5.5: Effusion and Diffusion of Gases
- Page ID
- 370748
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Define and explain effusion and diffusion
- State Graham’s law and use it to compute relevant gas properties
If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule
In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure \(\PageIndex{1}\)). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure \(\PageIndex{1}\). The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:
\[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{6.8.2}\]
Multiplying both sides by 2 and rearranging give
\[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{6.8.3}\]
Taking the square root of both sides gives
\[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{6.8.4}\]
Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. This shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.
Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure \(\PageIndex{3}\) for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds.
The lightest gases have a wider distribution of speeds and the highest average speeds.
Note
Molecules with lower masses have a wider distribution of speeds and a higher average speed.
Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure \(\PageIndex{4}\).
The average distance traveled by a molecule between collisions is the mean free path. The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10−8 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 1010 m (about 6 million miles).
Note
The denser the gas, the shorter the mean free path.
If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure \(\PageIndex{2}\)). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:
\[\textrm{rate of effusion}∝\dfrac{1}{\sqrt{ℳ}} \nonumber \]
This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:
\[\dfrac{\textrm{rate of effusion of B}}{\textrm{rate of effusion of A}}=\dfrac{\sqrt{ℳ_\ce{A}}}{\sqrt{ℳ_\ce{B}}} \nonumber \]
Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.
Solution
From Graham’s law, we have:
\[\dfrac{\textrm{rate of effusion of hydrogen}}{\textrm{rate of effusion of oxygen}}=\mathrm{\dfrac{\sqrt{1.43\cancel{g\: L^{−1}}}}{\sqrt{0.0899\cancel{g\: L^{−1}}}}=\dfrac{1.20}{0.300}=\dfrac{4}{1}} \nonumber \]
Using molar masses:
\[\dfrac{\textrm{rate of effusion of hydrogen}}{\textrm{rate of effusion of oxygen}}=\mathrm{\dfrac{32\cancel{g\: mol^{−1}}}{2\cancel{g\: mol^{−1}}}=\dfrac{\sqrt{16}}{\sqrt{1}}=\dfrac{4}{1}} \nonumber \]
Hydrogen effuses four times as rapidly as oxygen.
At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Using the same apparatus at the same temperature and pressure, at what rate will sulfur dioxide effuse?
- Answer
-
52 mL/s
Here’s another example, making the point about how determining times differs from determining rates.
It takes 243 s for 4.46 × 10−5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 × 10−5 mol Ne to effuse?
Solution
It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:
\[\textrm{rate of effusion}=\dfrac{\textrm{amount of gas transferred}}{\textrm{time}}\nonumber \]
and combine it with Graham’s law:
\[\dfrac{\textrm{rate of effusion of gas Xe}}{\textrm{rate of effusion of gas Ne}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}\nonumber \]
To get:
\[\dfrac{\dfrac{\textrm{amount of Xe transferred}}{\textrm{time for Xe}}}{\dfrac{\textrm{amount of Ne transferred}}{\textrm{time for Ne}}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}\nonumber \]
Noting that amount of A = amount of B, and solving for time for Ne:
\[\dfrac{\dfrac{\cancel{\textrm{amount of Xe}}}{\textrm{time for Xe}}}{\dfrac{\cancel{\textrm{amount of Ne}}}{\textrm{time for Ne}}}=\dfrac{\textrm{time for Ne}}{\textrm{time for Xe}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}}=\dfrac{\sqrt{ℳ_\ce{Ne}}}{\sqrt{ℳ_\ce{Xe}}} \nonumber \]
and substitute values:
\[\mathrm{\dfrac{time\: for\: Ne}{243\:s}=\sqrt{\dfrac{20.2\cancel{g\: mol}}{131.3\cancel{g\: mol}}}=0.392}\nonumber \]
Finally, solve for the desired quantity:
\[\mathrm{time\: for\: Ne=0.392×243\:s=95.3\:s}\nonumber \]
Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for \(\ce{Xe}\), which means the time of effusion for Ne will be smaller than that for Xe.
A party balloon filled with helium deflates to \(\dfrac{2}{3}\) of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to \(\dfrac{1}{2}\) of its original volume?
- Answer
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32 h
Finally, here is one more example showing how to calculate molar mass from effusion rate data.
An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?
Solution
From Graham’s law, we have:
\[\mathrm{\dfrac{rate\: of\: effusion\: of\: Unknown}{rate\: of\: effusion\: of\: CO_2}}={\sqrt{ℳ_\mathrm{CO_2}}}{\sqrt{ℳ_{Unknown}}} \nonumber \]
Plug in known data:
\[\dfrac{1.66}{1}=\dfrac{\sqrt{44.0\:\ce{g/mol}}}{\sqrt{ℳ_{Unknown}}} \nonumber \]
Solve:
\[ℳ_{Unknown}=\mathrm{\dfrac{44.0\:g/mol}{(1.66)^2}=16.0\:g/mol} \nonumber \]
The gas could well be CH4, the only gas with this molar mass.
Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.
- Answer
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163 g/mol
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF6, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235UF6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238UF6 molecules. The gas that has passed through the barrier is slightly enriched in 235UF6 and the residual gas is slightly depleted. The small difference in molecular weights between 235UF6 and 238UF6 only about 0.4% enrichment, is achieved in one diffuser (Figure \(\PageIndex{4}\)). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
The large scale separation of gaseous 235UF6 from 238UF6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF6.
Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.
Graham’s law of Diffusion and Effusion: https://youtu.be/9HO-qgh-iGI
Summary
Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law).
Key Equations
- \(\textrm{rate of diffusion}=\dfrac{\textrm{amount of gas passing through an area}}{\textrm{unit of time}}\)
- \(\dfrac{\textrm{rate of effusion of gas B}}{\textrm{rate of effusion of gas A}}=\dfrac{\sqrt{m_B}}{\sqrt{m_A}}=\dfrac{\sqrt{ℳ_B}}{\sqrt{ℳ_A}}\)
Summary
- diffusion
- movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)
- effusion
- transfer of gaseous atoms or molecules from a container to a vacuum through very small openings
- Graham’s law of effusion
- rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses
- mean free path
- average distance a molecule travels between collisions
- rate of diffusion
- amount of gas diffusing through a given area over a given time