Question 1 of 7: Knudsen and molecular diffusion of N₂ through a capillary
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 04-Chem-A3 (Mass Transfer Operations). Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Every property datum (diffusivities, Henry’s constants, solubilities, packing constants) is supplied in the question or its data table, so each answer is self-contained.
Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — Knudsen/molecular diffusion, convective mass-transfer coefficients, gas absorption in packed towers; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed., Wiley) — boundary-layer analogies and dimensional analysis; Bird, Stewart & Lightfoot, Transport Phenomena (2nd ed., Wiley) — Chapman–Enskog kinetic theory; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — Sherwood–Holloway packed-tower correlation; supporting data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 1: Knudsen and molecular diffusion of N₂ through a capillary (Part A — equal value)
Given. An open capillary connects a pure-N₂ reservoir ($x_{A1}=1$) to a pure-He reservoir ($x_{A2}=0$); the transport is isobaric counter-diffusion in which, by Graham’s law, $N_B/N_A=-\sqrt{M_A/M_B}$. As the total pressure falls, the mean free path grows relative to the pore radius and the mechanism shifts from ordinary (molecular) diffusion toward Knudsen (wall-collision) diffusion.
Quantity
Value
Temperature $T$
298 K
Capillary length $L$
0.10 m
Pore diameter (radius $r$)
$10\ \mu$m ($r=5\times10^{-6}$ m)
Binary diffusivity $D_{AB}$ at 1 atm
$6.98\times10^{-5}$ m²/s
Molar masses $M_A$ (N₂), $M_B$ (He)
0.02802, 0.004003 kg/mol
End compositions $x_{A1},\ x_{A2}$
1.0, 0
Find. Knudsen diffusivities; the steady-state flux at 0.001, 0.1, 10 atm; the log–log flux–pressure plot with both limiting asymptotes; and the two limiting-region fluxes.
Figure 1. Combined-regime flux (solid) with the Knudsen ($N_A\propto P$, slope +1) and molecular ($N_A\to$ const) asymptotes. The three computed operating points at 0.001, 0.1 and 10 atm trace the transition from wall-collision to bulk-collision control.
Approach. Knudsen diffusivity depends only on pore radius, $T$ and $M$ (not $P$); the combined-regime flux uses the series-resistance form of the open-capillary equation with the Graham-law ratio $\alpha=1-\sqrt{M_A/M_B}$; the two asymptotes fall out by letting $D_{AB}\!\to\!\infty$ (Knudsen limit) or $D_{K}\!\to\!\infty$ (molecular limit).
(a) Knudsen diffusivities. With the mean molecular speed $\bar v=\sqrt{8RT/\pi M}$, $$D_K=\frac{2}{3}\,r\,\bar v=\frac{2}{3}\,r\sqrt{\frac{8RT}{\pi M}}.$$ For N₂: $D_{KA}=\tfrac{2}{3}(5\times10^{-6})\sqrt{8(8.314)(298)/[\pi(0.02802)]}=\boxed{1.58\times10^{-3}\ \text{m}^2/\text{s}}$; for He: $D_{KB}=4.19\times10^{-3}$ m²/s. Because $r$, $T$ and $M$ are fixed, these values are identical at 0.001, 0.1 and 10 atm — Knudsen diffusion is pressure-independent.
Graham-law flux ratio. In an open, isobaric system the two species counter-diffuse with $N_B/N_A=-\sqrt{M_A/M_B}$, so $$\alpha=1-\sqrt{M_A/M_B}=1-\sqrt{0.02802/0.004003}=-1.646.$$
(b) Combined-regime flux. The transition-regime flux for the open capillary is $$N_A=\frac{D_{AB}P}{\alpha RTL}\,\ln\!\frac{1-\alpha x_{A2}+D_{AB}/D_{KA}}{1-\alpha x_{A1}+D_{AB}/D_{KA}},$$ where the molecular diffusivity scales as $D_{AB}(P)=D_{AB,1\text{atm}}/P_{\text{atm}}$ and the product $D_{AB}P$ is constant. Evaluating: at $P=0.001$ atm $N_A=\boxed{6.23\times10^{-4}}$; at $0.1$ atm $N_A=1.32\times10^{-2}$; at $10$ atm $N_A=1.68\times10^{-2}$ mol/m²·s. The flux rises steeply at low $P$ then saturates.
(c)–(d) Log–log plot. Plotted in Figure 1: at low pressure the points lie on the Knudsen asymptote of unit slope ($N_A\propto P$); at high pressure they flatten onto the horizontal molecular asymptote. The 0.1-atm point sits in the knee of the transition.
(e) Limiting-region fluxes. Pure Knudsen flux $N_A=D_{KA}P(x_{A1}-x_{A2})/(RTL)$ gives $6.47\times10^{-4}$ at 0.001 atm and $6.47\times10^{-2}$ at 0.1 atm (linear in $P$). The molecular limit ($D_K\to\infty$) is $$N_A=\frac{D_{AB}P}{\alpha RTL}\ln\frac{1-\alpha x_{A2}}{1-\alpha x_{A1}}=\boxed{1.69\times10^{-2}\ \text{mol/m}^2\text{s}},$$ a constant, since $D_{AB}P$ is fixed.
