Helium is found to diffuse four times more rapidly than an unknown gas. What is the approximate molar mass of the unknown gas?

Jul 28, 2017

Around $\text{64 g/mol}$.

You can start from any expression for the speed. The RMS speed is a fine choice:

${v}_{R M S} = \sqrt{\frac{3 R T}{M}}$,

where $R$ and $T$ are from the ideal gas law, and $M$ is the molar mass of the gas in $\text{kg/mol}$.

The ratio of two speeds $v$ is directly proportional to the ratio of the gas diffusion rates $z$ (the $3 R T$ cancels out):

$\frac{{v}_{R M S , B}}{{v}_{R M S , A}} = \textcolor{g r e e n}{{z}_{B} / {z}_{A} = \sqrt{{M}_{A} / {M}_{B}}}$

which is Graham's law of diffusion; the gas with more mass per particle diffuses more slowly. (In this case, the molar mass can now be in $\text{g/mol}$ and it won't matter.)

Since helium is known to diffuse $4$ times as fast as unknown gas $B$, then off the top of my head, helium is probably ${4}^{2} = 16$ times as light (one-sixteenth the molar mass of $B$).

Let's check mathematically.

${z}_{B} / {z}_{A} = \sqrt{{M}_{A} / {M}_{B}}$

If we assign helium as $A$, then the unknown gas is $B$ and ${z}_{B} / {z}_{A} = \frac{1}{4}$.

$\implies \frac{1}{4} = \sqrt{\frac{4.0026}{M} _ B}$

$\implies \frac{1}{16} = \frac{4.0026}{M} _ B$

$\implies \textcolor{b l u e}{{M}_{B}} = 4.0026 \times 16 \approx \textcolor{b l u e}{\text{64 g/mol}}$

So the unknown gas has about $\boldsymbol{16}$ times higher of a molar mass.