Water at the bottom of a narrow tube 0.200 m deep evaporates at about 10.8 g m⁻² h⁻¹ at 25 °C. Ether at 20 °C, in a tube half as deep, loses about 853 g m⁻² h⁻¹, some 79 times faster (Worked Examples 2 and 3). In both cases a vapour leaves a liquid surface and diffuses through air. When the vapour is dilute, Fick’s law alone predicts the rate. When it is concentrated, the escaping vapour sets the whole gas moving, and that bulk flow carries extra vapour along with it.

In this lesson you will separate the diffusion flux from the total flux, then solve the two classic steady cases: equimolar counterdiffusion, where the two gases trade places, and diffusion through a stagnant film, where one gas cannot move (Figure 1).

1. Fick’s law$J_A = -c\,D_{AB}\,\dfrac{dx_A}{dz}$diffusion flux
2. +add bulk flow
3. Total flux$N_A = J_A$
   $+\;x_A\,(N_A + N_B)$what crosses a fixed plane
4. B?what does B do?
5. Two cases$N_B = -N_A$
   or $N_B = 0$B moves back: straight line; B stagnant: log form

Learning Objectives

After this lesson, you should be able to:

1. Write Fick’s law for a binary gas and explain the difference between the diffusion flux and the total flux.
2. Calculate the steady flux for equimolar counterdiffusion between two reservoirs.
3. Calculate the steady flux of a vapour through a stagnant gas, with the log form or the log-mean pressure.
4. Decide when bulk flow can be ignored, and choose the right flux model for a physical situation.

Key Ideas

1. Fick’s Law

In a binary gas mixture of A and B at uniform temperature and total pressure, A diffuses from high to low mole fraction. Fick’s first law gives the diffusion flux:

Here $c = \frac{P}{RT}$ is the total molar concentration, $D_{AB}$ is the diffusivity and $z$ is the distance along the path. The minus sign says that A moves down its gradient. Use SI units throughout: $P$ in Pa, $T$ in K, and $D_{AB}$ in m²/s, which is the value in cm²/s times $10^{-4}$. For air at 25 °C and 101.325 kPa, $c = 40.88\,\text{mol}\,\text{m}^{-3}$.

The problems in this lesson take $D_{AB}$ from Perry’s Table 2-141, which lists values at 0 °C and 1 atm, and scale them to the working temperature as $T^{1.75}$. For water vapour in air, 0.220 cm²/s at 0 °C becomes 0.256 cm²/s at 25 °C.

Key idea: Fick’s law gives the diffusion flux $J_A$, driven by the mole-fraction gradient: $c$ from the ideal gas law with $T$ in K, and $D_{AB}$ in m²/s.

2. Two Fluxes: J and N

$J_A$ is measured relative to the mixture, which may itself be moving. A plane fixed to the apparatus sees the total flux $N_A$, which adds the A carried by the net motion of the mixture (Perry’s Eq. 5-190, Figure 2):

The last term is the bulk flow: the mole fraction of A times the net molar flux of the mixture. Because $J_A + J_B = 0$ in a binary mixture, bulk flow appears only when $N_A$ and $N_B$ do not cancel. Perry’s Handbook notes that this term “may account for as much as 10 percent of the flux” and is frequently ignored. That is true only while A is dilute; Key Idea 5 shows how quickly it grows as $x_A$ rises.

$N_A$ = $J_A$ diffusion + $x_A\,(N_A + N_B)$ bulk flow

Equimolar counterdiffusion, $N_B = -N_A$: $x_A\,(N_A + N_B) = 0$, so the total flux is pure diffusion

Stagnant B, $N_B = 0$: $N_A = \dfrac{J_A}{1-x_A}$, always larger than the diffusion flux

Dilute A, $x_A \ll 1$: $N_A \approx J_A$ whatever B does

Key idea: $J_A$ is diffusion relative to the moving mixture; $N_A$ is what crosses a fixed plane. They are equal when $N_A + N_B = 0$, and nearly equal when A is dilute.

