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First Law of Thermodynamics

SO2 and CH4 a...

Question

$S O_{2}$ and $C H_{4}$ are introduced in a vessel in the molar ratio $1 : 2$ . The ratio of molecules of two gases present in the container when their rate of effusion becomes equal is :

A

1 : 2

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B

2 : 1

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C

4 : 1

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D

1 : 4

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Solution

The correct option is B $2 : 1$
Rate of effusion for a gas $(r) \propto \frac{P}{\sqrt{M}}$ , where $P$ is pressure and $M$ is molar mass of gas.
Since the rate of effusion for both gases is equal so,
$\frac{P_{1}}{\sqrt{M_{1}}} = \frac{P_{2}}{\sqrt{M_{2}}}$
$\Rightarrow \frac{P_{1}}{P_{2}} = \frac{\sqrt{M_{1}}}{\sqrt{M_{2}}}$
During effusion all conditions are identical so $P r e s s u r e o f g a s (P) \propto M o l e s o f g a s (n) \propto M o l e c u l e s o f g a s (N)$
$= \frac{N_{1}}{N_{2}} \Rightarrow \frac{N_{S O_{2}}}{N_{C H_{4}}} = \sqrt{\frac{M_{S O_{2}}}{M_{C H_{4}}}} = \sqrt{\frac{64}{16}} = \frac{2}{1}$

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