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Dalton's Law of Partial Pressures

In a 13L ve...

Question

In a $13 L$ vessel initially following reaction occur $C (s) + S_{2} (g) ⇌ {C S}_{2} (g)$ by $12 g$ $C$ , $64 g$ $S_{2}$ , $76 g$ ${C S}_{2}$ at $1027^{o} C$ temperature then total pressure is:

200 R

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158 R

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100 R

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79 R

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Solution

The correct option is A $200 R$
$C (s) + S_{2} (g) ⇌ C S_{2} (g)$
Number of moles of $C = \frac{12 g}{12 g} = 1$
Number of moles of $S O_{2} = \frac{64}{64} = 1$
Number of moles of $C S_{2} = \frac{76}{76} = 1$
Total number of gaseous moles= $1 (S_{2}) + 1 (C S_{2}) = 2$ moles
From ideal gas equation, $P V = n R T$
$P = \frac{n R T}{V}$
$= \frac{2 \times R \times (273 + 1027)}{13}$
$\Rightarrow \frac{2 R (1300)}{13}$
$= 200 R$

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