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Gay Lussac's Law

A container h...

Question

A container holds $22.4$ litre of a gas at $1$ atmospheric pressure and at $0^{\circ} C$ . The gas consists of a mixture of argon, oxygen and sulphur dioxide in which:
(a) Partial pressure of $S O_{2} =$ (Partial pressure $O_{2}) +$ (Partial pressure of $A r$ ).
(b) Partial pressure of $O_{2} = 2 \times$ partial pressure of $A r$ .
Calculate the density of the gas mixture under these conditions.

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Solution

Volume of gas mixture $= 22.4 L$
Total pressure = $1 a t m$
$P a r t i a l P r e s s u r e_{(S O_{2})} = P a r t i a l P r e s s u r e_{(O_{2})} + P a r t i a l P r e s s u r e_{(A r)}$
$= 2 P a r t i a l P r e s s u r e_{(A r)} + P a r t i a l P r e s s u r e_{(A r)}$
$T o t a l P r e s s u r e = 6 P a r t i a l P r e s s u r e_{(A r)}$
Mole Fraction:
$χ_{A r} = \frac{1}{6}$
$χ_{O_{2}} = \frac{1}{3}$
$χ_{S O_{2}} = \frac{1}{2}$

For total number of moles:
$P \times V = n \times R \times T$
where, $n = \frac{m}{M}$
$n_{T} = 1$

For $A r m = 6.4 g m$
For $O_{2} m = 10.67 g m$
For $S O_{2} m = 32 g m$

Total mass $= 6.4 + 10.67 + 32$
$= 49.07$

$d e n s i t y = \frac{m a s s}{v o l u m e}$
$= \frac{49.07}{22.4}$

$= 2$

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