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Question: A \(4\;:\;\;1\;\;molar\) mixture of \(He\) and \(C{H_4}\) gases is contained in a vessel at \(20\;ba...

A 4  :    1    molar4\;:\;\;1\;\;molar mixture of HeHe and CH4C{H_4} gases is contained in a vessel at 20  bar20\;bar pressure. Due to a hole in the vessel, the gas mixture leaks out. the ratio of number of HeHe to CH4C{H_4} in the mixture effusing out initially will be
A.8  :    18\;:\;\;1
B.1  :    81\;:\;\;8
C.1  :    41\;:\;\;4
D.4  :    14\;:\;\;1

Explanation

Solution

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
Rate1Rate2  =  M2M1\dfrac{{Rat{e_1}}}{{Rat{e_2}}}\; = \;\sqrt {\dfrac{{{M_2}}}{{{M_1}}}}
where Rate1Rat{e_1} is rate of effusion of first gas, Rate2Rat{e_2}is rate of effusion of second gas, M1{M_1}is the molar mass of the first gas and M2{M_2}is the molar mass of the second gas. Let’s write the given values
Number of moles of helium, nHe  =  45{{\text{n}}_{He}}\; = \;\dfrac{4}{5}
Number of moles of methane, nCH4  =  15{{\text{n}}_{C{H_4}}}\; = \;\dfrac{1}{5}
Total pressure, P  =  20  bar{\text{P}}\; = \;20\;bar
Since we have two gases, we need to calculate the pressure of both gases
  PHe  =  4  5  ×  20  =  16  bar\;{P_{He}}\; = \dfrac{{\;4\;}}{5}\; \times \;20\; = \;16\;bar
  PCH4  =  1  5  ×  20  =  4  bar\;{P_{C{H_4}}}\; = \dfrac{{\;1\;}}{5}\; \times \;20\; = \;4\;bar
now graham’s formula for rate calculation becomes
RateHeRateCH4  =  PHePCH4  MCH4MHe\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
RateHeRateCH4  =  164164\dfrac{{Rat{e_{He}}}}{{Rat{e_{C{H_4}}}}}\; = \;\dfrac{{16}}{4}\sqrt {\dfrac{{16}}{4}}
RateHeRateCH4  =  81\dfrac{{Rat{e_{He}}}}{{Rat{e_{C{H_4}}}}}\; = \;\dfrac{8}{1}
So, the ratio of gases effusing out initially is 8  :    18\;:\;\;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.