Gas molecules are in constant motion and continually collide with the walls of their container. If a small opening is present, some molecules may escape through that opening. This process is called effusion. A familiar example is a helium-filled balloon that gradually loses gas and shrinks over time.
The rate at which a gas effuses depends on how frequently its molecules collide with the opening. According to the Kinetic Molecular Theory, lighter molecules move faster on average than heavier molecules at the same temperature. Because faster molecules strike the opening more frequently, they effuse more rapidly.
Graham's Law relates the rate of effusion of a gas to its molar mass:
\[ \text{Rate of Effusion} \propto \frac{1}{\sqrt{MW}} \]
This relationship can be used to compare two gases directly:
\[ \frac{\text{Rate}_A} {\text{Rate}_B} = \sqrt{ \frac{MW_B} {MW_A} } \]
Since the rate is inversely proportional to the square root of the molar mass, lighter gases effuse more rapidly than heavier gases. For example, helium effuses much faster than nitrogen because helium atoms have a much smaller molar mass.
The square-root dependence is important. A gas with half the molar mass of another gas does not effuse twice as fast. Instead, it effuses
\[ \sqrt{2} \approx 1.41 \]
times faster.
Effusion and diffusion are related but distinct processes.
Both processes are influenced by molecular speed and therefore by molecular mass. As a result, Graham's Law is often used as an approximation for comparing both effusion and diffusion rates.
The origin of Graham's Law can be traced directly to the Maxwell-Boltzmann distribution and the expression for the average molecular speed:
\[ v_{ave} \propto \sqrt{ \frac{T} {MW} } \]
At constant temperature, lighter molecules move faster and therefore reach an opening or diffuse through a gas mixture more rapidly than heavier molecules.
Big picture: Graham's Law connects molecular motion to observable behavior. The same molecular speeds predicted by the Maxwell-Boltzmann distribution explain why lighter gases diffuse and effuse more rapidly than heavier gases.
One important application of Graham's Law and the Kinetic Molecular Theory is the measurement of vapor pressures using a Knudsen cell. A Knudsen cell consists of a small chamber containing a substance and a tiny orifice through which vapor molecules are allowed to escape. By measuring the amount of material lost over a known period of time, it is possible to determine the vapor pressure of the substance.
The operation of the Knudsen cell relies on the fact that molecules in a thermalized gas are continually colliding with the walls of the container. Any molecule that would have collided with the portion of the wall occupied by the orifice instead escapes from the cell. As a result, the rate at which mass is lost is directly related to the frequency with which molecules strike the opening, which in turn depends on the vapor pressure of the substance.
For the Knudsen model to be valid, the orifice must be sufficiently small that molecules escape without undergoing significant collisions within the opening itself. Under these conditions, the effusion process can be described using the predictions of the Kinetic Molecular Theory.
The vapor pressure of the substance can be related to the mass lost through the orifice by
\[ p = \frac{g} {A\,\Delta t} \sqrt{ \frac{2\pi RT} {MW} } \]
where
| Symbol | Meaning |
|---|---|
| \(p\) | Vapor pressure |
| \(g\) | Mass lost through the orifice |
| \(A\) | Area of the effusion orifice |
| \(\Delta t\) | Time allowed for effusion |
| \(R\) | Ideal gas constant |
| \(T\) | Temperature (K) |
| \(MW\) | Molar mass of the vapor species |
The equation shows that larger vapor pressures produce larger rates of mass loss. By carefully measuring the mass lost from the cell and knowing the temperature, orifice area, and molar mass of the vapor, the vapor pressure can be determined experimentally.
Many substances have vapor pressures that are too small to measure accurately using ordinary pressure gauges. The Knudsen cell provides a sensitive method for determining these vapor pressures by converting a pressure measurement into a mass-loss measurement.
The technique is particularly useful for solids and low-volatility liquids, where only a small fraction of the material exists in the gas phase. Because the experiment is based on molecular effusion, it provides a direct connection between the Kinetic Molecular Theory and experimentally measurable properties.
Big picture: The Knudsen cell uses the molecular collisions predicted by the Kinetic Molecular Theory to measure vapor pressure. By observing how rapidly molecules escape through a tiny opening, we can determine the pressure exerted by the vapor inside the cell.
Uranium can be enriched in \(^{235}U\) by allowing gaseous uranium hexafluoride, \(UF_6\), to diffuse through a system of nozzles. Calculate the ratio of the diffusion rate of \(^{235}UF_6\) relative to \(^{238}UF_6\).
Graham's Law gives
\[ \frac{\text{Rate}_{235}} {\text{Rate}_{238}} = \left( \frac{MW_{238}} {MW_{235}} \right)^{1/2} \]
First calculate the molar masses:
\[ MW(^{235}UF_6) = 235 + 6(19.00) = 349.00\ \mathrm{g\,mol^{-1}} \]
\[ MW(^{238}UF_6) = 238 + 6(19.00) = 352.00\ \mathrm{g\,mol^{-1}} \]
Substitute into Graham's Law:
\[ \frac{\text{Rate}_{235}} {\text{Rate}_{238}} = \left( \frac{352.00} {349.00} \right)^{1/2} \]
\[ \frac{\text{Rate}_{235}} {\text{Rate}_{238}} = 1.0043 \]
Therefore, \(^{235}UF_6\) effuses only about 1.004 times faster than \(^{238}UF_6\).
