23-Chem-A2 Unit Operations and Separation Processes · December 2018
Question 6 of 6: Adsorption Isotherm Selection and Surface Area
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-CHEM-A2, Unit Operations and Separation Processes — December 2018, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 continuous-thickener sizing, A2 series-pipe friction pressure drop, A3 packed/fluidized-bed pressure drop & power; Part B (Separation Processes) — B1 flash vs. differential distillation, B2 batch-drying time, B3 adsorption isotherm fitting & surface area. The rubric asks for two problems per part; all six are worked in full below.
Reference texts (subject). McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 2 (5th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (§6 fluid & particle mechanics, §12 drying, §16 adsorption, §18 solid–liquid separation).
Question B3: Adsorption Isotherm Selection and Surface Area (25 points)
Given. Ten pressure–uptake pairs (above); ethane $M=30.07$ g/mol; liquid density $354.9$ kg/m³ ($0.3549$ g/cm³); $N_A=6.023\times10^{23}$; STP molar volume $22{,}414$ cm³/mol.
Find. (a) which isotherm (Langmuir or Freundlich) fits the data, by linearising both and comparing $R^2$; (b) the total specific surface area of the sieve from the monolayer capacity.
Figure B3 — The uptake clearly saturates toward a monolayer plateau; the Langmuir linear plot $P/q$ vs. $P$ is straighter ($R^2=0.998$) than the Freundlich $\ln q$ vs. $\ln P$ ($R^2=0.988$). The Langmuir slope gives the monolayer capacity $q_m$.
Approach. Linearise each model, fit by least squares, and compare $R^2$; then convert the winning model's monolayer capacity $q_m$ to molecules per gram and multiply by the ethane molecular cross-sectional area (from the liquid density) to get the surface area.
(a) Langmuir linearisation. $q=\dfrac{q_m K P}{1+KP}\Rightarrow\dfrac{P}{q}=\dfrac{1}{q_m}P+\dfrac{1}{q_mK}$. A straight-line fit of $P/q$ vs. $P$ gives slope $1/q_m$ and intercept $1/(q_mK)$: $$R^2=0.9976,\quad q_m=\boxed{60.0\ \text{cm}^3\text{STP/g}},\quad K=5.6\times10^{-3}\ \text{mmHg}^{-1}.$$
Model selection. Both are respectable, but Langmuir fits better ($R^2=0.9976>0.9884$) and — decisively — the data plateau at high pressure (uptake approaches ~46 and extrapolates to $q_m\approx60$), the physical signature of monolayer saturation that Freundlich (a power law with no ceiling) cannot represent. Langmuir isotherm governs, with $q_m=60.0$ cm³STP/g.
(b) Molecules adsorbed at monolayer. Convert the monolayer capacity to moles then molecules per gram: $$n_m=\frac{q_m}{22{,}414}=\frac{60.0}{22{,}414}=2.68\times10^{-3}\ \text{mol/g},\quad N=n_mN_A=1.61\times10^{21}\ \text{molecules/g}.$$
Molecular cross-sectional area (from liquid density). The area per molecule is $a_m=1.091\left(\dfrac{M}{\rho_L N_A}\right)^{2/3}$: $$a_m=1.091\left(\frac{30.07}{0.3549\,(6.023\times10^{23})}\right)^{2/3}=2.95\times10^{-15}\ \text{cm}^2\ (0.295\ \text{nm}^2).$$
Total specific surface area. $$S=N\,a_m=1.61\times10^{21}(2.95\times10^{-15})=4.76\times10^{6}\ \text{cm}^2/\text{g}=\boxed{476\ \text{m}^2/\text{g}}.$$ This is the right order for a 5A molecular sieve, confirming the monolayer interpretation.
