Question

A 4:1 molar mixture of $He$ and $C{H}_4$ is contained in a vessel at 20 $bar$ pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition ratio of the mixture effusing out initially?

• A. $\frac{r_{He}}{r_{C{H}_4}}=8:1$.
• B. $\frac{r_{He}}{r_{C{H}_4}}=1:8$.
• C. $\frac{r_{He}}{r_{C{H}_4}}=1:4$.
• D. None of these

Answer 1

Answer: (A)

$\frac{r_{He}}{r_{C{H}_4}}=8:1$.

$Pressure=20bar$

$Molarratio=4:1\left(HeandC{H}_4\right)$

$P_{He}=\frac{4}{5}\times 20=16bar$

$P_{C{H}_4}=\frac{1}{5}\times 20=4bar$

$\frac{r_1}{r_2}=\frac{P_1}{P_2}\times \sqrt{\frac{M_2}{M_1}}$

$\frac{r_{He}}{r_{C{H}_4}}=\frac{P_{He}}{P_{C{H}_4}}\times \sqrt{\frac{M_{C{H}_4}}{M_{He}}}=\frac{16}{4}\times \sqrt{\frac{16}{4}}=8$

Therefore, the ratio of moles of helium and methane coming out initially is 8:1.