3. Equimolar Counterdiffusion

Suppose the total pressure is uniform and every molecule of A that moves one way is replaced by a molecule of B moving the other way, as in two large bulbs joined by a tube. Then $N_B = -N_A$, the bulk-flow term is zero and $N_A = J_A$. At steady state $N_A$ is the same at every $z$, so with constant $c$ and $D_{AB}$ the mole fraction falls in a straight line. Integrating Fick’s law across a path of length $\Delta z$ gives Perry’s Eq. 5-197:

For ideal gases $x_A = \frac{p_A}{P}$, so the driving force can be written with partial pressures: $x_{A1}-x_{A2} = \frac{p_{A1}-p_{A2}}{P}$.

Key idea: in equimolar counterdiffusion there is no bulk flow: $N_A = J_A = -N_B$, the profile is a straight line, and the flux is proportional to the mole-fraction difference.

Worked Example 1: Equimolar Counterdiffusion Between Two Bulbs

Problem: Two large bulbs are joined by a straight tube 0.200 m long. Both hold water vapour (A) mixed with air (B) at 25 °C and a total pressure of 101.325 kPa. The partial pressure of water vapour is 2.40 kPa in bulb 1 and 0.600 kPa in bulb 2. Air diffuses back at the same molar rate. Find the steady molar flux $N_A$ and the partial pressure of water vapour halfway along the tube. Use $D_{AB} = 0.256\,\text{cm}^{2}\,\text{s}^{-1}$, the Table 2-141 value of 0.220 cm²/s at 0 °C scaled as $T^{1.75}$.

Step 1. Put the data in SI units. Use $T$ in K and $P$ in Pa.

Step 2. Find the mole-fraction difference.

Step 3. Substitute. There is no bulk flow, so $N_A = J_A$.

Step 4. Mid-length partial pressure. The profile is a straight line, so the midpoint is the average of the two ends.

Final Answer: $N_A \approx 9.29 \times 10^{-5}\,\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$, with $N_B = -N_A$; $p_A = 1.50\,\text{kPa}$ at mid-length

Check: leaving $D_{AB}$ in cm²/s gives 0.929 mol m⁻² s⁻¹, 10 000 times too large, and using 25 instead of 298.15 in $c$ gives $1.11 \times 10^{-3}$ mol m⁻² s⁻¹, about 12 times too large. A diffusion flux in a gas over a path of a few centimetres is typically $10^{-5}$ to $10^{-3}$ mol m⁻² s⁻¹.

Your turn: equimolar counterdiffusion between two bulbs

Same steps as Worked Example 1: convert $T$ to K and $D_{AB}$ to m²/s, find $c$, divide the partial-pressure difference by $P$, then substitute. Enter the flux as, for example, 1.23e-5.

Your turn: equimolar counterdiffusion at another pressure

One step more: these bulbs are not at 101.325 kPa. A gas diffusivity varies as $\frac{1}{P}$, so scale the tabulated $D_{AB}$ to the actual pressure first, and use that same pressure in $c$ and in the mole fractions.

4. Diffusion Through a Stagnant Film

Now let liquid A evaporate at the bottom of a narrow tube into a gas B that does not dissolve in the liquid, with a stream of gas sweeping the vapour away at the top. This is the Stefan tube. B has nowhere to go at the liquid surface, so at steady state its net flux is zero: $N_B = 0$. B is not motionless at the molecular scale. It diffuses down toward the liquid, and the bulk flow carries it back up at exactly the same rate.

With $N_B = 0$, the flux equation becomes $N_A = J_A + x_A N_A$, so $N_A = \frac{J_A}{1-x_A}$. Integrating with $N_A$ constant along the tube gives Perry’s Eq. 5-198:

Writing $1-x_A = \frac{p_B}{P}$ turns this into an equivalent log-mean form, with $p_{BM}$ the log-mean partial pressure of B:

The ratio $\frac{P}{p_{BM}}$ is the bulk-flow factor: the stagnant-film flux is the straight-line (Fick-only) flux times this factor, which is always greater than 1. The profile also changes shape. Instead of a straight line, $p_A$ bows upward (Figure 3).

Key idea: when B is stagnant, $N_B = 0$ and bulk flow adds to diffusion. Use the natural log of $\frac{1-x_{A2}}{1-x_{A1}}$, or the Fick-only flux times $\frac{P}{p_{BM}}$.

