Chapter 8

Adsorption & Chromatography

The separating agent is a solid, the equilibrium is a curve rather than a line, and for the first time in this course nothing is at steady state.

Wankat, Ch. 17–18 4 interactive apps 3 worked examples Self-marking problem set Prerequisite: Ch. 6

By the end of this chapter you should be able to

Say what makes a good adsorbent, and name the main families.
Fit and use the linear, Langmuir and Freundlich isotherms, and say which fits which data.
Get the isosteric heat of adsorption from isotherms at several temperatures, and explain why temperature swing works.
Use the extended Langmuir isotherm and compute a selectivity.
Compute retention time, selectivity and resolution for a linear chromatographic column.
Explain the PSA and TSA cycles in terms of a working capacity.
(Going further) Predict whether a concentration front sharpens into a shock or spreads, and find the breakthrough time.

Separating on a surface

Every separation so far has created a second fluid phase — a vapour, a solvent, another liquid — and let the components choose between two bulk phases. Adsorption creates a second phase that is barely a phase at all: the surface of a porous solid. A gram of a good adsorbent carries several hundred to a couple of thousand square metres of internal area, and molecules stick to it selectively.

Figure 8.1 — A solute partitions between a fluid and the internal surface of a porous solid. The concentration in the fluid is (or a partial pressure ); the amount held on the solid is the loading , per gram or per litre of adsorbent. Unlike every earlier chapter, one of the two phases does not flow.

What is genuinely new here

Two things, and they are connected. The equilibrium is strongly curved — a surface saturates, so cannot rise for ever, and the isotherm bends over. And the solid does not flow, so a fixed bed can never reach steady state: it loads up, breaks through, and has to be regenerated. Adsorption is inherently a cyclic, transient operation, and that is why it is the last chapter of this course rather than the first.

8.1 Adsorbents

A good adsorbent needs five things at once, and the art is that they conflict:

Requirement	Why	What it costs you
Large surface area	Capacity. 500–2000 m²/g is normal, and it is all internal.	Small pores, which slow diffusion in and out.
High selectivity	It has to prefer one component appreciably over another.	Usually a stronger interaction, so a bigger regeneration bill.
Fast kinetics	The cycle is minutes; a slow solid wastes most of its capacity.	Larger pores and smaller particles — the opposite of the first line, and more pressure drop.
Mechanical strength and stability	Beds are pressurised, depressurised and heated thousands of times a year.	Binder, which is dead weight.
Low price	Beds are large; the inventory is the capital cost.	Everything above.

The families in industrial use are zeolites (crystalline aluminosilicates with pores of exactly one size — true molecular sieves), activated carbons (cheap, hydrophobic, enormous area), silica gel and activated alumina (drying), carbon molecular sieves (which separate by rate rather than by capacity), and the metal–organic frameworks and covalent organic frameworks now moving out of the laboratory.

The 2025 Nobel Prize in Chemistry

The 2025 prize went to Susumu Kitagawa, Richard Robson and Omar Yaghi for the development of metal–organic frameworks — crystalline solids assembled from metal nodes and organic linkers, with pore size, shape and chemistry all designed rather than found. Surface areas above 7000 m²/g have been reported. The prize is a direct endorsement of the subject of this chapter, and Chapter 1 links the lectures1 · Nobel.

The Nobel Prize in Chemistry 2025. NobelPrize.org, Nobel Prize Outreach. nobelprize.org/prizes/chemistry/2025

Adsorption is not absorption

One letter, two different operations. Absorption dissolves the solute into the bulk of a liquid — Chapter 6. Adsorption holds it on a surface. The clue is in the prefix, and the physical difference is that a surface has a finite number of sites, which is exactly why the isotherm saturates and Henry's law does not.

8.2 Where it is used

Separation	What is separated	Adsorbent	Process
Air separation	O₂ / N₂	Zeolites, carbon molecular sieves	PSA / VSA
Hydrogen from reformer off-gas	CO₂ / H₂ / CH₄	Zeolites, aluminas, carbons	PSA
Landfill-gas upgrading	CH₄ / CO₂	Zeolites, carbon molecular sieves	PSA with TSA
Gas drying	H₂O from anything	Silica gel, alumina, 3A zeolite	TSA
Xylene isomers	p-xylene from the others	Zeolites	Simulated moving bed
Optical isomers	One enantiomer from the other	Chiral stationary phases	Chromatography, SMB, SFC
Post-combustion capture	CO₂ / N₂	Zeolites, MOFs	PSA, TSA, hybrids
Direct air capture	CO₂ from 420 ppm	Amine-functionalised sorbents	Temperature–vacuum swing
Biomolecule purification	Peptides, monoclonal antibodies	Ion-exchange resins	Chromatography, SMB

Two things are worth noticing in that table. Adsorption owns the separations where distillation is impossible — you cannot distil oxygen from nitrogen at ambient temperature, and you certainly cannot distil one enantiomer from another. And chromatography, in its analytical form, is the most widely used measurement technique in chemistry; every GC and HPLC trace you will ever see is the subject of §8.7.

