How fast does a solvent vapor spread through air, or a dissolved gas through water? The answer sets the size of an absorber, the time a dryer needs and the dose a drug patch delivers. It depends on one property, the diffusivity $D_{AB}$, and very often nobody has measured it for your pair at your temperature and pressure.

In this lesson you will estimate gas diffusivities with the Fuller and Chapman-Enskog equations, move a measured value to a new temperature and pressure, and estimate liquid diffusivities with the Wilke-Chang equation (Figure 1).

1. Gas pairmeasured value at other conditions?
2. $\dfrac{T^{1.75}}{P}$rescale it, or estimate with Fuller or Chapman-Enskog
3. Gas estimateabout 0.1 cm²/s at 1 atm
4. $\dfrac{T}{\mu}$liquids: Wilke-Chang, or rescale
5. Liquid estimateabout 10⁻⁵ cm²/s

Learning Objectives

After this lesson, you should be able to:

1. Estimate the diffusivity of a gas pair at low pressure with the Fuller equation.
2. Rescale a measured gas diffusivity to a new temperature and pressure.
3. Estimate a gas diffusivity with Chapman-Enskog theory and the collision integral.
4. Estimate a liquid diffusivity at infinite dilution with the Wilke-Chang equation, and move a liquid diffusivity to a new temperature.

Key Ideas

1. How Big Is a Diffusivity?

The diffusivity in Fick’s law tells how quickly molecules of A spread through B. Its size depends strongly on the phase (Figure 2):

• Gases near room temperature and 1 atm: about 0.1 cm²/s. Ethanol vapor in air at 0 °C is 0.102 cm²/s (Perry’s Table 2-141).
• Liquids: about $10^{-5}$ cm²/s, some 10 000 times smaller. Ethanol in water at 25 °C is $1.28 \times 10^{-5}$ cm²/s, a measured value.

In a gas, molecules fly freely between collisions. Raising the pressure packs them closer and shortens that free path, so $D_{AB}$ is inversely proportional to $P$ up to a few tens of atm. Raising the temperature speeds the molecules up, so $D_{AB}$ rises steeply with $T$. For a binary gas pair, $D_{AB} = D_{BA}$: one value describes the pair.

Gases near 1 atm: $D_{AB}$ $\approx 0.1\,\text{cm}^{2}\,\text{s}^{-1}$, varies as $\dfrac{T^{1.75}}{P}$. Ethanol in air, 0 °C: $0.102\,\text{cm}^{2}\,\text{s}^{-1}$

Liquids: $D_{AB}$ $\approx 10^{-5}\,\text{cm}^{2}\,\text{s}^{-1}$, varies as $\dfrac{T}{\mu}$. Ethanol in water, 25 °C: $1.28 \times 10^{-5}\,\text{cm}^{2}\,\text{s}^{-1}$

Key idea: a gas diffusivity is about 0.1 cm²/s at 1 atm and varies as $\frac{1}{P}$; a liquid diffusivity is about $10^{-5}$ cm²/s and hardly depends on pressure.

2. The Fuller Equation for Gases

The working correlation for a gas pair at low pressure is the Fuller (Fuller-Schettler-Giddings) equation, Perry’s Eq. 5-202:

with $D_{AB}$ in cm²/s, $T$ in K, $P$ in atm and the molar masses in g/mol. Each diffusion volume $\Sigma v$ is a sum of tabulated increments (Perry’s Table 5-13): C 16.5, H 1.98, O 5.48, with whole-molecule values for simple gases such as air (20.1). Light molecules and small diffusion volumes give a large $D_{AB}$ (Figure 3). Perry’s gives an average error of about 5% for this equation.

$D_{AB} = \dfrac{0.001\,T^{1.75}\sqrt{\dfrac{1}{M_A} + \dfrac{1}{M_B}}}{P\left(\sqrt[3]{\Sigma v_A} + \sqrt[3]{\Sigma v_B}\right)^{2}}$

$T^{1.75}$: temperature in K; a hotter gas diffuses faster

$\sqrt{\dfrac{1}{M_A} + \dfrac{1}{M_B}}$: molar masses in g/mol, in the numerator; light gases diffuse faster

$P$: pressure in atm, in the denominator

$\left(\sqrt[3]{\Sigma v_A} + \sqrt[3]{\Sigma v_B}\right)^{2}$: diffusion volumes from Table 5-13; large molecules diffuse slower

Key idea: add the Table 5-13 increments to get each $\Sigma v$, use $T$ in K and $P$ in atm, and keep the molar-mass term $\sqrt{\frac{1}{M_A} + \frac{1}{M_B}}$ in the numerator.

