Question

A flask of $2d{m}^3$ capacity contains $O_2$ at $101.325kPa$ and $300K$. The gas pressure is reduced to $0.1Pa$. Assuming ideal behaviour, answer the following:
i) What will be the volume of the gas which is left behind?
ii) What amount of $O_2$ and $t$ he corresponding number of molecules are left behind in the flask?
iii) In now $2g$ of $N_2$ is introduced, what will be the pressure of the flask

Answer 1

Given $V_1=2d{m}^3$ $P_t=101.325k{P}_a,P_2=0.1P_a,T=300K$

(i) Volume of $O_2$ left behind will be the same i.e $2d{m}^3$

(ii) The amount of $O_2$ left behind

$n=\frac{P_2{V}_1}{RT}=\frac{0.1\times {10}^{-3}k{P}_a\times 2d{m}^3}{8.314\times k{P}_{a}d{m}^3/k/mol\times 300}$

$=8.019\times {10}^{-8}mol$

(iii) $2g$ of $N_2=\frac{1}{14}mol$

Total amount of gases in the flask

$=\frac{1}{14}+8.125\times {10}^{-8}\simeq \frac{1}{14}mol$

Thus the pressure of the flask is given by,

$P=\frac{nRT}{V}$

$=[\left(\frac{1}{14}\right)\times (8.315k{P}_{a}d{m}^3{k}^{-1}mo{l}^{-1}\times 300K\left)\right]/2d{m}^3$

$=89.08k{P}_a$ .