Liquefaction of gases — Linde, Claude processes (concept): Question bank

Prerequisites worth re-reading if a group stings: Critical constants ($T_c, P_c, V_c$), van der Waals equation, Real gases and compressibility factor Z, Enthalpy and constant-H processes, First law of thermodynamics, Intermolecular forces.

Before you start: the symbols and pictures this bank assumes

Read these callouts once so that every reveal below is meaningful. Nothing here is stated without a definition and a picture.

Formula Deriving the inversion condition step by step

What we want: the boundary where $μ_{J T} = 0$ . Why: that boundary separates cooling from heating.

From the identity $μ_{J T} = \frac{1}{C _{p}} [T (\frac{\partial V}{\partial T})_{P} - V]$ and $C_{p} > 0$ , setting $μ_{J T} = 0$ gives the inversion condition $T (\frac{\partial V}{\partial T})_{P} = V .$
Put in the gas. For a van der Waals gas at low pressure we may expand the equation of state to first order in $P$ , giving a molar volume $V_{m} \approx \frac{R T}{P} + b - \frac{a}{R T} .$ Why this form: $R T / P$ is the ideal part, $+ b$ adds the space molecules occupy, and $- a / (R T)$ is the shrinkage from attraction.
Differentiate at constant $P$ : $(\frac{\partial V _{m}}{\partial T})_{P} = \frac{R}{P} + \frac{a}{R T ^{2}} .$
Form the bracket $T (\frac{\partial V _{m}}{\partial T})_{P} - V_{m}$ : $(\frac{R T}{P} + \frac{a}{R T}) - (\frac{R T}{P} + b - \frac{a}{R T}) = \frac{2 a}{R T} - b .$ The $R T / P$ terms cancel — that is why the answer is pressure-free in this limit.
Set it to zero ( $μ_{J T} = 0$ ): $\frac{2 a}{R T} - b = 0 \Rightarrow$ $T_{i, max} = \frac{2 a}{R b} .$ This is the maximum inversion temperature — the top of the inversion curve, where its upper branch meets the $P = 0$ axis. So $2 a / (R b)$ is the highest temperature at which any throttling can still cool the gas; at nonzero pressures the actual crossing temperatures sit below it. Treat it as a ceiling, not a fixed operating point.

True or false — justify

An ideal gas cools when throttled through a porous plug.

False. Throttling is constant-enthalpy with no external work; for an ideal gas

(\partial U / \partial V)_{T} = a / V_{m}^{2} = 0

, so nothing is spent pulling molecules apart,

μ_{J T} = 0

and

Δ T = 0

. The cooling is purely a real-gas effect.

Above its critical temperature a gas can still be liquefied if you use enough pressure.

False. By the very definition of

T_{c}

, above it thermal kinetic energy always overwhelms attraction; any pressure just gives a very dense gas, never a liquid.

The Joule–Thomson effect keeps temperature constant because enthalpy is constant.

False. It is the enthalpy

H = U + P V

that stays constant, not the temperature — that constant-

H

constraint (see figure s01) is precisely what forces

T

to change as internal energy is spent pulling molecules apart.

Cooling by throttling happens only when the gas starts inside the inversion curve.

True. Inside the curve

μ_{J T} > 0

, so a pressure drop gives a temperature drop; cross to the outside and the sign flips and the gas warms (figure s04).

The Claude process would still cool an ideal gas.

True. Its extra cooling comes from adiabatic work-doing expansion: with

q = 0

and

w < 0

(gas does work on surroundings),

Δ U = w < 0

, so

T

falls even with no intermolecular attractions.

Compression alone is what liquefies air in the Linde process.

False. Compression heats the gas and is cooled back to start conditions; the actual temperature descent to below

T_{c}

comes from repeated JT throttling stacked by the counter-current heat exchanger.

Helium has a very high inversion temperature.

False. Helium's attractions (

a

) are extremely weak, so its maximum inversion temperature

2 a / (R b) \approx 40

K is very low; at room temperature it lies outside the inversion curve and would warm on throttling.

A single throttling step can cool a gas all the way to liquefaction.

False. One pass gives only a small

Δ T

; the counter-current heat exchanger's regenerative feedback accumulates many small drops into liquefaction.

The Joule–Thomson coefficient

μ_{J T}

has the same sign for a gas at all conditions.

False. It is positive inside the inversion curve (cooling) and negative outside it (heating), passing through zero along the inversion curve — and at a fixed pressure it changes sign at two temperatures, not one.

Spot the error

"Since gases cool when they expand, any expanding gas liquefies more easily."

The blanket "gases cool on expansion" is only true for adiabatic work-doing expansion; a throttled ideal gas does not cool at all, and a real gas outside its inversion curve actually warms.

"The Linde and Claude processes both rely on van der Waals attractions to cool."

Linde's cooling is purely the real-gas JT effect (

a \neq = 0

); Claude's extra cooling comes from work-doing expansion (

Δ U = w < 0

), which occurs even for an ideal gas — so Claude does not need attractions.

"To liquefy hydrogen at 300 K, just throttle it hard enough."