Try it: estimate a gas pair

Pick two gases, then move the temperature and pressure sliders. Watch two things: on log axes the temperature line has a slope of 1.75, and $D_{AB} P$ does not change when only the pressure changes.

Open the gas diffusivity simulator full screen

Worked Example 1: Fuller, Ethanol Vapor in Air

Problem: Estimate $D_{AB}$ for ethanol (A, $\mathrm{C_2\!H_6\!O}$, 46.07 g/mol) in air (B, 28.96 g/mol) (a) at 273.15 K and 1 atm, and compare with the measured 0.102 cm²/s; (b) at 298.15 K and 1 atm; (c) at 298.15 K and 2 atm.

Step 1. Diffusion volumes from Table 5-13: two C, six H and one O for ethanol; 20.1 for air.

Step 2. The molar-mass term and the volume term.

Step 3. Substitute for each set of conditions.

Final Answer: (a) $0.106\,\text{cm}^{2}\,\text{s}^{-1}$; (b) $0.123\,\text{cm}^{2}\,\text{s}^{-1}$; (c) $0.0617\,\text{cm}^{2}\,\text{s}^{-1}$

Check: (a) is about 4% above the measured 0.102 cm²/s, inside the typical error of the equation. Doubling the pressure in (c) halves the value from (b), because $D_{AB} P$ is constant at fixed $T$.

Your turn: a Fuller estimate

Same steps as Worked Example 1: build both diffusion volumes, evaluate the two terms to 4 significant figures, then substitute with $T$ in K.

Worked Example 2: Rescaling a Measured Gas Diffusivity

Problem: Perry’s Table 2-141 gives $D_{AB} = 0.102$ cm²/s for ethanol in air at 273.15 K and 1 atm. Estimate the value at 298.15 K and 1 atm from this measurement.

Step 1. Choose the rule. The Fuller equation varies as $\frac{T^{1.75}}{P}$, so a measured value can be moved with the same powers:

Step 2. Substitute, with both temperatures in K.

Final Answer: $D_{AB} \approx 0.119\,\text{cm}^{2}\,\text{s}^{-1}$ at 298.15 K and 1 atm

Compare with the Fuller estimate at the same conditions, 0.123 cm²/s from Worked Example 1(b). The Fuller value is about 4% higher at both temperatures, because both use $T^{1.75}$. When a measured value exists, rescaling it is usually the better estimate.

Your turn: rescale a measured value

Same idea as Worked Example 2: convert both temperatures to K, raise their ratio to the power 1.75, and divide by the pressure ratio.

3. Chapman-Enskog Theory for Gases

Kinetic theory gives the form behind the Fuller equation (Perry’s Sec. 5, with Lennard-Jones parameters):

with $\sigma_{AB}$ in Å. The pair parameters come from the pure-gas values by the combining rules, and the collision integral $\Omega_D$ comes from the reduced temperature $T^{*}$ by the Neufeld fit (Figure 4):

When no Lennard-Jones values are tabulated, Perry’s Table 5-12 estimates them from critical properties. Because $\Omega_D$ falls as $T$ rises, the effective temperature exponent drifts between about 1.66 and 2.0, which is why a single $T^{1.75}$ rule is only approximate over a wide range.

1. Pair parameters$\sigma_{AB} = \dfrac{\sigma_A + \sigma_B}{2}$, $\dfrac{\varepsilon_{AB}}{k_B} = \sqrt{\dfrac{\varepsilon_A}{k_B}\,\dfrac{\varepsilon_B}{k_B}}$combining rules
2. $T^{*} = \dfrac{k_B T}{\varepsilon_{AB}}$Step 2: reduced temperature
3. Collision integral$2.454 \rightarrow \Omega_D = 1.006$; $3.594 \rightarrow \Omega_D = 0.9075$Neufeld fit, at each T
4. $\Omega_D$Step 3: substitute
5. Diffusivity$D_{AB} = \dfrac{0.001858\,T^{1.5}\sqrt{\dfrac{1}{M_A} + \dfrac{1}{M_B}}}{P\,\sigma_{AB}^{2}\,\Omega_D}$in cm²/s

Key idea: combine the pure-gas $\sigma$ and $\frac{\varepsilon}{k_B}$, find $T^{*}$ and $\Omega_D$ at the new temperature, then substitute; recompute $\Omega_D$ every time $T$ changes.