Try it: the Stefan tube and bulk flow

Start with the flux condition set to stagnant B. Raise $x_{A1}$ at the bottom of the tube and watch the probe split the total flux into its diffusion and bulk-flow parts: the bulk-flow share grows quickly once $x_{A1}$ passes about 0.2. Then switch to equimolar counterdiffusion and see the bulk-flow term vanish and the profile become a straight line.

Open the Stefan tube and bulk flow simulator full screen

Worked Example 2: Water Evaporating Through Stagnant Air

Problem: Water at the bottom of a narrow vertical tube evaporates into air at 25 °C and 101.325 kPa. The water surface stays 0.200 m below the open top, where a stream of dry air keeps the vapour partial pressure at zero. The vapour pressure of water at 25 °C is 3.1699 kPa (Perry’s Table 2-5), and $D_{AB} = 0.256\,\text{cm}^{2}\,\text{s}^{-1}$ as in Worked Example 1. Find the steady molar flux of water vapour, the evaporation rate in g m⁻² h⁻¹, and the bulk-flow factor.

Step 1. Concentration and diffusivity. The conditions match Worked Example 1: $c = 40.88\,\text{mol}\,\text{m}^{-3}$ and $D_{AB} = 0.256 \times 10^{-4}\,\text{m}^{2}\,\text{s}^{-1}$.

Step 2. The mole fractions at the two ends. The vapour at the liquid surface is saturated; at the top it is zero.

Step 3. Substitute into the stagnant-film flux.

Step 4. Bulk-flow factor. The partial pressure of air is $p_{B1} = 101.325-3.1699 = 98.1551\,\text{kPa}$ at the water surface and $p_{B2} = 101.325\,\text{kPa}$ at the top.

Step 5. Evaporation rate. Multiply by the molar mass of water, 18.02 g mol⁻¹, and by 3600 s h⁻¹.

Final Answer: $N_A \approx 1.66 \times 10^{-4}\,\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$, about 10.8 g m⁻² h⁻¹; bulk-flow factor 1.016

Check: the Fick-only flux is $1.637 \times 10^{-4}\,\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$, so bulk flow adds only 1.6%. Water at 25 °C has $x_{A1} = 0.03128$: the vapour is dilute, and Perry’s 10 percent remark holds comfortably.

Your turn: evaporation through a stagnant gas

Same steps as Worked Example 2: find $x_{A1}$ from the vapour pressure given in the problem, evaluate the log term, then substitute.

5. When Is Bulk Flow Negligible?

With $x_{A2} = 0$, the ratio of the stagnant-film flux to the Fick-only flux depends only on the mole fraction at the liquid surface:

Figure 4 plots it. The factor is 1.05 at $x_{A1} = 0.10$ and reaches 1.10, a 10 percent correction, at about $x_{A1} = 0.18$. Beyond that it climbs steeply: 1.19 at 0.30 and 1.72 at 0.70. Perry’s remark that bulk flow adds at most about 10 percent is therefore a statement about dilute A. A volatile liquid near its boiling point, or any concentrated vapour, needs the full log form.

Key idea: bulk flow adds less than 10 percent only while $x_{A1}$ is below about 0.18. For a concentrated vapour, ignoring it underestimates the flux badly.

6. Which Model Applies?

The deciding question is what happens to gas B at the boundaries:

• Equimolar counterdiffusion: B moves back as fast as A moves forward, so $N_B = -N_A$. Typical cases are two bulbs or reservoirs joined by a tube at uniform pressure.
• Stagnant film: B cannot cross one boundary, so $N_B = 0$. Typical cases are a liquid evaporating into an insoluble gas, or a gas absorbed by a liquid that does not take up the carrier gas.
• Dilute A: when $x_A$ stays below a few percent, both models give nearly the same flux and the simple Fick form is enough.

Key idea: ask whether B has a source or a sink. If it is exchanged one for one, use equimolar counterdiffusion; if it has nowhere to go, use the stagnant-film result.

Worked Example 3: Evaporation of Diethyl Ether, a Concentrated Vapour

Problem: Diethyl ether (A) sits at the bottom of a narrow vertical tube at 20 °C and 101.325 kPa. Its surface is 0.100 m below the open top, where dry air sweeps the vapour away. Air does not dissolve in the liquid. Find the steady evaporation mass flux $n_A$, and find how far off the Fick-only estimate would be. Data: vapour pressure 59.04 kPa at 20 °C (Perry’s Table 2-8 equation); $D_{AB} = 0.0880\,\text{cm}^{2}\,\text{s}^{-1}$, the Table 2-141 value of 0.0778 cm²/s at 0 °C scaled as $T^{1.75}$; $M_A = 74.12\,\text{g}\,\text{mol}^{-1}$.

