CHE 318 Lecture 03

Molecular Diffusion in Gases: General Solutions

Author

Dr. Tian Tian

Published

January 9, 2026

Note

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Recap

Governing equation for diffusion binary mixture gas systems \[ N_{A} = -c_T D_{AB} \dfrac{d x_A}{dz} + x_A(N_A + N_B) \]

Case 1: equimolar counter diffusion (EMCD)
- $N_A + N_B = 0$
- $N_A = -c_T D_{AB} \dfrac{d x_A}{d z}$
- Linear $x_A$, $c_A$ and $p_A$ profiles
- $N_A = \dfrac{D_{AB}}{RT(z_2 - z_1)}(p_{A1} - p_{A2})$

Case 2: diffusion through stagnant B
- $N_B = 0$
- $N_A(1 - x_A) = -c_T D_{AB}\dfrac{d x_A}{dz}$
- Usually in log-mean pressure form
- $N_A = \dfrac{D_{AB}}{RT(z_2-z_1)} \dfrac{p_T}{p_{Bm}} \left(p_{A1} - p_{A2}\right)$

Office Hour

Wednesday 11:15 - 12:15
DICE 12-245
Can take place in other meeting rooms if many people are attending
Please use seminar time wisely for assignment questions!

Learning Outcomes

After today’s lecture, you will be able to:

Derive general solution for mass transport in binary gas mixture
Define the conditions for each cases we have studied so far
Formulate governing equations for various case systems
Apply general solution to defined problem
Analyze pressure / concentration profile

Deriving the General Solution (I)

General case for steady state (S.S) transport

$N_B = k N_A$ ($k$ is a constant. why?)
Only $x_A$ (or $c_A$, $p_A$) dependent on $z$, not $N_A$ or $N_B$ 👉 separation of variables!

\[\begin{align} N_A - x_A(N_A + N_B) &= -D_{AB}c_T \dfrac{d x_A}{dz}\\ \frac{dz}{D_{AB} c_T} &= -\dfrac{dx_A}{N_A - x_A (N_A + N_B)} \\ \int_{z_1}^{z_2} \frac{dz}{D_{AB} c_T} &= \int_{x_{A1}}^{x_{A2}} -\dfrac{dx_A}{N_A - x_A (N_A + N_B)} \end{align}\]

Deriving the General Solution (II)

L.H.S.

\[ \int_{z_1}^{z_2} \frac{dz}{D_{AB} c_T} = \frac{(z_2 - z_1)}{D_{AB} c_T} \]

R.H.S.

Using substitution $u = N_A - x_A (N_A + N_B)$, $d u = -(N_A + N_B)\,dx_A$

\[\begin{align} \int_{x_{A1}}^{x_{A2}} -\dfrac{dx_A}{N_A - x_A (N_A + N_B)} &= - \cdot - \frac{1}{N_A + N_B} \ln\!\left[N_A - x_A (N_A + N_B)\right] \bigg\vert_{x_{A1}}^{x_{A2}} \\ &= \frac{1}{N_A + N_B} \ln\!\left[\dfrac{ N_A - x_{A2} (N_A + N_B)} {N_A - x_{A1} (N_A + N_B)} \right] \end{align}\]

Deriving the General Solution (III)

We want to solve for $N_A$

\[\begin{align} \frac{(z_2 - z_1)}{c_T D_{AB}} &= \frac{1}{N_A + N_B} \ln\!\left[\dfrac{ N_A - x_{A2} (N_A + N_B)} {N_A - x_{A1} (N_A + N_B)} \right] \\ N_A &= \frac{c_T D_{AB}}{(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ N_A - x_{A2} (N_A + N_B)} {N_A - x_{A1} (N_A + N_B)} \right] \end{align}\]

We can gather all $N_A / (N_A + N_B)$ in the R.H.S since it’s a constant

\[\begin{align} \boxed{ N_A = \frac{ c_T D_{AB}}{(z_2 - z_1)} \left(\frac{N_A}{N_A + N_B}\right) \% \ln\!\left[ \dfrac{\frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1}} \right] } \end{align}\]

This is the general solution for binary gas transport.
Also works for $N_A = -N_B$ (EMCD), with a little bit trick

Relation Between Gen. Sol. and EMCD

The general solution is not applicable $N_B = -N_A$, but we can prove $N_B / N_A \to -1$ reduces to the EMCD equation.