8.3 Adsorption equilibrium — the isotherm

At fixed temperature, the loading on the solid is a function of the fluid concentration alone. That function is the isotherm. Three forms cover most of what you will meet.

8.1

For a gas, replace by the partial pressure . The linear form is the low-concentration limit of Langmuir — the Henry's law of adsorption — and it is the only one that makes the mathematics of §8.7 easy. It is not the limit of Freundlich: , which for goes to infinity as , so a Freundlich isotherm has no Henry's-law region at all. All four of the isotherms fitted in App 1 have between 1.26 and 1.72, so this is not a corner case — it is the normal one. Langmuir comes from a physical picture: a fixed number of identical sites, one molecule each, no interaction between neighbours. Its two parameters mean something — is the monolayer capacity and measures how strongly the molecule sticks. Freundlich is empirical, with no saturation limit, and it usually fits heterogeneous surfaces such as activated carbon rather better than Langmuir does.

Figure 8.2 — The shapes. Langmuir and Freundlich are favourable: steep at low concentration, so a small partial pressure still gives a useful loading. The linear isotherm is the borderline case. An unfavourable isotherm curves the other way, and §8.9 shows that the shape decides whether a front in a bed sharpens or smears.

Fitting is not the same as understanding

Three parameters will fit almost any five points. What matters is whether the parameters are physically sensible and whether they extrapolate. A Langmuir fit that returns a five times the largest measured loading has told you nothing except that you measured only the initial slope. App 1 reports the fits and lets you see when this is happening.

App 1

Fitting an isotherm

Real data: benzene vapour on silica gel at four temperatures, from Shen and Smith (1968). Fit all three isotherms and compare. Watch how the linear model fails badly at 70 °C, where the surface is nearly saturated, and recovers at 130 °C, where the same pressures only reach the foot of the curve. Switch the axes to logarithmic to see what the Freundlich model is really claiming.

Temperature

Axes

Show

The data and the fits

Benzene on silica gel, 832 m²/g.

Residuals

Measured minus fitted, as a percentage.

Langmuir q_sat

—

Langmuir K

—

Freundlich n

—

R² linear

—

R² Langmuir

—

R² Freundlich

—

✓—

8.4 Temperature and the heat of adsorption

Adsorption is exothermic — a molecule on a surface has given up freedom, so it must have given up energy too. Raise the temperature and the loading falls, which is the whole basis of thermal regeneration. Quantitatively, at fixed loading,

8.2

which is the Clausius–Clapeyron equation with the adsorbed phase playing the part of the condensate. Mind the sign convention. is the enthalpy change of the adsorbing molecule, gas → surface, and it is negative because adsorption is exothermic. The isosteric heat is defined as the heat released, , so it is a positive number. Isosteric means "at constant loading", which is why the measurement has to be made by cross-plotting several isotherms rather than from any one of them.

Where the sign comes from

Equate the chemical potentials of the gas and the adsorbed molecule and differentiate along a line of constant loading. With and at fixed ,

using from . Since , this is equation (8.2). A negative slope on the van 't Hoff plot therefore gives a positive : raising at fixed loading needs a higher pressure, which is exactly why heating regenerates the bed.

How big should it be?

Compare it with the heat of vaporisation of the same molecule. If is similar, the molecule is held about as tightly as it is in its own liquid — ordinary physisorption. If it is two or three times larger, the surface is unusually attractive and regeneration will be expensive. If it is five times larger, a bond has formed: that is chemisorption, and you may not get the molecule back at all.

App 2

The isosteric heat

Pick a loading, and the app finds the pressure that gives it at each of the four temperatures, plots against , and turns the slope into a heat of adsorption. The line is straight, which is itself the evidence that the heat is roughly independent of loading over this range. Compare the answer with benzene's heat of vaporisation.

