Chemistry - Ideal Gas Question with Solution | TestHub

Given that,$\mathrm{V}=2\mathrm{L},{\mathrm{T}}_1=50+273=323\mathrm{K},{\mathrm{P}}_1=3\mathrm{a}\mathrm{t}\mathrm{m}$
On heating$\mathrm{V}=2\mathrm{L},{\mathrm{T}}_2=200+273=473,{\mathrm{P}}_2=?$
Using P-T law,$\frac{{\mathrm{P}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2}{{\mathrm{T}}_2}$
${\mathrm{P}}_2=\frac{3\times 473}{323}=4.39$atm
and$\mathrm{n}=\frac{\mathrm{P}\mathrm{V}}{\mathrm{R}\mathrm{T}}=\frac{3\times 2}{0.0821\times 323}=0.226$
Now value is opened till the pressure is maintained at 1 atm.
Thus, at constant V and T,$\mathrm{P}\propto \mathrm{n}$
So$4.39\propto 0.226$
So$1\propto {\mathrm{n}}_{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}$
So$\mathrm{n}=\frac{0.226}{4.39}=0.052$
So moles escaped out$=0.226-0.052=0.174$
So fraction of moles escaped out$=\frac{0.174}{0.226}=0.77$