NivaarExam PrepOfficial exam papers ↗

23-Chem-A3 Heat and Mass Transfer · December 2016

Question 6 of 7: BET Surface Area from a Nitrogen Isotherm

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Format: open-book, three-hour paper; seven questions in three parts (Part A Q1–2, Part B Q3–4, Part C Q5–7). A candidate answers one from A, one from B and two from C (four questions, equal value). All seven are solved here as a complete study resource.

Reference texts: R.E. Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — film theory, wetted-wall and flat-plate convective mass transfer, packed-tower absorption, distillation; Coulson & Richardson, Chemical Engineering Vol. 1 (6th ed.) and Vol. 2 (5th ed., Richardson, Harker & Backhurst) — diffusion, drying/evaporation, absorption with reaction, adsorption, membrane separations; C.J. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — Sherwood-number correlations, McCabe–Thiele with side streams; property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 6: BET Surface Area from a Nitrogen Isotherm (Part C — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. N₂ adsorption at 77 K (cm³ liquid per kg solid) versus relative pressure; liquid N₂ density 808 kg/m³; N₂ cross-sectional area $\sigma=0.162$ nm². Particles: 15 nm spheres, 2290 kg/m³ (for a geometric cross-check).

$P/P^{\circ}$0.10.20.30.40.5
cm³ liq N₂/kg66.775.283.993.4108.4

Find. the specific (and hence total) surface area from the BET monolayer capacity.

00.20.40.60.81060120180240300360N₂ isotherm 77 Krelative pressure P/P°adsorbed N₂ (cm³ liq / kg)
Figure 6 — Type-II N₂ isotherm at 77 K; the BET transform of the 0.1–0.3 region gives the monolayer capacity.

Approach. Linearise with the BET equation over $P/P^{\circ}=0.1$–0.3, extract the monolayer volume, convert to moles then molecules, and multiply by the molecular cross-section.

  1. BET linear plot. With $x=P/P^{\circ}$ and $V$ the adsorbed volume, $\dfrac{x}{V(1-x)}=\dfrac{1}{V_mC}+\dfrac{C-1}{V_mC}x$. Regressing the three points (0.1–0.3) gives slope $=1.72\times10^{-2}$ and a negligibly small intercept (large $C$), so $V_m\approx1/\text{slope}$.
  2. Monolayer capacity. $$V_m=\frac{1}{\text{slope}+\text{intercept}}=58.4\ \text{cm}^3\ \text{liq N}_2/\text{kg}.$$
  3. Convert to moles per kg. $n_m=\dfrac{V_m\rho_{L}}{M_{N_2}}=\dfrac{58.4\times10^{-6}(808)}{0.028014}=1.68\ \text{mol/kg}.$
  4. Surface area. $$S=n_mN_A\sigma=1.68(6.022\times10^{23})(0.162\times10^{-18})=\boxed{1.64\times10^{5}\ \text{m}^2/\text{kg}\ (164\ \text{m}^2/\text{g})}.$$
  5. Geometric cross-check. For 15 nm spheres, $S_{geo}=\dfrac{6}{\rho_p d}=\dfrac{6}{2290(15\times10^{-9})}=1.75\times10^{5}$ m²/kg — within 7% of the BET value, confirming the particles are essentially non-porous 15 nm spheres.
QuantityResult
Monolayer capacity $V_m$58.4 cm³ liq N₂/kg (1.68 mol/kg)
BET specific area$1.64\times10^{5}$ m²/kg (164 m²/g)
Geometric area (15 nm spheres)$1.75\times10^{5}$ m²/kg (175 m²/g)