Are the effusion rates of carbon dioxide and propane the same?

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Are the effusion rates of carbon dioxide and propane the same?

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Answer 1 · 2017-04-20T19:09:49

They are close... not quite the same.

From the root-mean-square speeds, one could derive the rate of effusion.

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

The rate of effusion $z$ $z$ is directly proportional to the velocity. Thus, $\frac{z_{A}}{z_{B}} = \frac{v_{A}}{v_{B}}$ $z_A/z_B = v_(A)/(v_(B))$ , and we have Graham's law of effusion:

$\frac{z_{A}}{z_{B}} = \frac{\sqrt{\frac{3 R T}{M_{A}}}}{\sqrt{\frac{3 R T}{M_{B}}}}$ $bb(z_A/z_B) = sqrt((3RT)/M_A)/sqrt((3RT)/M_B)$

$= \sqrt{\frac{M_{B}}{M_{A}}}$ $= bb(sqrt(M_B/M_A))$

Hence, if we choose ${CO}_{2}$ $"CO"_2$ to be $A$ $A$ , we have that

$\frac{z_{C O_{2}}}{z_{C_{3} H_{8}}} = \sqrt{\frac{M_{C O_{2}}}{M_{C_{3} H_{8}}}}$ $z_(CO_2)/z_(C_3H_8) = sqrt(M_(CO_2)/M_(C_3H_8))$

$= \sqrt{\frac{44.01 g/mol}{44.1 g/mol}}$ $= sqrt("44.01 g/mol"/"44.1 g/mol")$

$= 0.9979$ $= 0.9979$

So they're almost the same, but propane effuses slightly faster.

$z_{C O_{2}} = 0.9979 z_{C_{3} H_{8}}$ $z_(CO_2) = 0.9979z_(C_3H_8)$

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Are the effusion rates of carbon dioxide and propane the same?

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