Question

A $4\;:\;\;1\;\;molar$ mixture of $He$ and $C{H_4}$ gases is contained in a vessel at $20\;bar$ pressure. Due to a hole in the vessel, the gas mixture leaks out. the ratio of number of $He$ to $C{H_4}$ in the mixture effusing out initially will be
A.$8\;:\;\;1$
B.$1\;:\;\;8$
C.$1\;:\;\;4$
D.$4\;:\;\;1$

Answer 1

Hint: We are given two components of gas which are undergoing effusion. Let us first understand what effusion is. Effusion is a process in which the gas leaks or air escapes through a hole whose diameter is less than the mean free path of molecules. Collisions are negligible so all the molecules pass through the hole comfortably.

Complete answer:
We will use Graham's Law for the calculation of the required ratio. Let’s see what graham law says at constant temperature and pressure atoms having less molar mass will effuse faster than the atoms having more molar mass in other words rate of effusion is inversely proportional to the square root of the molar mass of the effusing gas. The formula is written as
$\dfrac{{Rat{e_1}}}{{Rat{e_2}}}\; = \;\sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
where $Rat{e_1}$ is rate of effusion of first gas, $Rat{e_2}$is rate of effusion of second gas, ${M_1}$is the molar mass of the first gas and ${M_2}$is the molar mass of the second gas. Let’s write the given values
Number of moles of helium, ${{\text{n}}_{He}}\; = \;\dfrac{4}{5}$
Number of moles of methane, ${{\text{n}}_{C{H_4}}}\; = \;\dfrac{1}{5}$
Total pressure, ${\text{P}}\; = \;20\;bar$
 Since we have two gases, we need to calculate the pressure of both gases
$\;{P_{He}}\; = \dfrac{{\;4\;}}{5}\; \times \;20\; = \;16\;bar$
$\;{P_{C{H_4}}}\; = \dfrac{{\;1\;}}{5}\; \times \;20\; = \;4\;bar$
now graham’s formula for rate calculation becomes
$\dfrac{{Rat{e_{He}}}}{{Rat{e_{C{H_4}}}}}\; = \;\dfrac{{{P_{He}}}}{{{P_{C{H_4}}}}}\;\sqrt {\dfrac{{{M_{C{H_4}}}}}{{{M_{He}}}}} $
Putting values in above formula we get
$\dfrac{{Rat{e_{He}}}}{{Rat{e_{C{H_4}}}}}\; = \;\dfrac{{16}}{4}\sqrt {\dfrac{{16}}{4}} $
$\dfrac{{Rat{e_{He}}}}{{Rat{e_{C{H_4}}}}}\; = \;\dfrac{8}{1}$
So, the ratio of gases effusing out initially is $8\;:\;\;1$.
Hence option (A) is the correct answer.

Note:
Graham’s formula for rate calculation changes when there is pressure mentioned as both gases will have some pressure. so, the reciprocal of the square root of mass is multiplied by the pressure of the gas to obtain the rate of effusion.