Question

If a $2$ litre flask of $N_2$ at ${20}^o$C and $70$cm P is connected with a $3$ litre of another flask of $O_2$ at the same temperature and $100$ cm P. What will be the final pressure after the gases have thoroughly mixed at the same temperature as before? Also calculate the mole$\%$ of each gas in the resulting mixture. The volume of stopcock may be neglected.

Answer 1

Solution:-

$1atm=\text{76 cm Hg}$

$\Rightarrow \text{1 cm Hg}=\frac{1}{76}atm$

$\therefore P_{N_2}=\frac{70}{76}atm\&P_{O_2}=\frac{100}{76}atm$

${\left(P_f\right)}_{mix}=\frac{P_{N_2}{V}_{N_2}+P_{O_2}{V}_{O_2}}{V_{N_2}+V_{O_2}}=\frac{70\times 2+100\times 3}{5}=88\text{cm Hg}=1.157atm$

$\therefore {\left(P_f\right)}_{mix}=1.157atm$

For mole $\%$-

As we know that,

$PV=nRT$

$\Rightarrow n=\frac{PV}{RT}$

For $N_2$-

$T=293K$

$P=\frac{70}{76}atm$

$V=2$

$R=0.821$

$\therefore n=\frac{\frac{70}{76}\times 2}{0.821\times 293}=0.00766\text{moles}$

For $O_2$-

$T=293K$

$P=\frac{100}{76}atm$

$V=3$

$R=0.821$

$\therefore n=\frac{\frac{100}{76}\times 3}{0.821\times 293}=0.0164\text{moles}$

$\therefore \text{mole}\%\text{of}{N}_2=\frac{\text{moles of}{N}_2}{\text{total no. of moles}}\times 100=\frac{0.00766}{0.00766+0.0164}\times 100=31.84\%$

$\text{mole}\%\text{of}{O}_2=\frac{\text{moles of}{O}_2}{\text{total no. of moles}}\times 100=\frac{0.0164}{0.00766+0.0164}\times 100=68.16\%$