A vessel contains equimolar mixture of He and $C H_{4}$ gases. If a pinhole made in the vessel, then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Average molecular weight of gaseous mixture effused out at the beginning is 8

The average molecular weight of gaseous mixture remaining in the vessel increases with time

When the average molecular weight of gaseous mixture remained in the vessel crosses 12, then net rate of effusion of $C H_{4}$ is more than that of He

Rate of effusion of $C H_{4} a n d H e,$ do not change with time

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At beginning $\frac{r_{H e}}{r_{C H_{4}}} = \frac{P_{H e}}{P_{C H_{4}}} \sqrt{\frac{M_{C H_{4}}}{M_{H e}}} = \frac{1}{1} \sqrt{\frac{16}{4}} = \frac{2}{1}$
a) $M_{m i x} = 4 \times \frac{2}{3} + 16 \times \frac{1}{3} = 8$
b) Since lighter gas rate of effusion is more, average M.W of gas left in the vessel increases
c) $\frac{r H e}{r C H_{4}} = \frac{P_{H e}}{P_{C H_{4}}} \sqrt{\frac{M_{C H_{4}}}{M_{H e}}} \Rightarrow 1 = \frac{n_{H e}}{n_{C H_{4}}} \sqrt{\frac{16}{4}} \Rightarrow \frac{n_{H e}}{n_{C H_{4}}} = \frac{1}{2}$
Average M.W $= 4 \times \frac{1}{3} + 16 \times \frac{2}{3} = \frac{36}{3} = 12$