Loading q, ×10⁻⁵ mol/g 20.0

Compare with

The four isotherms

The horizontal line is the loading you chose.

The van 't Hoff plot

ln p against 1/T at that fixed loading.

Isosteric heat q_st

—kJ/mol

Slope

—K

Points used

—

Ratio to ΔH_vap

—

✓—

8.5 More than one adsorbate

A real feed has at least two components, and they compete for the same sites. The natural extension of the Langmuir picture — same sites, still one molecule each, still no interaction — is the extended Langmuir isotherm:

8.3

Note what the denominator does: every component reduces every loading. Add a strongly held impurity and the capacity for your product collapses, which is why a PSA unit almost always has a guard layer in front of it. The selectivity follows:

8.4

and this is the number that does for adsorption what relative volatility2a §2.7 does for distillation and what does for absorption. At low concentration it is just the ratio of the two Henry slopes, which is the form used in §8.7.

8.6 Fixed beds

In practice you do not stir an adsorbent into a tank. You pack it into a column and push the fluid through.

Figure 8.3 — A fixed bed part way through its cycle. The upstream end is saturated, the downstream end is untouched, and between them is a mass-transfer zone that travels along the bed. When it reaches the exit, solute appears in the product — breakthrough — and the bed must be taken off line and regenerated.

A solute balance on a slice of the bed, ignoring dispersion, gives the equation that governs everything that follows:

8.5

with the bed voidage and the interstitial fluid velocity. If the solid is in local equilibrium with the fluid, , the chain rule turns this into a statement about how fast a given concentration travels:

8.6

Read Equation 8.6 before going on

It says a concentration moves at the fluid velocity divided by a retardation factor, and the retardation is set by the local slope of the isotherm. A steeply held solute (large ) crawls; a weakly held one nearly keeps up with the fluid. Everything else in this chapter — retention times, separation, breakthrough, whether a front is sharp or smeared — is Equation 8.6 applied to different isotherm shapes.

8.7 Linear chromatography

At low enough concentration every isotherm is linear, , so is a constant and every concentration of a given component travels at the same speed. A narrow pulse injected at the inlet stays narrow and comes out at a definite time:

8.7

is the time for an unretained tracer — the void volume divided by the flow — and is the phase ratio. Two components with different come out at different times, and that is chromatography. The three numbers that describe the result are

8.8

the capacity factor, the selectivity and the resolution, where is the baseline width of a peak. Two peaks are counted as separated at .

Where the peak width comes from

Equilibrium theory says a pulse stays a pulse. Real peaks broaden, because mass transfer is not instantaneous and because the fluid does not all travel at the same speed. The usual bookkeeping is to pretend the column is a cascade of equilibrium stages — theoretical plates, the same fiction as a distillation tray — which gives . A good analytical column has in the tens of thousands. This course does not compute from first principles; App 3 takes it as an input, which is exactly what a chromatographer does with a real column.

App 3

The chromatogram

A real analytical column: 125 mm long, 4.6 mm bore, silica gel, 70 % void. Two components with Langmuir constants differing by only 37 %. Set the flow and watch the two peaks separate — and watch what it costs you in time. Then drop the plate count and see the same separation disappear into one lump.

Flow rate 1.00 cm³/min

Component A K_A, L/g 0.0787

Component B K_B, L/g 0.1079

Theoretical plates 2000

Bed voidage ε 0.70

The chromatogram

Detector signal against time.

Resolution against flow

Slower is better — up to a point you cannot afford.

Void time t₀

—min

t_R of A

—min

t_R of B

—min

Selectivity α

—

Capacity factor k′ of A

—

Resolution

—

✓—

8.8 Cyclic processes

A bed loads up and must then be emptied. There are only two levers, and each names a process.

Figure 8.4 — Two isotherms and the working capacity between them. Adsorb where the isotherm is high — cold, or at high partial pressure. Regenerate where it is low — hot, or at low pressure. The useful capacity of the solid is not ; it is the difference, and a solid with a huge capacity but a flat swing is useless.

Cycle	The lever	Time scale	Suits
PSA / VSA	Pressure. Adsorb at pressure, blow down to atmosphere or to vacuum.	Seconds to minutes — pressure changes fast.	Bulk gas separations: air, hydrogen, landfill gas.
TSA	Temperature. Adsorb cold, regenerate with hot gas or steam.	Hours — you have to heat and cool a bed of solid.	Trace removal, drying, strongly held solutes.