Let $s = N_A / (N_A + N_B)$, so $s \to \infty$ when $N_B/N_A \to -1$
Use the Taylor expansion that $\lim_{u \to 0} \ln(1-u) = -u$

\[\begin{align} \lim_{s \to \infty} \left(\frac{N_A}{N_A + N_B}\right) \ln\!\left[ \dfrac{ \frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1} } \right] &= s \ln\!\left(\frac{s - x_2}{s - x_1}\right) \\ &= s \ln\!\left(\frac{1 - x_2/s}{1 - x_1/s}\right) \\ &= s [(-\frac{x_2}{s}) - (-\frac{x_1}{s})] \\ &= x_1 - x_2 \end{align}\]

This is the EMCD result!

Three Forms of Gass Mass Transport Equations (I)

In $c_A$ form

Case 1: EMCD
- $N_A + N_B = 0$ \[ N_A = \dfrac{D_{AB}}{(z_2 - z_1)}(c_{A1} - c_{A2}) \]
Case 2: stagnant B
- $N_B = 0$
\[ N_A = \frac{D_{AB} c_T }{(z_2 - z_1)} \ln\! \left( \frac{1 - \frac{c_{A2}}{c_T} } {1-\frac{c_{A1}}{c_T}} \right) \]

General solution
- $N_B = k N_A$

\[ N_A = \frac{D_{AB} c_T}{(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ \frac{N_A}{N_A + N_B} - \frac{c_{A2}}{c_T}} {\frac{N_A}{N_A + N_B} - \frac{c_{A1}}{c_T}} \right] \]

Three Forms of Gass Mass Transport Equations (II)

In $x_A$ form

Case 1: EMCD
- $N_A + N_B = 0$ \[ N_A = \dfrac{D_{AB} c_T}{(z_2 - z_1)}(x_{A1} - x_{A2}) \]
Case 2: stagnant B
- $N_B = 0$
\[ N_A = \frac{D_{AB} c_T }{(z_2 - z_1)} \ln\! \left( \frac{1 - x_{A2} } {1- x_{A1}} \right) \]

General solution
- $N_B = \lambda N_A$

\[ N_A = \frac{D_{AB} c_T}{(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ \frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1}} \right] \]

Three Forms of Gass Mass Transport Equations (III)

In $p_A$ form (ideal gas)

Case 1: EMCD
- $N_A + N_B = 0$ \[ N_A = \dfrac{D_{AB}}{RT(z_2 - z_1)}(p_{A1} - p_{A2}) \]
Case 2: stagnant B
- $N_B = 0$
\[ N_A = \frac{D_{AB} p_T}{RT(z_2 - z_1)} \ln\! \left( \frac{p_T - p_{A2}} {p_T - p_{A1}} \right) \]

General solution
- $N_B = \lambda N_A$

\[ N_A = \frac{D_{AB} p_T}{RT(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ \frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1}} \right] \]

General Solution – $x_A(z)$ Profile

In all the cases we study the S.S. condition:

\[\frac{d N_A}{d z} = 0\]

For the general case $s = N_A / (N_A + N_B)$, we have

\[\begin{align} \frac{dN_A}{dz} &= \frac{d}{dz} \left[ \frac{D_{AB} c_T}{s - x_A} \frac{d x_A}{d z}\right] = 0\\ \frac{D_{AB} c_T}{s - x_A} \frac{d x_A}{d z} &= \mathrm{Const} \end{align}\]

$x_A(z)$ has a general solution with exponential form (with $K_1$ and $K_2$ being constants)

\[\begin{align} \boxed{ x_{A} = s - K_1 e^{K_2 z} } \end{align}\]

$x_A(z)$ Profile

Notes on the Relation Between $N_A$ and $N_B$ (1)

Using steady state condition, we need to know relation between $N_A$ and $N_B$ before solving the general EQ.

It depends on the system setup and mass balance.

Let’s consider the same chemical reaction of hydrogen dissociation by a solid catalst

\[\begin{align} \text{H}_2\ (A, \text{gas}) \rightarrow 2\text{H}^*\ (B, \text{gas}) \end{align}\]

Case 1: reaction through a solid catalyst at bottom of a tube
- The flux of A and B have opposite signs
- $N_B = -2 N_A$

Notes on the Relation Between $N_A$ and $N_B$ (2)

Case 2: gas flow through a solid catalyst inside a tube
- The flux of A and B have the same sign
- $N_B = 2 N_A$

Take home question: - How do the flux directions affect the general solution?

