Let's start it with a question :

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 .


$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.

  • $\begingroup$ Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of essential parts. // Optionally, here are formatting guides for texts and formulas/equations/expressions. $\endgroup$
    – Poutnik
    Oct 17 at 13:03
  • $\begingroup$ @Poutnik I've made edits accordingly. $\endgroup$
    – TPL
    Oct 17 at 13:31
  • $\begingroup$ You can also use $\frac{R_\text{He}}{R_\text{mix}}$ for $\frac{R_\text{He}}{R_\text{mix}}$ or $\dfrac{R_\text{He}}{R_\text{mix}}$ for $\dfrac{R_\text{He}}{R_\text{mix}}$ $\endgroup$
    – Poutnik
    Oct 17 at 13:50

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.