Analyzing Effusion Rates in Gas Mixtures: A Comparative Approach to He and CH4 and problems

Question

Let's start it with a question :

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Suppose there is a mixture of $\ce{He}$ and $\ce{CH4}$ Gas in a container of volume $V$ and a certain temperature with a small hole at a wall of the container, the ratio of moles of $\ce{He}$ to $\ce{CH4}$ is $3:1$. What would be the ratio of rate of effusion of helium( $R_He $) to the rate of effusion of the whole gaseous mixture ($R_mi_x)$) ? ( assuming ideal gases and Molar Mass of the gaseous mixture (avg) is 7 g/mol)

Although it can be resolved by applying the rules on He gas and the gaseous mixture directly, I choose to solve it using a different method.

So, We know that

$(R_A/R_B) = (n_a /n_b)$ $\sqrt{M_B/M_A} $

where molar masses $(M_A and M_B)$ and the number of moles $(n_A and n_B)$ of each gas .

So

$R_\ce{He}/ R_\ce{CH4} = 6/1 $

Now, Let at the initial moment,

$R_\ce{He}$ = $6x$ moles/sec and

$R_\ce{CH4}$ = $x$ moles/sec ( where x is any +ve number)

Now, in a very small time interval "$dt$",

Mass effused by $\ce{He}$ = $4 (6x)(dt)$

Mass effused by $\ce{CH4}$ = $16(x)(dt)$

$\ce{Total mass effused = 4(6x)(dt) + 16(x)(dt) = 40x (dt) gm}$

and Molar Mass of the gaseous mixture(avg) (let's call it Gas B) is 7 g/mol

so The moles of the Gas B (gaseous mixture) effused in dt are simply = $40xdt/7$

Therefore, Rate of Effusion of the gaseous mixture must be $40x/7$ moles/sec

and Hence,

$R_\ce{He}/R_\ce{mix} = 6x/(40x/7) = 42/40$

right ?

but this answer doesn't matches with the given answer key :

$R_\ce{He}/R_\ce{mix} = \sqrt{7}$ * $(3/8)$ and which is the correct answer

I get what they did, which makes sense (they directly applied the ratio law to $\ce{He}$ and the mixture) but My doubt is what's wrong with solving with above approach ? Am I missing something ?

I'd value any response that attempts to increase my understanding in this.

Answer 1

The supposedly "correct" answer is actually wrong. The answer key claims that the helium effuses at $3\sqrt{7}/8 = 0.99$ times the rate of the overall mixture. Applying the same logic to methane, we would calculate that methane effuses at a rate of $\sqrt{7}/16 = 0.17$ times the rate of the overall mixture. This is nonsense. These are the only two gases in the mixture, so the rates of effusion should sum to $1.0$ times the rate of the mixture. But $0.99 + 0.17 > 1.0$, so this method can't be valid.

When there are nonlinear effects (for example, in diffusion or effusion), don't expect a mixture to behave like a simple weighted average of its components.

Your proposed answer is also wrong, though much of your reasoning is correct. You made an error with the units. You calculated the helium effuses at a rate of $6x\ \mathrm{mol/s}$. The overall mixture effused at a rate of $(40/7)x\ \mathrm{g/s}$. You then divided the first number by the second, giving an answer in mol/g, which is not the pure molar ratio you're looking for.

The problem is actually quite a bit simpler than that. You identified that in this mixture, the molar rate of effusion of helium is 6 times that of methane. In other words, for every 7 molecules effusing, 6 are helium and 1 is methane. Therefore, the molar rate of effusion of helium is 6/7 that of the overall mixture.