Measuring Diffusivity $D_{AB}$ In Gases

Several experimental methods exist for gases, evaporated liquid and sublimated solids (see Geankoplis Ch 6.2E)
We will introduce the two-bulb method for gases
Typical setup see Duncan & Toor AIChE J. 1962, 8, 38-41
Which case should we apply?

Two-Bulb Experiment: Conditions

Conditions for the two-bulb experiment

Initially, pure A in bulb 1 and pure B and bulb 2
$t=0$, the valve is opened slowly to minimize convection
$t=t_e$, the valve is closed and the composition in 1 and 2 are analyzed
The valve open time is much shorter than $t_2 - t_1$
The tube is a very thin capillary 👉 no convection
Constant $p_T$ and $T$ throughout experiment

Two-Bulb Experiment: Pseudo Steady State

We can apply:

EMCD condition (no convection)
Pseudo-steady state (P.S.S) assumption
- At each time step, we assume $N_A$ or $J_{Az}^*$ is constant (linear $c_A$ profile)
- But net $N_A$ flux accumulates A molecules in bulb 2
- $c_A$ in bulb 2 changes over time

Two-Bulb Experiment: Governing Equations

We use the mass balance of the system

\[\begin{align*} \text{[Mass In]} &= \text{[Accumulation]} \\ S \cdot J_{Az}^* &= V_2 \frac{d c_{A2}}{dt} \\ D_{AB} \frac{c_{A1} - c_{A2}}{L} \cdot S &= V_2 \frac{d c_{A2}}{dt} \end{align*}\]

Since the total pressure $p_T$ is constant, the average concentration of $c_A$ in the system, $c_{A, av}$ is also constant

\[\begin{align*} V_T c_{A, av} &= (V_1 + V_2)c_{A, av} \\ &= V_1c_{A1}(t=0) + V_2c_{A2}(t=0) \\ &= V_1c_{A1}(t\neq0) + V_2c_{A2}(t\neq0) \end{align*}\]

Two-Bulb Experiment: Solutions

We are interested in $c_{A2}$, so substitute $c_{A1}$ with relation to $c_{A, av}$:

\[ c_{A1} = \frac{V_T c_{A, av} - V_2 c_{A2}}{V_1} \]

After the substitution, rearrangement and integration we get:

\[\begin{align*} \ln\!\left[ \frac{c_{A, av} - c_{A2}(t=t_e)}{c_{A, av} - c_{A2}(t=0)} \right] &= -\frac{D_{AB} V_T S}{V_1 V_2 L} \cdot t \\ \end{align*}\]

We can extract $D_{AB}$ usingt the slope of the plot!

Some Typical Diffusivity Measurements

Summary

General solution for $N_B = k N_A$
Like stagnant B case, general solution take a logarithm form for $N_A$
When $x_{A1}$ and $x_{A_2}$ are determined, the flux $N_A$ is controlled by the relation between $N_B$ and $N_A$
The ratio between $N_B$ and $N_A$ depends on the system and setup
Diffusivity $D_{AB}$ can be determined by EMCD through two-bulb setup

Preview of Next Class: Theories for predicting $D_{AB}$

Kinetic theory

\[ D_{AB} = \frac{1}{3} \overline{u} \lambda \]

Chapman-Enskog theory

\[ D_{AB} \propto \frac{T^{\frac{3}{2}}}{P} \left(\frac{1}{m_A} + \frac{1}{m_B} \right)^{\frac{1}{2}} \]

Fuller method

\[ D_{AB} \propto \frac{T^{1.75}}{P} \left(\frac{1}{m_A} + \frac{1}{m_B} \right)^{\frac{1}{2}} \]

