Question 7 of 7: Adsorption of acetone vapour on activated carbon
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. EGBC 04-CHEM-A3 Mass Transfer Operations, May 2016 — three-hour open-book exam. Seven questions in three parts: Part A (Q1–2, molecular diffusion), Part B (Q3–4, film mass transfer), Part C (Q5–7, staged and equilibrium separations). The candidate answers one of Q1–2, one of Q3–4 and two of Q5–7 (four questions of equal value). All seven are worked in full below.
Reference texts. J.M. Coulson & J.F. Richardson, Chemical Engineering Vol. 1 (Fluid Flow, Heat and Mass Transfer) and Vol. 2 (Particle Technology & Separation Processes) — the source of all seven problems; C.J. Geankoplis, Transport Processes and Separation Process Principles; R.E. Treybal, Mass-Transfer Operations; Bird, Stewart & Lightfoot, Transport Phenomena; Perry’s Chemical Engineers’ Handbook.
Question 7 — Adsorption of acetone vapour on activated carbon (25 marks)
The isotherm pressures are printed “N/m²” but are taken as kN/m²: only then are they consistent with $p^{*}=37.9\text{ kN/m}^2$ and the stated 40%/5% relative saturations (40% → $0.40\times37.9=15.2\text{ kN/m}^2$, which lies inside the 0–90 table range). Read as N/m², the table would cover only $\sim0.24\%$ saturation and the problem could not be solved.
Find. (a) carbon mass to bring $s$ from 40% to 5%; (b) equilibrium $s$ when 1.6 kg carbon is added.
$p_A$ (kN/m²)
0
5
10
30
50
90
$w$ (kg/kg)
0
0.14
0.19
0.27
0.31
0.35
Freundlich fit $w=0.088\,p^{0.317}$ to the acetone-on-carbon data, with the 40% and 5% relative-saturation operating pressures marked.
Approach. Convert relative saturation to acetone partial pressure and gas-phase moles via the ideal-gas law; the air amount is fixed, so an acetone mass balance between the gas and the adsorbed phase (loading read from the fitted isotherm) gives the carbon mass in (a) and the equilibrium pressure in (b).
Acetone initially in the gas. $p_{A,0}=0.40(37.9)=15.2\text{ kN/m}^2$; $n=\dfrac{p_{A,0}V}{RT}=\dfrac{15\,160(1)}{8.314(303)}=6.02\text{ mol}=0.349\text{ kg}.$
Acetone left at 5% saturation. $p_{A,f}=0.05(37.9)=1.90\text{ kN/m}^2\Rightarrow n_f=0.75\text{ mol}=0.044\text{ kg}$, so $0.305\text{ kg}$ must be adsorbed.
Carbon mass (part a). The carbon equilibrates with the final gas ($p_{A,f}=1.90\text{ kN/m}^2$), giving loading $w=0.088(1.90)^{0.317}=0.108\text{ kg/kg}$. Hence
$$m_C=\frac{\text{acetone adsorbed}}{w}=\frac{0.305}{0.108}=\boxed{2.8\ \text{kg activated carbon.}}$$
The final pressure of 1.90 kN/m² lies below the lowest tabulated point (5 kN/m²), so this loading is read from the fitted curve, which follows the steep rise the data show near the origin. A straight line from the origin to the 5 kN/m² point would give only 0.053 kg/kg and about 5.8 kg of carbon, but it ignores that curvature. Part (a) is therefore sensitive to how the isotherm is drawn below 5 kN/m²; the fitted curve gives the best estimate.
Only 1.6 kg carbon added (part b). Total acetone (0.349 kg) splits between gas and carbon at the unknown pressure $p$ (kN/m²):
$$0.349=\underbrace{\frac{p\cdot10^3\,V}{RT}M_A}_{\text{gas}}+\underbrace{1.6\,(0.088\,p^{0.317})}_{\text{adsorbed}} .$$
Solving, $p=5.0\text{ kN/m}^2$, so
$$s=\frac{p}{p^{*}}=\frac{5.0}{37.9}=\boxed{0.13\;(\approx13\%).}$$
With less carbon than part (a) required, the space is only partly cleaned — the saturation settles between the 40% start and the 5% target, as expected.
