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Question

Chemistry Question on Ideal gas equation

A vessel contains equal volumes of SO2\text{S}{{\text{O}}_{\text{2}}} and CH4.\text{C}{{\text{H}}_{4}}. Through a small hole the gases effused into vacuum. After 200 seconds the total volume is reduced to half. What is the ratio of SO2\text{S}{{\text{O}}_{\text{2}}} and CH4\text{C}{{\text{H}}_{\text{4}}} remaining in the vessel?

A

1:02

B

2:01

C

1:01

D

1:04

Answer

2:01

Explanation

Solution

According to Grahams law of diffusion, rSO2rCH4=MCH4MSO2\frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{{{M}_{C{{H}_{4}}}}}{{{M}_{S{{O}_{2}}}}}} rSO2rCH4=1664=14=12\frac{{{r}_{S{{O}_{2}}}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\frac{16}{64}}=\sqrt{\frac{1}{4}}=\frac{1}{2} rSO2:rCH4=1:2{{r}_{S{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=1:2 SO2\text{S}{{\text{O}}_{\text{2}}} and CH4\text{C}{{\text{H}}_{4}} effused in the ratio 1 : 2. So, the amount remains in the vessel will be in the ratio as SO2:CH4=2:1S{{O}_{2}}:C{{H}_{4}}=2:1