Footnotes

Geankoplis Table 6.2-1↩︎

--- title: "CHE 318 Lecture 03" subtitle: "Molecular Diffusion in Gases: General Solutions" author: "Dr. Tian Tian" date: "2026-01-09" format: html: {} revealjs: output-file: slides.html pdf: output-file: L03.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L03.pdf) - Handwritten notes 👉 [Open in pdf](./public/L03_annotated.pdf) ::: ::: ## Recap - **Governing equation** for diffusion binary mixture gas systems $$ N_{A} = -c_T D_{AB} \dfrac{d x_A}{dz} + x_A(N_A + N_B) $$ ::: {.columns} ::: {.column width="50%"} - Case 1: equimolar counter diffusion (EMCD) - $N_A + N_B = 0$ - $N_A = -c_T D_{AB} \dfrac{d x_A}{d z}$ - Linear $x_A$, $c_A$ and $p_A$ profiles - $N_A = \dfrac{D_{AB}}{RT(z_2 - z_1)}(p_{A1} - p_{A2})$ ::: ::: {.column width="50%"} - Case 2: diffusion through stagnant B - $N_B = 0$ - $N_A(1 - x_A) = -c_T D_{AB}\dfrac{d x_A}{dz}$ - Usually in log-mean pressure form - $N_A = \dfrac{D_{AB}}{RT(z_2-z_1)} \dfrac{p_T}{p_{Bm}} \left(p_{A1} - p_{A2}\right) $ ::: ::: --- ## Office Hour - Wednesday 11:15 - 12:15 - DICE 12-245 - Can take place in other meeting rooms if many people are attending - Please use seminar time wisely for assignment questions! ## Learning Outcomes {.center} After today's lecture, you will be able to: - **Derive** general solution for mass transport in binary gas mixture - **Define** the conditions for each cases we have studied so far - **Formulate** governing equations for various case systems - **Apply** general solution to defined problem - **Analyze** pressure / concentration profile ## Deriving the General Solution (I) General case for steady state (S.S) transport - $N_B = k N_A$ ($k$ is a constant. *why?*) - Only $x_A$ (or $c_A$, $p_A$) dependent on $z$, not $N_A$ or $N_B$ 👉 separation of variables! ```{=tex} \begin{align} N_A - x_A(N_A + N_B) &= -D_{AB}c_T \dfrac{d x_A}{dz}\\ \frac{dz}{D_{AB} c_T} &= -\dfrac{dx_A}{N_A - x_A (N_A + N_B)} \\ \int_{z_1}^{z_2} \frac{dz}{D_{AB} c_T} &= \int_{x_{A1}}^{x_{A2}} -\dfrac{dx_A}{N_A - x_A (N_A + N_B)} \end{align} ``` ## Deriving the General Solution (II) - L.H.S. $$ \int_{z_1}^{z_2} \frac{dz}{D_{AB} c_T} = \frac{(z_2 - z_1)}{D_{AB} c_T} $$ - R.H.S. Using substitution $u = N_A - x_A (N_A + N_B)$, $d u = -(N_A + N_B)\,dx_A$ ```{=tex} \begin{align} \int_{x_{A1}}^{x_{A2}} -\dfrac{dx_A}{N_A - x_A (N_A + N_B)} &= - \cdot - \frac{1}{N_A + N_B} \ln\!\left[N_A - x_A (N_A + N_B)\right] \bigg\vert_{x_{A1}}^{x_{A2}} \\ &= \frac{1}{N_A + N_B} \ln\!\left[\dfrac{ N_A - x_{A2} (N_A + N_B)} {N_A - x_{A1} (N_A + N_B)} \right] \end{align} ``` ## Deriving the General Solution (III) We want to solve for $N_A$ ```{=tex} \begin{align} \frac{(z_2 - z_1)}{c_T D_{AB}} &= \frac{1}{N_A + N_B} \ln\!\left[\dfrac{ N_A - x_{A2} (N_A + N_B)} {N_A - x_{A1} (N_A + N_B)} \right] \\ N_A &= \frac{c_T D_{AB}}{(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ N_A - x_{A2} (N_A + N_B)} {N_A - x_{A1} (N_A + N_B)} \right] \end{align} ``` We can gather all $N_A / (N_A + N_B)$ in the R.H.S since it's a constant ```{=tex} \begin{align} \boxed{ N_A = \frac{ c_T D_{AB}}{(z_2 - z_1)} \left(\frac{N_A}{N_A + N_B}\right) \% \ln\!