Q. how to calculate rate of effusion

Q: Temperature cancels in comparative Graham calculations. Absolute effusion scales with the average molecular speed, which depends on T:

\[ \bar{v}=\sqrt{\frac{8R T}{\pi M}} \] So flux \propto\sqrt{T}.

Q: How is Graham's law derived from kinetic theory?

Rates scale with average molecular speed. From kinetic theory \bar{v}\propto\sqrt{1/M}. Substituting speeds into relative rates gives

Answer

Graham’s law says the effusion rate \(r\) is inversely proportional to the square root of the molar mass \(M\). For two gases this gives

\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]

Equivalently \(r \propto \frac{1}{\sqrt{M}}\) so to calculate a rate compute \(r = K \frac{1}{\sqrt{M}}\) or compare two gases with the ratio above.

Detailed Explanation

Effusion means the escape of gas molecules through a very small hole into a vacuum or much lower pressure region. The rate of effusion is proportional to the typical molecular speed given by kinetic theory. To calculate an effusion rate use Graham’s law which follows from the kinetic theory relations for molecular speed and the proportionality between rate and speed. I will derive the result step by step and show how to use it in practice.

Step 1, write the expression for a representative molecular speed. One commonly used measure is the mean speed. According to kinetic theory the mean molecular speed is given by the formula \(\bar{v}=\sqrt{\frac{8 R T}{\pi M}}\). In this expression \(R\) is the gas constant, \(T\) is the absolute temperature, and \(M\) is the molar mass in consistent units.

Step 2, relate effusion rate to molecular speed. The effusion rate \(r\) is proportional to the mean speed, so \(r\propto \bar{v}\). Combining this proportionality with the mean speed formula shows that the effusion rate is inversely proportional to the square root of the molar mass. In symbols the proportionality becomes \(r\propto \dfrac{1}{\sqrt{M}}\).

Step 3, form the ratio for two different gases. Comparing gas 1 and gas 2 eliminates the proportionality constant and temperature dependent factors if the temperature is the same for both gases. The result is Graham’s law written as the ratio of rates

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Chemistry FAQs

What is Graham's law of effusion?

Graham's law relates effusion rates to molar masses.

\[ \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \] Here \(r\) is effusion rate and \(M\) is molar mass. Lighter gases effuse faster.

How do I compare effusion rates of two gases?

Take the square root of the inverse molar mass ratio. Compute \(r_1/r_2=\sqrt{M_2/M_1}\). gas with smaller \(M\) has the larger rate. No absolute temperature needed if both gases share the same \(T\).

Which units should I use for molar mass?

For ratios you may use g mol^-1 because units cancel. For absolute formulas with \(R\) use kg mol^-1 so units match \(R=8.314\) J mol^-1 K^-1. Consistency matters.

How does temperature affect effusion?

Temperature cancels in comparative Graham calculations. Absolute effusion scales with the average molecular speed, which depends on \(T\):

\[ \bar{v}=\sqrt{\frac{8R T}{\pi M}} \] So flux \(\propto\sqrt{T}\).

How do I compute the absolute number flux through small hole?

Number of molecules per time is

\[ \dot{N}=\tfrac{1}{4}\,n\,\bar{v}\,A \] with \(n\) number density, \(A\) hole area, and \(\bar{v}=\sqrt{\frac{8R T}{\pi M}}\).

How long does it take to evacuate container by effusion?

Molecules decay exponentially for effusion into vacuum:

\[ N(t)=N_0 e^{-t/\tau},\quad \tau=\frac{4V}{A\bar{v}} \] Here \(V\) is container volume and \(A\) hole area. Use \(\bar{v}\) as above.

When is the effusion model valid instead of diffusion?

Effusion applies when the hole diameter is much smaller than the mean free path, written as diameter < mean free path. Then collisions at the hole are negligible and molecular flow dominates.

How is Graham's law derived from kinetic theory?

Rates scale with average molecular speed. From kinetic theory \(\bar{v}\propto\sqrt{1/M}\). Substituting speeds into relative rates gives

\[ \frac{r_1}{r_2}=\frac{\bar{v}_1}{\bar{v}_2}=\sqrt{\frac{M_2}{M_1}} \] which is Graham's law.

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