_1} = 50 + 273 = 323\,K,$$     $${P_1} = 3\,atm$$
$${\text{On heating}}\,\,V = 2L,$$     $${T_2} = 200 + 273 = 473,{P_2} = ?$$
$${\text{Using }}P - T{\text{ law,}}\,\,\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}$$
$$\eqalign{ & {P_2} = \frac{{3 \times 473}}{{323}} \cr & \,\,\,\,\,\,\,\, = 4.39\,atm \cr} $$
$$\eqalign{ & {\text{and}}\,\,n = \frac{{PV}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3 \times 2}}{{0.0821 \times 323}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.226 \cr} $$
Now valve is opened till the pressure is maintained at $$1\,atm.$$
Thus, at constant $$V$$ and $$T,P \propto n$$
$$\eqalign{ & \therefore {P_2} \propto n\,\,\,\,\therefore 1 \propto {n_{{\text{left}}}} \cr & \therefore \frac{{{P_2}}}{1} = \frac{n}{{{n_{{\text{left}}}}}}\,\,\,\therefore \,\,{n_{{\text{left}}}} = \frac{{0.226}}{{4.39}} = 0.052 \cr} $$
$$\therefore {\text{Moles escaped out }}$$    $$ = 0.226 - 0.052 = 0.174$$
$$\therefore {\text{Fraction of moles escaped out}}$$       $$ = \frac{{0.174}}{{0.226}} = 0.77$$