\left[ \dfrac{\frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1}} \right] } \end{align} ``` - This is the general solution for binary gas transport. - Also works for $N_A = -N_B$ (EMCD), with a little bit trick ## Relation Between Gen. Sol. and EMCD The general solution is not applicable $N_B = -N_A$, but we can prove $N_B / N_A \to -1$ reduces to the EMCD equation. - Let $s = N_A / (N_A + N_B)$, so $s \to \infty$ when $N_B/N_A \to -1$ - Use the Taylor expansion that $\lim_{u \to 0} \ln(1-u) = -u$ ```{=tex} \begin{align} \lim_{s \to \infty} \left(\frac{N_A}{N_A + N_B}\right) \ln\!\left[ \dfrac{ \frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1} } \right] &= s \ln\!\left(\frac{s - x_2}{s - x_1}\right) \\ &= s \ln\!\left(\frac{1 - x_2/s}{1 - x_1/s}\right) \\ &= s [(-\frac{x_2}{s}) - (-\frac{x_1}{s})] \\ &= x_1 - x_2 \end{align} ``` This **is** the EMCD result! ## Three Forms of Gass Mass Transport Equations (I) In $c_A$ form ::: {.columns} ::: {.column width="40%"} - Case 1: EMCD - $N_A + N_B = 0$ $$ N_A = \dfrac{D_{AB}}{(z_2 - z_1)}(c_{A1} - c_{A2}) $$ - Case 2: stagnant B - $N_B = 0$ $$ N_A = \frac{D_{AB} c_T }{(z_2 - z_1)} \ln\! \left( \frac{1 - \frac{c_{A2}}{c_T} } {1-\frac{c_{A1}}{c_T}} \right) $$ ::: ::: {.column width="60%"} - General solution - $N_B = k N_A$ $$ N_A = \frac{D_{AB} c_T}{(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ \frac{N_A}{N_A + N_B} - \frac{c_{A2}}{c_T}} {\frac{N_A}{N_A + N_B} - \frac{c_{A1}}{c_T}} \right] $$ ::: ::: ## Three Forms of Gass Mass Transport Equations (II) In $x_A$ form ::: {.columns} ::: {.column width="40%"} - Case 1: EMCD - $N_A + N_B = 0$ $$ N_A = \dfrac{D_{AB} c_T}{(z_2 - z_1)}(x_{A1} - x_{A2}) $$ - Case 2: stagnant B - $N_B = 0$ $$ N_A = \frac{D_{AB} c_T }{(z_2 - z_1)} \ln\! \left( \frac{1 - x_{A2} } {1- x_{A1}} \right) $$ ::: ::: {.column width="60%"} - General solution - $N_B = \lambda N_A$ $$ N_A = \frac{D_{AB} c_T}{(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ \frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1}} \right] $$ ::: ::: ## Three Forms of Gass Mass Transport Equations (III) In $p_A$ form (ideal gas) ::: {.columns} ::: {.column width="40%"} - Case 1: EMCD - $N_A + N_B = 0$ $$ N_A = \dfrac{D_{AB}}{RT(z_2 - z_1)}(p_{A1} - p_{A2}) $$ - Case 2: stagnant B - $N_B = 0$ $$ N_A = \frac{D_{AB} p_T}{RT(z_2 - z_1)} \ln\! \left( \frac{p_T - p_{A2}} {p_T - p_{A1}} \right) $$ ::: ::: {.column width="60%"} - General solution - $N_B = \lambda N_A$ $$ N_A = \frac{D_{AB} p_T}{RT(z_2 - z_1)} \frac{N_A}{N_A + N_B} \ln\!\left[\dfrac{ \frac{N_A}{N_A + N_B} - x_{A2}} {\frac{N_A}{N_A + N_B} - x_{A1}} \right] $$ ::: ::: ## General Solution -- $x_A(z)$ Profile In all the cases we study the S.S. condition: $$\frac{d N_A}{d z} = 0$$ For the general case $s = N_A / (N_A + N_B)$, we have ```{=tex} \begin{align} \frac{dN_A}{dz} &= \frac{d}{dz} \left[ \frac{D_{AB} c_T}{s - x_A} \frac{d x_A}{d z}\right] = 0\\ \frac{D_{AB} c_T}{s - x_A} \frac{d x_A}{d z} &= \mathrm{Const} \end{align} ``` $x_A(z)$ has a general solution with exponential form (with $K_1$ and $K_2$ being constants) ```{=tex} \begin{align} \boxed{ x_{A} = s - K_1 e^{K_2 z} } \end{align} ``` --- ## $x_A(z)$ Profile ```{=html} <iframe width="100%" height="800" src="../../scripts/L03_xa_profile.html" title="Webpage example"></iframe> ``` --- ## Notes on the Relation Between $N_A$ and $N_B$ (1) Using steady state condition, we need to know relation between $N_A$ and $N_B$ *before* solving the general EQ. It depends on the system setup and mass balance. Let's consider the same chemical reaction of hydrogen dissociation by a solid catalst ```{=tex} \begin{align} \text{H}_2\ (A, \text{gas}) \rightarrow 2\text{H}^*\ (B, \text{gas}) \end{align} ``` - Case 1: reaction through a solid catalyst at bottom of a tube - The flux of A and B have **opposite** signs - $N_B = -2 N_A$ ## Notes on the Relation Between $N_A$ and $N_B$ (2) - Case 2: gas flow through a solid catalyst inside a tube - The flux of A and B have the **same** sign - $N_B = 2 N_A$ Take home question: - How do the flux directions affect the general solution? --- ## Measuring Diffusivity $D_{AB}$ In Gases - Several experimental methods exist for gases, evaporated liquid and sublimated solids (see *Geankoplis* Ch 6.2E) - We will introduce the two-bulb method for gases - Typical setup see Duncan & Toor *AIChE J.* 1962, 8, 38-41 - Which case should we apply? --- ## Two-Bulb Experiment: Conditions Conditions for the two-bulb experiment - Initially, pure A in bulb 1 and pure B and bulb 2 - $t=0$, the valve is opened slowly to minimize convection - $t=t_e$, the valve is closed and the composition in 1 and 2 are analyzed - The valve open time is much shorter than $t_2 - t_1$ - The tube is a very thin capillary 👉 no convection - Constant $p_T$ and $T$ throughout experiment --- ## Two-Bulb Experiment: Pseudo Steady State We can apply: - EMCD condition (no convection) - *Pseudo*-steady state (P.S.S) assumption - At each time step, we assume $N_A$ or $J_{Az}^*$ is constant (linear $c_A$ profile) - But net $N_A$ flux accumulates A molecules in bulb 2 - $c_A$ in bulb 2 **changes** over time --- ## Two-Bulb Experiment: Governing Equations We use the mass balance of the system ```{=tex} \begin{align*} \text{[Mass In]} &= \text{[Accumulation]} \\ S \cdot J_{Az}^* &= V_2 \frac{d c_{A2}}{dt} \\ D_{AB} \frac{c_{A1} - c_{A2}}{L} \cdot S &= V_2 \frac{d c_{A2}}{dt} \end{align*} ``` Since the total pressure $p_T$ is constant, the average concentration of $c_A$ in the system, $c_{A, av}$ is also constant ```{=tex} \begin{align*} V_T c_{A, av} &= (V_1 + V_2)c_{A, av} \\ &= V_1c_{A1}(t=0) + V_2c_{A2}(t=0) \\ &= V_1c_{A1}(t\neq0) + V_2c_{A2}(t\neq0) \end{align*} ``` --- ## Two-Bulb Experiment: Solutions We are interested in $c_{A2}$, so substitute $c_{A1}$ with relation to $c_{A, av}$: $$ c_{A1} = \frac{V_T c_{A, av} - V_2 c_{A2}}{V_1} $$ After the substitution, rearrangement and integration we get: ```{=tex} \begin{align*} \ln\!\left[ \frac{c_{A, av} - c_{A2}(t=t_e)}{c_{A, av} - c_{A2}(t=0)} \right] &= -\frac{D_{AB} V_T S}{V_1 V_2 L} \cdot t \\ \end{align*} ``` We can extract $D_{AB}$ usingt the slope of the plot! --- ## Some Typical Diffusivity Measurements ![](public/geankoplis-gas-diff.png){width=700} ^[Geankoplis Table 6.2-1] --- ## Summary - General solution for $N_B = k N_A$ - Like stagnant B case, general solution take a **logarithm** form for $N_A$ - When $x_{A1}$ and $x_{A_2}$ are determined, the flux $N_A$ is controlled by the relation between $N_B$ and $N_A$ - The ratio between $N_B$ and $N_A$ depends on the system and setup - Diffusivity $D_{AB}$ can be determined by EMCD through two-bulb setup --- ## Preview of Next Class: Theories for predicting $D_{AB}$ - Kinetic theory $$ D_{AB} = \frac{1}{3} \overline{u} \lambda $$ - Chapman-Enskog theory $$ D_{AB} \propto \frac{T^{\frac{3}{2}}}{P} \left(\frac{1}{m_A} + \frac{1}{m_B} \right)^{\frac{1}{2}} $$ - Fuller method $$ D_{AB} \propto \frac{T^{1.75}}{P} \left(\frac{1}{m_A} + \frac{1}{m_B} \right)^{\frac{1}{2}} $$