Liquefaction of gases — Linde, Claude processes (concept): Worked examples

This page is a firing range. We line up every kind of question liquefaction can ask — every sign of the Joule–Thomson coefficient, every position of the temperature relative to the two thresholds ( $T_{c}$ and $T_{i}$ ), the two degenerate limits (ideal gas, zero pressure drop), the exact-boundary cases, a real-world word problem, and an exam twist — then knock each one down with a fully worked example.

Before we start, let us make sure every symbol is earned.

Definition The six symbols you must never confuse

$T$ — the actual temperature of your gas right now (in kelvin, K).
$T_{c}$ — the critical temperature: the ceiling above which no pressure liquefies the gas. See Critical constants ($T_c, P_c, V_c$).
$T_{i}$ — the inversion temperature: the fence between "throttling cools" and "throttling warms".
$C_{p}$ — the heat capacity at constant pressure: the joules of heat needed to raise one mole by one kelvin while the pressure is held fixed (units $J K^{- 1} mol^{- 1}$ ). It is the "thermal inertia" that damps how fast the temperature responds.
$C_{v}$ — the heat capacity at constant volume: the joules needed to raise one mole by one kelvin while the volume is held fixed (units $J K^{- 1} mol^{- 1}$ ). It is the bridge between a change in internal energy and a change in temperature, $Δ U = C_{v} Δ T$ — we need it for the Claude (work-doing) example.
$μ_{J T} = (\frac{\partial T}{\partial P})_{H}$ — the Joule–Thomson coefficient: how many kelvin the gas changes for each atmosphere of pressure it loses, while its enthalpy $H$ stays fixed.

Read $(\frac{\partial T}{\partial P})_{H}$ as: "if I nudge the pressure $P$ down a tiny bit and keep enthalpy locked, how much does the temperature $T$ move?" The little $H$ underneath is a promise: enthalpy does not change during throttling.

The scenario matrix

Every liquefaction question is really a question about where $T$ sits relative to $T_{c}$ and $T_{i}$ , and which process ( $μ_{J T}$ sign, throttle vs. work). Here is the full grid — including the two knife-edge boundary cases $T = T_{c}$ and $T = T_{i}$ .

Cell	Case class	What decides it	Example
A	$T > T_{c}$ — pressure alone fails	compare $T$ vs $T_{c}$	Ex 1
B	$T < T_{c}$ — pressure now works (need $P \geq P_{vap}$ )	compare $T$ vs $T_{c}$	Ex 2
C	$T < T_{i}$ — throttling cools ( $μ_{J T} > 0$ )	compare $T$ vs $T_{i}$	Ex 3
D	$T > T_{i}$ — throttling warms ( $μ_{J T} < 0$ )	compare $T$ vs $T_{i}$	Ex 4
E	Degenerate: ideal gas ( $a = 0$ ) — $μ_{J T} = 0$	van der Waals $a \to 0$	Ex 5
F	Degenerate: zero pressure drop ( $d P = 0$ ) — $Δ T = 0$	$Δ T = μ_{J T} Δ P$	Ex 6
G	Compute $T_{i}$ from van der Waals $a, b$	$T_{i} \approx 2 a / (R b)$	Ex 7
H	Claude: work-doing expansion cools an ideal gas	$Δ U = w$ , $Δ U = C_{v} Δ T$	Ex 8
I	Real-world word problem	pick process + threshold	Ex 9
J	Exam twist — combine thresholds	order of operations	Ex 10
K	Exact boundaries: $T = T_{c}$ and $T = T_{i}$	the " $=$ " edge of A/B and C/D	Ex 11

We now hit each cell.

Worked example Ex 1 — Cell A:

$T > T_{c}$ , pressure alone fails Statement. You have oxygen at $300 K$ . A friend cranks the pressure to $500 atm$ hoping to squeeze it into a liquid. Will it work? ( $T_{c} (O_{2}) = 155 K$ .)

Forecast: Guess now — does 500 atm liquefy it, yes or no?

Compare $T$ with $T_{c}$ . $T = 300 K$ , $T_{c} = 155 K$ , so $T > T_{c}$ .
- Why this step? Liquefaction needs intermolecular attractions to beat thermal jiggling. Above $T_{c}$ the molecules are too fast — see Intermolecular forces. Pressure just makes a denser gas, never a liquid.
Conclusion: No pressure — not 500, not 5000 atm — liquefies $O_{2}$ at 300 K. You must first cool below 155 K.

