If a 2 litre flask on $N_{2}$ at $20^{0} C$ and 70 cm of P is connected with a 3 litre of another flask of $O_{2}$ at the same temperature and 100 cm of 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.

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Solution

For this type of problem the total moles of each is conserved
$n_{1} + n_{2} = n_{f}$ $n_{1} =$ initial mole of $N_{2}$
$\Rightarrow \frac{P_{1} V_{1}}{R T_{1}} + \frac{P_{2} V_{2}}{R T_{2}} = \frac{P_{f} V_{f}}{R T_{f}}$ $n_{2} =$ initial mole of $O_{2}$
But $T_{1} = T_{2} = T_{f}$ $n_{f} =$ total moles of mixtur $(O_{2} + N_{2})$
Therefore, ${∵ n = \frac{P V}{R T}}$
$P_{1} V_{1} + P_{2} V_{2} = P_{f} (V_{f})$ $P_{1} =$ Pressure of flask of $H_{2}$
$\Rightarrow P_{f} = \frac{P_{1} V_{1} + P_{2} V_{2}}{V_{f}}$ $P_{2} =$ Pressure of flask of $O_{2}$
$V_{1} =$ initial volume of flask $N_{2}$
$= \frac{70 c m \times 2 + 100 \times 3}{(2 + 3) l t}$ $V_{2} =$ Volume of flask $O_{2}$
$= \frac{440 l t}{5 l t} c m$ of $P$
$= 88 c m$ of $P$
final pressure $= 88 c m$ of $P$ (ans)
$n_{1} = \frac{P_{1} V_{1}}{293} = \frac{70 \times 2}{293} = 0.48 \frac{c m o f P \times l t}{K} n_{2} = \frac{P_{2} V_{2}}{293} = \frac{300}{293} = 1.02 \frac{c m o f P \times l t}{K}$
% of $N_{2} = \frac{n_{1}}{n_{1} + n_{2}} \times 100 = \frac{0.48}{1.503} \times 100 = 0.319 \times 100$
$= 31.9$ %(ans)
% of $O_{2} = \frac{n_{2}}{n_{1} + n_{2}} \times 100 = \frac{1.02}{1.503} \times 100 = 67.86$ % (ans)

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