Step 1. Total concentration at 293.15 K.

Step 2. The mole fraction and the log term.

Step 3. Molar flux (keep 4 s.f.).

Step 4. Mass flux.

Step 5. Compare with the Fick-only flux.

Final Answer: $n_A \approx 0.237\,\text{g}\,\text{m}^{-2}\,\text{s}^{-1}$ ($N_A \approx 3.20 \times 10^{-3}\,\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$)

Check: bulk flow raises the flux by half. The Fick-only estimate is 33.3% low, far beyond Perry’s 10 percent, because $x_{A1} = 0.5827$ is not dilute. Halfway up the tube the partial pressure of ether is 35.9 kPa, against 29.52 kPa on a straight line (Figure 3).

Your turn: evaporation of a volatile liquid

Same steps as Worked Example 3: the vapour is concentrated, so keep the full log term, find the molar flux to 4 s.f., then multiply by $M_A$ to get a mass flux.

Your turn: the bulk-flow factor

A different route to the same correction: find $p_{B1}$ and $p_{B2}$, take their log mean $p_{BM}$, and divide the total pressure by it. No flux is needed, and the factor has no units.

Common Mistakes

Common mistake: Leaving $D_{AB}$ in cm²/s. Multiply by $10^{-4}$ to get m²/s before you combine it with $c$ in mol m⁻³. In Worked Example 1 the slip gives 0.929 mol m⁻² s⁻¹, 10 000 times too large.

Common mistake: Using °C, or kPa, in the ideal gas law. $c = \frac{P}{RT}$ needs $T$ in K and $P$ in Pa when $R = 8.314$ J mol⁻¹ K⁻¹. With 25 in place of 298.15 the flux comes out about 12 times too large; with $P$ in kPa it comes out 1000 times too small.

Common mistake: Not scaling $D_{AB}$ with pressure. Tabulated gas diffusivities are at 1 atm. At another total pressure, multiply by $\frac{101.325\,\text{kPa}}{P}$ before you use it.

Common mistake: Using the base-10 log. The stagnant-film result uses the natural log. $\log_{10}$ gives a flux 2.303 times too small.

Common mistake: Ignoring bulk flow for a concentrated vapour. The Fick-only flux is fine for dilute A, but in Worked Example 3 it is 33.3% low. Check $x_{A1}$ first: above about 0.18, bulk flow adds more than 10 percent.

Common mistake: Turning the bulk-flow factor upside down. $\frac{P}{p_{BM}}$ is always greater than 1, because $p_{BM}$ is less than $P$. If your factor is below 1, you divided the wrong way.

Common mistake: Choosing the wrong model. Ask what B does. Bulbs at uniform pressure exchange molecules one for one; an insoluble gas over a liquid is stagnant.

Common mistake: Reporting a molar flux when a mass flux is asked for. Multiply $N_A$ in mol m⁻² s⁻¹ by $M_A$ in g mol⁻¹ to get $n_A$ in g m⁻² s⁻¹.

Common mistake: Rounding too early. Keep 4 significant figures in $c$, the mole fractions and the log term; round only the final answer to 3.

Key Takeaway

The total flux is diffusion plus bulk flow, $N_A = J_A + x_A(N_A + N_B)$. When the two gases trade places one for one, bulk flow vanishes and the flux is linear in the mole-fraction difference. When the second gas is stagnant, bulk flow adds to diffusion and the flux carries a natural log, or equivalently the factor $\frac{P}{p_{BM}}$. That factor is close to 1 for a dilute vapour and large for a concentrated one.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson, including problems where you must decide which model applies: 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 flux equations (Eqs. 5-190, 5-197 and 5-198); Section 2 for the gas diffusivities (Table 2-141), the vapour pressures (Tables 2-5 and 2-8) and the molar masses (Table 2-32); Section 1 for the gas constant. The practice problems use illustrative conditions and are audited; see the lesson audit log.