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

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

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Answer 1 · 2017-07-28T05:22:41

Around $64 g/mol$ $"64 g/mol"$ .

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

$v_{R M S} = \sqrt{\frac{3 R T}{M}}$ $v_(RMS) = sqrt((3RT)/M)$ ,

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

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

$\frac{v_{R M S, B}}{v_{R M S, A}} = \frac{z_{B}}{z_{A}} = \sqrt{\frac{M_{A}}{M_{B}}}$ $(v_(RMS,B))/(v_(RMS,A)) = color(green)(z_B/z_A = sqrt(M_A/M_B))$

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 $g/mol$ $"g/mol"$ and it won't matter.)

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

Let's check mathematically.

$\frac{z_{B}}{z_{A}} = \sqrt{\frac{M_{A}}{M_{B}}}$ $z_B/z_A = sqrt(M_A/M_B)$

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

$\Rightarrow \frac{1}{4} = \sqrt{\frac{4.0026}{M_{B}}}$ $=> 1/4 = sqrt(4.0026/M_B)$

$\Rightarrow \frac{1}{16} = \frac{4.0026}{M_{B}}$ $=> 1/16 = 4.0026/M_B$

$\Rightarrow M_{B} = 4.0026 \times 16 \approx 64 g/mol$ $=> color(blue)(M_B) = 4.0026 xx 16 ~~ color(blue)("64 g/mol")$

So the unknown gas has about $16$ $bb16$ times higher of a molar mass.

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

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