Question

If a 2 litre flask of $N_2$ at ${20}^0$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 throughly mixed at the same temperature as before? Also calculate the mole % of each gas in the resulting mixture. The volume stopcock may be neglected.

Answer 1

Number of mole of $N_2$ can be calculated as

$P_1=70cm$, $T=293K$

$V_1=2L$

$PV=nRT$

$1atm=760mm$ of $Hg=76cm$

$\Rightarrow 70cm=\frac{70}{76}atm$

$\Rightarrow n_1=\frac{\frac{70}{76}atm\times 2L}{0.0821atmL{K}^{-1}mo{l}^{-1}\times 293K}=\frac{140}{76RT}$ mole

$\Rightarrow n_1=0.0765$ moles

Number of mole of $O_2$ can be calculated as

$P_2=100cm=\frac{100}{76}atm$

$V_2=3L,T=293K$

$n_2=\frac{\frac{100atm}{76}\times 3L}{0.0821atmL{K}^{-1}\times 293Kmo{l}^{-1}}=\frac{300}{76RT}mol$

$n_2=0.164$ mole

After mixing,

Final moles, $n=n_1+n_2$

Final volume= $2L+3L=5L$

Final pressure= $P=P_1+P_2$

$\frac{n_{1}RT}{V}+\frac{n_{2}RT}{V}$

$\Rightarrow P=\frac{140}{76RT}\times \frac{RT}{5}+\frac{300}{76RT}\times \frac{RT}{5}$

$\Rightarrow P=\frac{140+300}{76\times 55}=1.157atm$

$\Rightarrow P=88cm\Rightarrow$ final pressure

Mole % of $N_2=\frac{n_1}{n}\times 100=\frac{140}{440}\times 100=31.82$ %

Mole % of $O_2=\frac{300}{440}\times 100=68.18$%