Figure 8.5 — The simplest pressure-swing cycle. One bed is on line adsorbing at pressure while the other is depressurised and purged with a slip stream of product; then they swap. Two beds are the minimum for continuous production, and real units use four to twelve, with pressure-equalisation steps between them to recover the work that would otherwise be blown to atmosphere.

Why PSA is fast and TSA is slow

You can change the pressure of a bed in seconds; you cannot change its temperature in seconds, because a packed bed is a large lump of solid with a large heat capacity and poor conductivity. That single fact fixes the cycle time, and the cycle time fixes the throughput per kilogram of adsorbent. It is why oxygen concentrators are PSA and dryers are TSA.

8.9 Going further — shocks and spreading waves

Beyond the examinable core

This section goes past what the course assesses. It is here because it explains something the earlier sections leave hanging: why a real breakthrough curve is sharp when you adsorb and smeared when you regenerate, and because the answer is one line of Equation 8.6.

Equation 8.6 gives the velocity of a concentration. For a favourable isotherm, falls as rises, so high concentrations travel faster than low ones. Now consider the two things a bed does.

Figure 8.6 — The same isotherm, the two directions. Adsorption: the feed step sends fast high-concentration characteristics after slow low-concentration ones. They would cross, which is physically impossible, so they collapse into a shock — a self-sharpening front. Desorption: the concentrations leave in the opposite order, so the characteristics fan apart and the wave spreads without limit.

A shock travels at a velocity set by the chord of the isotherm rather than its tangent:

8.9

The asymmetry, and what it costs

Adsorption is self-sharpening and desorption is self-spreading, for one and the same solid. So the loading step is efficient — the bed is nearly fully used when it breaks through — while the regeneration step is not, and the tail of the desorption wave is what forces you to over-purge. In every real cycle, regeneration is the expensive half, exactly as it was for the absorber and its stripper6 §6.1.

App 4

Breakthrough

A bed of m-xylene on an adsorbent, with a genuinely favourable Langmuir isotherm. Load it and the front sharpens into a shock; clean it and the front spreads. The left panel draws the characteristics in position and time; the right one gives the curve you would actually measure at the exit. Flip the isotherm to unfavourable and everything swaps over.

Operation

Isotherm

Feed C₀, g/L 0.0200

Bed length, m 0.40

Interstitial velocity, mm/s 1.00

Characteristics

Each line carries one concentration through the bed.

What you measure at the exit

Concentration leaving the bed against time.

Front type

—

Breakthrough

—min

Fully spent / clean

—min

Bed used at breakthrough

—%

Retardation

—

Solute held, per m² of bed

—g/m²

✓—

8.10 Worked examples

Worked example 8.1Which isotherm, and what does it mean?

Shen and Smith measured benzene adsorption on silica gel (832 m²/g, pore volume 0.43 cm³/g, mean pore diameter 22 Å) at four temperatures. At 90 °C the data are

p, atm	5×10⁻⁴	1×10⁻³	2×10⁻³	5×10⁻³	1×10⁻²	2×10⁻²
q, ×10⁻⁵ mol/g	6.7	11.2	18.0	33.0	51.0	78.0

Fit all three isotherms and say which describes the data, and what the parameters imply.

Work it yourself first, then open

The fits (least squares on ):

linear: q = 4310 p, R² = 0.889

Langmuir: q_sat = 133, K = 67.6 atm⁻¹, R² = 0.993

Freundlich: q = 944 p^1/1.572, R² = 0.9996

Freundlich wins here, and by a clear margin. That is not an accident: silica gel is an energetically heterogeneous surface, with a spread of site strengths, and Freundlich is the model that assumes exactly that. Langmuir assumes identical sites and is only slightly worse; linear is not a serious candidate at these loadings.

What the parameters say. The Langmuir of 133×10⁻⁵ mol/g is 1.7 times the largest measured loading — so the fit is extrapolating well beyond the data, and that number should not be quoted as a monolayer capacity. Sanity-check it anyway: 133×10⁻⁵ mol/g of benzene, at roughly 0.30 nm² per molecule flat on the surface, covers

133×10⁻⁵ × 6.022×10²³ × 0.30×10⁻¹⁸ = 240 m²/g

against a measured 832 m²/g — so a "monolayer" by this fit covers under a third of the surface. Consistent with a heterogeneous solid whose strongest sites fill first.

