Question #133a5

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Question #133a5

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Answer 1 · 2016-02-24T13:28:15

${471 m s}^{- 1}$ $"471 m s"^(-1)$

Explanation:

The thing to remember about a gas' rate of effusion is that it's inversely proportional to the square root of its molar mass - this is known as Graham's Law

$rate of effusion \propto a \frac{1}{\sqrt{molar mass}}$ $color(blue)("rate of effusion " prop color(white)(a)1/sqrt("molar mass"))$

Effusion is simply the term used to describe how gas molecules escape through a small hole. This means that the rate of effusion will depend on how "massive" the gas molecules are.

More specifically, the fact that the rate of effusion is inversely proportional to the square root of the mass of its particles tells you that the heavier a gas particle is, the slower it will effuse, i.e. escape through a small hole in its container.

Similarly, lighter gas particles will effuse at a faster rate when compared with heavier gas particles.

So, look up the molar masses of neon, $Ne$ $"Ne"$ , and butane, $C_{4} H_{10}$ $"C"_4"H"_10$ and write the rate of effusion for each gas

$For Ne: {rate}_{N e} \propto \frac{1}{\sqrt{{20.18 g mol}^{- 1}}}$ $"For Ne: " "rate"_(Ne) prop 1/sqrt("20.18 g mol"^(-1))$

${For C}_{4} H_{10} : {rate}_{butane} \propto \frac{1}{\sqrt{{58.12 g mol}^{- 1}}}$ $"For C"_4"H"_10: " rate"_text(butane) prop 1/sqrt("58.12 g mol"^(-1))$

This means that you can compare the rates of effusion for the two gases by dividing these two expressions

$\frac{{rate}_{N e}}{{rate}_{butane}} = \frac{1}{\sqrt{{20.18 g mol}^{- 1}}} \cdot \frac{\sqrt{{58.12 g mol}^{- 1}}}{1}$ $"rate"_(Ne)/"rate"_text(butane) = 1/sqrt("20.18 g mol"^(-1)) * sqrt("58.12 g mol"^(-1))/1$

$"rate"_(Ne)/"rate"_text(butane) = sqrt( (58.12 color(red)(cancel(color(black)("g mol"^(-1)))))/(20.18color(red)(cancel(color(black)("g mol"^(-1)))))) = sqrt(58.12/20.18)$

$"rate"_(Ne)/"rate"_text(butane) = 1.6971$

So, this tells you that neon will effuse at a rate that is $1.6971$ faster than the rate of effusion of butane.

Since you know that the rate of effusion of neon is equal to $"800 m s"^(-1)$ , you can say that the rate of effusion of butane will be equal to

$"rate"_text(butane) = "rate"_text(Ne)/1.6971$

$"rate"_text(butane) = "800. m s"^(-1)/1.6971 = "471.39 m s"^(-1)$

Rounded to three sig figs, the number of sig figs you have for the rate of effusion of neon, the answer will be

$"rate"_"butane" = color(green)("471 m s"^(-1))$

So, does this result make sense?

Since butane has heavier molecules, it makes sense that it will effuse at a slower rate compared with neon.

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Question #133a5

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