Question

A 4:1 mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of the mixture effusing out initially is:

• A. 8 : 1
• B. 8 : 3
• C. 4 : 1
• D. 1 : 1

Answer 1

The correct option is A 8 : 1

Pressure of helium = 8 bar $C{H}_4=2\text{bar}$
According to Graham's law
$\frac{r_1}{r_2}=\frac{P_1}{P_2}\sqrt{\frac{M_2}{M_1}}$
$\frac{r_{He}}{r_{C{H}_4}}=\frac{P_{He}}{P_{C{H}_4}}\sqrt{\frac{M_{C{H}_4}}{M_{He}}}$
$\frac{r_{He}}{r_{C{H}_4}}=\frac{8}{2}\sqrt{\frac{16}{4}}=\frac{8}{1}=8:1$