Worked Example 3: Chapman-Enskog, Oxygen in Nitrogen

Problem: Estimate $D_{AB}$ for oxygen (A, 31.9988 g/mol) in nitrogen (B, 28.0134 g/mol) (a) at 273.15 K and 1 atm, and compare with the measured 0.181 cm²/s (Table 2-141); (b) at 400 K and 5 atm. Lennard-Jones parameters from the Table 5-12 rule: oxygen $\sigma = 3.433$ Å, $\frac{\varepsilon}{k_B} = 122.93$ K; nitrogen $\sigma = 3.657$ Å, $\frac{\varepsilon}{k_B} = 100.69$ K.

Step 1. Combine for the pair.

Step 2. (a) Collision integral and diffusivity at 273.15 K.

Step 3. (b) Recompute at 400 K and 5 atm. $T^{*}$ changes, so $\Omega_D$ does too.

Final Answer: (a) $0.172\,\text{cm}^{2}\,\text{s}^{-1}$; (b) $0.0674\,\text{cm}^{2}\,\text{s}^{-1}$

Check: (a) is about 5% below the measured 0.181 cm²/s. Scaling (a) by $\frac{T^{1.75}}{P}$ instead of recomputing $\Omega_D$ gives 0.0669 cm²/s, close here but not the same. The simpler critical-point rule of Table 5-12 for $\sigma$ and $\frac{\varepsilon}{k_B}$ gives about 0.162 cm²/s, 10% low: the choice of Lennard-Jones parameters dominates the error.

Your turn: a Chapman-Enskog estimate

Same steps as Worked Example 3: find $T^{*}$ at the new temperature, evaluate $\Omega_D$ from the Neufeld fit, then substitute with $\sigma_{AB}^{2}$.

4. Diffusion in Liquids

Molecules in a liquid are packed together, so a solute moves by squeezing past its neighbors and the solvent’s viscosity controls the rate. For a large spherical solute of radius $r$, the Stokes-Einstein equation gives

For small solutes at infinite dilution, the working correlation is the Wilke-Chang equation:

with $D_{AB}$ in cm²/s, $T$ in K, the solvent viscosity $\mu$ in cP, the solvent molar mass $M_B$ in g/mol, and $V_A$ the molar volume of the solute at its normal boiling point in cm³/mol. The association factor $\varphi$ is 2.26 for water.¹ Both equations say that $\frac{D\mu}{T}$ is roughly constant for one solute and solvent, which moves a measured value to a new temperature (Figure 5):

¹ The original Wilke-Chang paper used 2.6 for water; Perry’s 9th edition gives the revised value 2.26, used throughout this lesson.

$D_{AB} = \dfrac{7.4 \times 10^{-8}\sqrt{\varphi M_B}\,T}{\mu\,V_A^{0.6}}$

$\sqrt{\varphi M_B}$: solvent association factor and molar mass; $\varphi = 2.26$ for water

$\mu$ in cP and $V_A^{0.6}$ in the denominator: a viscous solvent or a large solute slows diffusion

Temperature rule: $\dfrac{D\mu}{T}$ about constant, so $D_2 = D_1\,\dfrac{T_2}{T_1}\,\dfrac{\mu_1}{\mu_2}$

Key idea: in a liquid, $D$ is about $10^{-5}$ cm²/s and varies as $\frac{T}{\mu}$. Water’s viscosity falls steeply as it warms, so the viscosity ratio usually matters far more than the temperature ratio.

Worked Example 4: Wilke-Chang, Ethanol in Water

Problem: (a) Estimate $D_{AB}$ for ethanol (A) at infinite dilution in water (B) at 298.15 K, and compare with the measured $1.28 \times 10^{-5}$ cm²/s. (b) Move the measured value to 313.15 K. Data: $V_A = 62.70$ cm³/mol at the normal boiling point; $M_B = 18.02$ g/mol; $\varphi = 2.26$; water viscosity from Perry’s Table 2-135 with $\ln\mu$ interpolated linearly in $T$: 0.8924 cP at 298.15 K and 0.6545 cP at 313.15 K.

Step 1. Evaluate the two terms.

Step 2. (a) Substitute with $T$ in K and $\mu$ in cP.

