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Dalton's Law of Partial Pressures

A mixture of ...

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A mixture of $H_{2} O$ , $C O_{2}$ and $N_{2}$ are trapped in a glass apparatus with a volume of $0.731$ ml. The pressure of total mixture is $1.74$ mm of $H g$ at $23^{o} C$ . The sample is then transferred to a bulb which is in contact with dry ice ( ${- 75}^{o} C$ ) so that $H_{2} O (v)$ gets frozen. When the sample returned to normal temperature, pressure is reduced to $1.32$ mm of $H g .$ The sample is then transferred to a bulb which is in contact with liquid $N_{2} (- 95^{o} C)$ to freeze $C O_{2}$ . The sample contains $2.1 \times 10^{- x}$ moles of $N_{2}$ , which has a pressure of $0.53$ mm of $H g$ . Find the value of $x$ .

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Solution

Volume of container = $0.731$ ml

Temperature $= 23 + 273 = 296$ K

Initial pressure $= P_{H_{2} O} + P_{C O_{2}} + P_{N_{2}} = 1.74$ mm _________(i)

At ${- 75}^{o}$ $H_{2} O$ is frozen out and the mixture on returning at $23^{o}$ has pressure due to $C O_{2}$ and $N_{2}$ .

$P = P_{C O_{2}} + P_{N_{2}} = 1.32$ mm ____________(ii)

At ${- 95}^{o}$ $C O_{2}$ is also frozen out and the mixture on returning at $23^{o}$ has pressure due to $N_{2}$ alone.

$P_{N_{2}} = 0.53$ mm ____________(iii)

By Eqs (ii) and (iii), $P_{C O_{2}} = 1.32 - 0.53 = 0.79$ mm

By Eqs (i) and (ii), $P_{H_{2} O} = 1.74 - 1.32 = 0.42$ mm

Now using $P V = n R T$ for each gas separately

For $N_{2}$ ; $n =$ $\frac{P V}{R T}$ $= \frac{0.53 \times 0.731}{760 \times 0.0821 \times 296 \times 1000}$ $= 2.1 \times 10^{- 8}$

For $C O_{2}$ ; $n =$ $\frac{P V}{R T}$ $= \frac{0.79 \times 0.731}{760 \times 0.0821 \times 296 \times 1000}$ $= 3.1 \times 10^{- 8}$

For $H_{2} O$ ; $n =$ $\frac{P V}{R T}$ $= \frac{0.42 \times 0.731}{760 \times 0.0821 \times 296 \times 1000}$ $= 1.7 \times 10^{- 8}$

So, value of $x$ is $8$ .

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