And the pores. A 22 Å pore is about four benzene molecules wide. At the highest pressures the mechanism is probably no longer monolayer adsorption at all but pore filling, which is one more reason Langmuir struggles.

App 1, at 90 °C. Then switch to 70 °C and watch the linear fit collapse to R² = 0.78.

Worked example 8.2An analytical separation

A column 125 mm long and 4.6 mm in diameter is packed with silica gel; the void volume is 70 % of the empty-column volume. Two solutes follow Langmuir isotherms with g/L and , L/g. A very dilute pulse is injected and the mobile phase flows at 1.0 cm³/min. Find the retention times and the flow rate that would separate the peaks by 4.0 minutes.

Work it yourself first, then open

Column geometry.

V_col = π(0.23 cm)²(12.5 cm) = 2.0774 cm³ · V₀ = 0.70(2.0774) = 1.4542 cm³

t₀ = 1.4542/1.0 = 1.454 min — the unretained tracer

φ = (1 − 0.70)/0.70 = 0.42857

Henry slopes. "Very dilute" means the linear limit of the Langmuir isotherm, :

K_A′ = 89.32(0.0787) = 7.030 · K_B′ = 89.32(0.1079) = 9.638

selectivity α = 9.638/7.030 = 1.371

Retention times.

t_R,A = 1.4542[1 + 0.42857(7.030)] = 5.835 min

t_R,B = 1.4542[1 + 0.42857(9.638)] = 7.461 min

Δt = 1.625 min

A four-minute separation. Both retention times scale as 1/Q, so the gap does too:

Δt = V₀φ(K_B′ − K_A′)/Q = 1.6262/Q = 4.00 → Q = 0.406 cm³/min

The trade you just made. Slowing the flow by a factor of 2.46 widened the gap by 2.46 — and pushed the analysis time from 7.5 minutes to 18.4. That is the whole economics of chromatography: resolution is bought with time, and the only way out is a better selectivity, which means a different stationary phase.

One more question the same numbers answer. If the column were saturated with 2 g/L of A and you washed it with pure solvent, the last trace of A would leave when the characteristic reached the exit — which travels at exactly the dilute-pulse speed. So the column is clean after 5.835 min, the retention time of A.

App 3, with the default constants and 1.00 cm³/min.

Worked example 8.3Loading and cleaning the same bed

A bed 0.40 m long with a voidage of 0.50 holds an adsorbent for which , with in g/L of sorbent and in g/L of solution. The interstitial velocity is 1.0 mm/s. (a) The bed is clean and a feed of 0.020 g/L enters — when does it break through? (b) The bed is saturated at 0.020 g/L and pure solvent enters — when is it clean?

Work it yourself first, then open

Set-up. φ = (1 − 0.5)/0.5 = 1.000, and

dq/dC = 5.5/(1 + 29C)² → 5.500 at C = 0, 2.203 at C = 0.020

The slope falls as C rises, so this isotherm is favourable and concentrated fronts travel faster than dilute ones.

(a) Adsorption — a shock. Use the chord, not the tangent:

Δq/ΔC = q*(0.02)/0.02 = 0.069620/0.020 = 3.481

v_sh = 1.0×10⁻³/(1 + 3.481) = 2.232×10⁻⁴ m/s

t_b = 0.40/2.232×10⁻⁴ = 1792 s = 29.9 min

and because the front is a shock, the concentration at the exit jumps from zero to the feed value essentially at once. The bed is used almost completely.

(b) Desorption — a spreading wave. Now every concentration travels at its own tangent velocity:

v(C = 0.020) = 1.0×10⁻³/(1 + 2.203) = 3.122×10⁻⁴ m/s

v(C = 0.010) = 1.0×10⁻³/(1 + 3.305) = 2.323×10⁻⁴ m/s

v(C = 0) = 1.0×10⁻³/(1 + 5.500) = 1.538×10⁻⁴ m/s

so at t = 100 s the wave stretches from z = 0.0154 m (where C has fallen to zero) to z = 0.0312 m (where it is still at feed value) — already 16 mm wide after 100 seconds. The exit concentration falls to 0.010 g/L at

t = 0.40/2.323×10⁻⁴ = 1722 s = 28.7 min

and the last trace leaves only when the C → 0 characteristic arrives:

t = 0.40/1.538×10⁻⁴ = 2600 s = 43.3 min

Compare the two. Loading takes 29.9 minutes and delivers a sharp front. Cleaning takes 43.3 minutes — 45 % longer — and the last third of that time is spent chasing a tail that is nearly at zero concentration. Regeneration is the expensive half of every cycle, and this is the reason.

