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If we have a mixture of (ideal) gases, what is the rate of diffusion of the total mixture? Also, what are the rates of diffusion of the individual components?

My teacher said it was inversely proportional to the square root of the average molar mass, and that individual rates were proportional to the partial pressure divided by the sq. root of molar mass.

However, in this case, the rates of each individual component don't add up to the total rate.

i.e. If gases A and B make up a mixture and R is the rate of diffusion then:

$$ R_{total}=R_{A} + R_{b} $$ $$ \frac{P_{total}}{\sqrt{M_{avg}}}=\frac{P_{A}}{\sqrt{M_{A}}}+\frac{P_{B}}{\sqrt{M_{B}}}$$

Which clearly isn't true.

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  • $\begingroup$ Why they cannot be added up? They are ideal gases, They do not interact with each other. You just have to work with the differences of their partial pressures on each side. $\endgroup$
    – Poutnik
    Commented Apr 25 at 14:32
  • $\begingroup$ I've clarified my question $\endgroup$ Commented Apr 25 at 16:48
  • $\begingroup$ $$R_\text{tot} = R_\text{A} + R_\text{B} = C \cdot \left(\frac{\Delta p_\text{A} }{ \sqrt{M_\text{A}}} + \frac{\Delta p_\text{B} }{ \sqrt{M_\text{B}}} \right)$$ $\endgroup$
    – Poutnik
    Commented Apr 25 at 17:27
  • $\begingroup$ Note that there is the arithmetic mean, geometric mean and harmonic mean. The arithmetic mean, often meant implicitly when talking about an average, is not always applicable. $\endgroup$
    – Poutnik
    Commented Apr 26 at 7:58

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It depends on the definition of average (strictly speaking on the choice of mean), as there are many types. The expression provided can be rewritten in the form $$M_{avg}^{-1/2} = \chi_A M_A^{-1/2} + \chi_B M_B^{-1/2}$$ which is not an arithmetic mean of the form $$M_{avg} = \chi_A M_A + \chi_B M_B$$ but has the general form $$\textrm{prop}_{avg} = \chi_A \textrm{prop}_A + \chi_B \textrm{prop}_B$$ and allows the definition $$M_{avg} = (\chi_A M_A^{-1/2} + \chi_B M_B^{-1/2})^{-2}$$ which is the square of the harmonic mean of $M^{1/2}$.

The expression in the original post computes an arithmetic mean rate, but requires definition of an average mass using the above expression, which is equivalent to evaluating the harmonic mean of the square root of the mass.

Harmonic means arise naturally during computation of mean flow rates. A classic example is computation of an effective resistance for a circuit consisting of two parallel resistances:

$$I_{avg} = I_1+I_2\\ \implies \frac{V}{R_{avg}} = \frac{V}{R_1}+\frac{V}{R_2} \\ \implies R_{avg} = \left({R_1}^{-1}+{R_2}^{-1}\right)^{-1} \\\\$$

The inertial resistance to effusion is proportional to the square root of the mass.

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  • $\begingroup$ So does this mean that what my teacher said for components of a mixture is true but what he said for the entire mixture (he said it was the arithmetic mean) is false? $\endgroup$ Commented Apr 25 at 17:55
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    $\begingroup$ No, it's true (well, depends on exactly what was claimed). The expression you wrote is an arithmetic mean of the rate, but requires definition of an average mass using the expression in the answer, which is equivalent to evaluating the harmonic mean of the square root of the mass. $\endgroup$
    – Buck Thorn
    Commented Apr 25 at 18:01

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