Step 3. (b) Rescale the measured value with $\frac{D\mu}{T}$ constant.

Final Answer: (a) $1.32 \times 10^{-5}\,\text{cm}^{2}\,\text{s}^{-1}$; (b) $1.83 \times 10^{-5}\,\text{cm}^{2}\,\text{s}^{-1}$

Check: (a) is about 3% above the measured value. With the original $\varphi = 2.6$ the estimate would be $1.41 \times 10^{-5}$ cm²/s, about 10% high. In (b), warming from 25 °C to 40 °C raises $D$ by a factor of 1.432, about 43%, while $\frac{T_2}{T_1}$ alone is only 1.050: the falling viscosity does most of the work.

Your turn: a Wilke-Chang estimate

Same steps as Worked Example 4(a): evaluate $\sqrt{\varphi M_B}$ and $V_A^{0.6}$, then substitute with $T$ in K and $\mu$ in cP. Enter the answer as a number times $10^{-5}$.

Your turn: move a liquid diffusivity to a new temperature

Same idea as Worked Example 4(b): multiply by the temperature ratio and by the viscosity ratio $\frac{\mu_1}{\mu_2}$, with the old viscosity on top.

Your turn: rescale a measured value with interpolated viscosity

This one starts from the measured value for ethanol in water at 25 °C, as in Worked Example 4(b), with the water viscosities given in the problem.

Common Mistakes

Common mistake: Reading the molar-mass term upside down. Some tables write the molar-mass term with a reduced molar mass $M_{AB}$. Putting $\sqrt{M_{AB}}$ literally in the numerator gives 2.19 cm²/s for Worked Example 1(b), about 18 times too large. The molar masses enter as $\sqrt{\frac{1}{M_A} + \frac{1}{M_B}}$: light gases diffuse faster. Forgetting the square root or the square on the volume term also gives a wrong size.

Common mistake: Using °C, or the wrong power of T. Every temperature in these equations is in K. In Worked Example 2, $\left(\frac{25}{0}\right)$ cannot even be evaluated. Fuller uses $T^{1.75}$ and Chapman-Enskog $T^{1.5}$ with $\Omega_D$; do not mix them up.

Common mistake: Inverting or dropping the pressure. $D_{AB}$ for a gas varies as $\frac{1}{P}$. In Worked Example 1(c), multiplying by 2 instead of dividing gives a value four times too large.

Common mistake: Keeping the old collision integral. In Worked Example 3(b), $\Omega_D$ drops from 1.006 to 0.9075. Leaving it out, forgetting to square $\sigma_{AB}$, or using $\sigma_{AB}$ where $\sigma_{AB}^{2}$ belongs all change the answer.

Common mistake: Mishandling the Wilke-Chang terms. Use $V_A^{0.6}$, not $V_A$; the viscosity in the denominator, in cP; $\varphi = 2.26$ for water; and the square root of $\varphi M_B$.

Common mistake: Leaving out a ratio in the liquid temperature rule. In Worked Example 4(b), using only $\frac{\mu_1}{\mu_2}$ or only $\frac{T_2}{T_1}$ misses part of the change, and turning the viscosity ratio upside down makes $D$ fall when the water warms.

Common mistake: Slips in Stokes-Einstein. Keep the 6 in $6\pi\mu r$, convert cP to Pa·s (1 cP is $10^{-3}$ Pa·s) and nm to m before you divide, and convert the result from m²/s to cm²/s.

Key Takeaway

Gas diffusivities are about 0.1 cm²/s and vary as $\frac{T^{1.75}}{P}$: estimate them with the Fuller equation from Table 5-13 diffusion volumes, or with Chapman-Enskog theory and a collision integral recomputed at each temperature, and rescale a measured value whenever one exists. Liquid diffusivities are about $10^{-5}$ cm²/s and vary as $\frac{T}{\mu}$: estimate them with the Wilke-Chang equation and move them to a new temperature with $\frac{D\mu}{T}$ constant.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson, including a Stokes-Einstein estimate for a large solute: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Methods and data: Perry’s Chemical Engineers’ Handbook, 9th edition: Section 5 for the Fuller, Chapman-Enskog and Wilke-Chang equations, the diffusion volumes and the Lennard-Jones estimates; Section 2 for the water viscosities, the critical properties and the measured diffusivities. The practice problems use illustrative conditions and are audited; see the lesson audit log.