App 4: adsorb, then switch to regenerate, with C₀ = 0.020, L = 0.40 m, v = 1.00 mm/s.

8.11 Check your understanding

Five multiple-choice questions, two short problems and two long ones — 41 marks. Work them offline, then enter your numbers.

Marks earned

0 / 0

Multiple choice

Short problems

Long problems

Think about it

Why does adsorption have no equivalent of the McCabe–Thiele staircase?

Because there is no countercurrent cascade to draw. In distillation, absorption and extraction the two phases both flow, in opposite directions, and a steady state exists in which every stage has a fixed composition. Adsorption's second phase is a stationary solid, so nothing is at steady state and the independent variable is time, not stage number. The nearest equivalent is a simulated moving bed, which fakes countercurrent flow by periodically shifting the feed and product ports along a ring of fixed beds — and an SMB really can be analysed with something very like an operating diagram. That is a graduate topic; it is also how every tonne of p-xylene is made.

A vendor offers an adsorbent with twice the capacity at the same price. Should you buy it?

Not until you know its working capacity. Figure 8.4 is the whole answer: what you use each cycle is the difference between the loading at adsorption conditions and the loading at regeneration conditions. A solid that holds twice as much but holds on to it just as tightly when you drop the pressure has doubled your inventory cost and changed nothing else. The same logic explains why amine sorbents for direct air capture, which have enormous capacity at 420 ppm, are hard to regenerate: the very strength that lets them work at 420 ppm is what makes the swing expensive. Ask for the isotherms at both ends of the cycle, and ask for the kinetics too.

Chromatography separates things distillation cannot. What is it giving up?

Throughput, and dilution. A chromatographic column processes a pulse at a time and delivers each component dissolved in a large volume of mobile phase which then has to be evaporated. It is superb where the value per kilogram is enormous and the alternative is nothing at all — chiral pharmaceuticals, monoclonal antibodies — and hopeless where the value is a few dollars a tonne. The industrial answer is the simulated moving bed, which converts the batch pulse operation into a continuous one and cuts the solvent requirement by an order of magnitude. Between them, selectivity and throughput are the two axes on which every separation in this course sits.

Equation 8.6 assumes local equilibrium. What does a real bed do instead?

It lags. Mass transfer into a porous particle takes time, so the solid is never quite in equilibrium with the fluid around it, and the effect is to broaden every front — including the shock, which in a real bed is not a discontinuity but a mass-transfer zone of finite width. There are two competing effects and they are worth keeping separate in your head: equilibrium theory decides whether a front tries to sharpen or spread, and kinetics sets a floor on how sharp it can get. A favourable isotherm with slow kinetics gives a front whose width settles to a constant — the constant-pattern front — which is what the mass-transfer zone in Figure 8.3 actually is.

Where this chapter connects

The agent: an adsorbent is a mass separating agent1 §1.6 that happens to be a solid, and it carries the same debt — §8.8 is where you pay it.
The equilibrium: the linear isotherm is Henry's law6 §6.2 with a surface instead of a liquid; the selectivity of Equation 8.4 is relative volatility2a §2.7 under another name.
The cycle: adsorb cold and regenerate hot is the absorber and its stripper6 §6.1, run in time rather than in space.
Plates: the theoretical plate of §8.7 is the equilibrium stage of Chapter 4a4a §4.6, borrowed as a bookkeeping device for band broadening.
What is genuinely different: nothing here is at steady state, and the second phase does not move. That is why the answers are times rather than stage counts.

Summary & key equations

Equilibrium

Linear — the low-concentration limit of Langmuir; Freundlich with has infinite initial slope and no Henry's-law limit

Langmuir — identical sites, one molecule each

Freundlich — empirical; fits heterogeneous surfaces, never saturates, and for has as

Extended Langmuir — every component suppresses every other

Selectivity as

Isosteric heat with ; compare with

Fixed beds

Velocity of a concentration

Retention time, ,

Capacity factor

Resolution, ; separated at

Shock velocity — the chord, not the tangent

Which happensfavourable isotherm → shock on adsorption, spreading wave on regeneration; unfavourable swaps them

Cycles

Working capacitythe difference in loading between the two ends of the cycle — not

PSA / VSApressure is the lever; seconds to minutes; bulk gas separation

TSAtemperature is the lever; hours; trace removal and drying

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