Physical interpretation: The lighter isotope-containing molecule effuses slightly faster, but the difference is very small. This is why uranium enrichment by gaseous diffusion requires many repeated separation steps.
Two identical balloons are filled to a volume of \(1.00\ \mathrm{L}\) at the same temperature and pressure. Balloon A contains helium, while Balloon B contains an unknown gas.
After a certain amount of time, Balloon A has decreased in volume by \(0.200\ \mathrm{L}\), while Balloon B has decreased in volume by \(0.0603\ \mathrm{L}\).
Determine the molar mass of the gas in Balloon B.
According to Graham's Law,
\[ \frac{\text{Rate}_A} {\text{Rate}_B} = \sqrt{ \frac{MW_B} {MW_A} } \]
Since both balloons are observed over the same time interval, the rate of effusion is proportional to the volume lost:
\[ \frac{\text{Rate}_A} {\text{Rate}_B} = \frac{\Delta V_A} {\Delta V_B} \]
Therefore,
\[ \frac{\text{Rate}_A} {\text{Rate}_B} = \frac{ 0.200\ \mathrm{L} } { 0.0603\ \mathrm{L} } = 3.317 \]
Substituting this ratio into Graham's Law gives
\[ 3.317 = \sqrt{ \frac{MW_B} {MW_A} } \]
Squaring both sides:
\[ (3.333)^2 = \frac{MW_B} {MW_A} \]
\[ 11.002 = \frac{MW_B} {MW_A} \]
The molar mass of helium is
\[ MW_A = 4.003\ \mathrm{g\,mol^{-1}} \]
Solving for the molar mass of gas B:
\[ MW_B = (11.002) \left( 4.003\ \mathrm{g\,mol^{-1}} \right) \]
\[ MW_B = 44.04\ \mathrm{g\,mol^{-1}} \]
Therefore, the unknown gas has a molar mass of
\[ MW_B = 44.04\ \mathrm{g\,mol^{-1}} \]
Physical interpretation: Helium effuses much more rapidly than the unknown gas because helium atoms are much lighter. The calculated molar mass is close to that of carbon dioxide (\(44.01\ \mathrm{g\,mol^{-1}}\)), suggesting that the unknown gas could plausibly be \(CO_2\).
A sample of naphthalene, \(C_{10}H_8\), is placed in a Knudsen cell at \(45.0^\circ\mathrm{C}\). The orifice area is \(0.500\ \mathrm{mm^2}\). After \(1.00\ \mathrm{hr}\), the cell loses \(0.370\ \mathrm{g}\) of naphthalene vapor. Calculate the vapor pressure of naphthalene.
\[ p = \frac{g} {A\Delta t} \left( \frac{2\pi RT}{MW} \right)^{1/2} \]
First convert all quantities to SI units:
\[ g = 0.370\ \mathrm{g} = 3.70\times10^{-4}\ \mathrm{kg} \]
\[ A = 0.500\ \mathrm{mm^2} = 5.00\times10^{-7}\ \mathrm{m^2} \]
\[ \Delta t = 1.00\ \mathrm{hr} = 3600\ \mathrm{s} \]
\[ T = 45.0 + 273.15 = 318.15\ \mathrm{K} \]
\[ MW = 128.17\ \mathrm{g\,mol^{-1}} = 0.12817\ \mathrm{kg\,mol^{-1}} \]
Substitute into the Knudsen equation:
\[ p = \frac{ 3.70\times10^{-4}\ \mathrm{kg} } { \left(5.00\times10^{-7}\ \mathrm{m^2}\right) \left(3600\ \mathrm{s}\right) } \left( \frac{ 2\pi \left(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\right) \left(318.15\ \mathrm{K}\right) } { 0.12817\ \mathrm{kg\,mol^{-1}} } \right)^{1/2} \]
\[ p = 74.0\ \mathrm{Pa} \]
If desired, convert to Torr:
\[ 74.0\ \mathrm{Pa} \left( \frac{1\ \mathrm{Torr}} {133.322\ \mathrm{Pa}} \right) = 0.555\ \mathrm{Torr} \]
Therefore, the vapor pressure of naphthalene at \(45.0^\circ\mathrm{C}\) is
\[ p = 74.0\ \mathrm{Pa} = 0.555\ \mathrm{Torr} \]
Physical interpretation: A larger mass loss would indicate that more molecules escaped through the orifice per unit time, corresponding to a larger vapor pressure inside the cell.
Identify the problem type, then choose the correct answer.
Big picture: Effusion, diffusion, and vapor-pressure measurements all depend on molecular motion. The same molecular speeds predicted by the Maxwell-Boltzmann distribution explain why lighter gases move faster, effuse more rapidly, and collide with surfaces more frequently than heavier gases.