At 300 K hydrogen lies outside its inversion curve (its maximum inversion temperature is

\approx 200

K), so throttling warms it; it must first be pre-cooled (e.g. with liquid

N_{2}

) into the cooling region.

μ_{J T} > 0

means the gas heats up."

A positive

μ_{J T} = (\partial T / \partial P)_{H} > 0

means temperature rises with pressure; since throttling drops pressure

(d P < 0)

, it gives

d T < 0

— the gas cools. Positive is the useful, cooling case.

"Because no heat enters during throttling, the internal energy stays the same."

It is the enthalpy

H = U + P V

that is conserved, not

U

; internal energy falls as work is done against attractions (

(\partial U / \partial V)_{T} > 0

), which is exactly why the temperature drops.

"The inversion temperature is a single fixed number for each gas."

The inversion condition is a curve in the

P

–

T

plane; at a given pressure it is crossed at two temperatures (upper and lower branches), and

2 a / (R b)

is only its low-pressure maximum, not a universal operating value.

"Claude's turbine is just a fancier throttle valve doing the same physics."

A throttle does no external work (constant-

H

JT effect only); the turbine is an adiabatic work-doing (near-isentropic) expansion where the gas surrenders internal energy as work, so its cooling mechanism is different and stronger.

Why questions

Why does an ideal gas show zero JT effect while a real gas cools?

An ideal gas has

(\partial U / \partial V)_{T} = 0

, so no energy is needed to pull molecules apart on expansion; a real gas has

(\partial U / \partial V)_{T} = a / V_{m}^{2} > 0

, so it must spend kinetic energy stretching its attractive "springs" (figure s03), which lowers its temperature.

Why must you cool below

T_{c}

before compressing, not after?

Above

T_{c}

molecular kinetic energy always beats attraction, so compression only densifies the gas; cooling first lets attractions become able to win once the molecules are pushed together.

Why is the counter-current heat exchanger essential in Linde's design?

Each throttling gives only a tiny

Δ T

, so the cold exiting gas is routed back to pre-cool the incoming warm gas, ratcheting the temperature down cycle after cycle until liquefaction.

Why can the Claude process reach low temperatures faster than Linde?

It adds adiabatic work-doing expansion — a strong cooling (

Δ U = w < 0

) that exists even for near-ideal gases — on top of JT cooling, and it recovers some of the expansion work, making it more efficient.

Why does the bracket

T (\partial V / \partial T)_{P} - V

decide the sign of

μ_{J T}

Exactly,

T (\partial V / \partial T)_{P} - V = - (\partial H / \partial P)_{T}

, and

μ_{J T} = - \frac{1}{C _{p}} (\partial H / \partial P)_{T} = \frac{1}{C _{p}} [T (\partial V / \partial T)_{P} - V]

. Since

C_{p} > 0

, the sign of

μ_{J T}

is the sign of this bracket; when attractions make the bracket positive, expansion cools the gas.

Why are

H_{2}

and He described as the "hardest" gases to liquefy?

They have tiny critical temperatures (so they must be cooled to very low

T

) and low maximum inversion temperatures

2 a / (R b)

(so they warm on throttling near room

T

), requiring pre-cooling by another gas first.

Edge cases

What happens to

μ_{J T}

exactly on the inversion curve?

It is zero, so throttling produces no temperature change there — it is the boundary between the cooling region (inside,

μ_{J T} > 0

) and the heating region (outside,

μ_{J T} < 0

What is the temperature change if a truly ideal gas is throttled?

Exactly zero, because

(\partial U / \partial V)_{T} = 0

makes

μ_{J T} = 0

; only real-gas attractions (

a \neq = 0

) can produce any JT cooling.

What if you compress a gas already below

T_{c}

but never throttle or cool further?

Enough pressure at

T < T_{c}

can itself condense it, since attractions can now dominate; below the critical point pressure alone suffices to reach the liquid state.

At a working pressure, what if the gas sits just outside the upper inversion branch?

Then

μ_{J T} < 0

, so the pressure drop of throttling warms it, pushing it further from liquefaction; you must first cool it inside the inversion curve (below the upper crossing temperature).

At a working pressure, what happens below the lower inversion branch (

T < T_{i, low}

You are again outside the cooling region, so

μ_{J T} < 0

and throttling warms the gas — the dome of cooling is closed at the bottom too, as the purple shaded region and lower-branch marker in figure s04 show; you must operate between the two branches.

If a gas had

b \to 0

(point-like molecules) but nonzero

a

, what happens to its maximum inversion temperature?

Since

T_{i, max} = 2 a / (R b)

, letting

b \to 0

sends it to very large values, so such a gas would sit inside its inversion curve — and thus cool on throttling — across essentially all attainable temperatures.

Recall One-line self-test

What decides whether throttling cools or heats a real gas, and why is it not one number? Answer ::: Whether the gas lies inside its inversion curve ( $μ_{J T} > 0$ , cooling) or outside it — above the upper branch or below the lower branch — where $μ_{J T} < 0$ and it heats; the boundary is a $P$ – $T$ curve, so the crossing temperatures depend on the working pressure, with $2 a / (R b)$ its low-pressure maximum.