Verify: $300 - 155 = 145 K$ of over-temperature. Positive gap ⇒ liquefaction impossible by pressure alone. Units: kelvin vs kelvin, consistent. ✔

Worked example Ex 2 — Cell B:

$T < T_{c}$ , pressure now works (and how much, via Clausius–Clapeyron) Statement. The same oxygen is cooled to $140 K$ . Can pressure now liquefy it, and estimate the exact pressure. Use $T_{c} (O_{2}) = 155 K$ , $P_{c} = 50 atm$ , and molar enthalpy of vaporisation $Δ H_{vap} \approx 6820 J mol^{- 1}$ , $R = 8.314 J K^{- 1} mol^{- 1}$ .

Forecast: Above or below the ceiling — and does "any big pressure" do it, or a specific number?

Compare $T$ with $T_{c}$ . $T = 140 K < 155 K = T_{c}$ .
- Why this step? Below $T_{c}$ , attractions can dominate. Now compression can actually condense the gas.
Name the target pressure. At a fixed $T < T_{c}$ there is a specific saturation vapour pressure $P_{vap} (T)$ at which gas and liquid coexist. You must reach $P \geq P_{vap} (T)$ to get liquid; below it the substance stays gas.
- Why this step? Condensation is not "the more the merrier from zero." Nothing liquefies until you hit the phase boundary $P_{vap} (T)$ ; any higher pressure keeps it liquid. So "enough pressure" is a specific number, not an arbitrary huge one.
Compute it with the Clausius–Clapeyron equation. The pressure on the gas–liquid boundary obeys $ln \frac{P _{2}}{P _{1}} = - \frac{Δ H _{vap}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Anchor at the critical point $(T_{1}, P_{1}) = (155 K, 50 atm)$ and solve for $P_{2} = P_{vap} (140 K)$ : $ln \frac{P _{2}}{50} = - \frac{6820}{8.314} (\frac{1}{140} - \frac{1}{155}) = - 820.3 \times (0.0007143 - 0.0064516 corrected below) .$ Working it cleanly: $\frac{1}{140} - \frac{1}{155} = 0.0071429 - 0.0064516 = 0.0006913 K^{- 1}$ , and $Δ H_{vap} / R = 6820/8.314 = 820.3 K$ , so $ln \frac{P _{2}}{50} = - 820.3 \times 0.0006913 = - 0.5672 \Rightarrow P_{2} = 50 e^{- 0.5672} \approx 28.4 atm .$
- Why this step? Clausius–Clapeyron is exactly the tool that turns "there is some $P_{vap}$ " into a number, using how vaporisation enthalpy bends the boundary. (Antoine's equation is the more accurate empirical cousin; CC is the physics-first estimate.)
Conclusion: Raising $P$ to about $28 atm$ liquefies $O_{2}$ at $140 K$ .

Verify: $6820/8.314 = 820.3 K$ ; $\frac{1}{140} - \frac{1}{155} = 0.0006913$ ; product $= 0.5672$ ; $50 e^{- 0.5672} = 28.4 atm$ , which sensibly sits below $P_{c} = 50 atm$ (vapour pressure must be under the critical pressure). ✔

Worked example Ex 4 — Cell D:

$T > T_{i}$ , throttling warms Statement. Hydrogen at $300 K$ is thrown straight through a throttle. Given $T_{i} (H_{2}) \approx 200 K$ , what happens — and what is the fix?

Forecast: Does raw hydrogen cool or warm on throttling?

Compare $T$ with $T_{i}$ . $T = 300 K > 200 K = T_{i}$ .
- Why this step? Above $T_{i}$ , $μ_{J T} < 0$ : the repulsive/kinetic term wins, and expansion warms the gas.
Direction: $Δ T = μ_{J T} Δ P$ with $μ_{J T} < 0$ and $Δ P < 0$ ⇒ $Δ T > 0$ — warming. Throttling hydrogen at 300 K is worse than useless.
Fix: pre-cool with liquid nitrogen to $\approx 77 K$ . Now $77 K < 200 K$ , so $μ_{J T} > 0$ and throttling finally cools.

Verify: Before fix: $300 > 200$ ⇒ warms. After fix: $77 < 200$ ⇒ cools. Sign flip confirmed. ✔ This is the whole reason He and $H_{2}$ are the hardest gases to liquefy.

See the picture below — it carries a fact the algebra does not. The figure plots $μ_{J T}$ (the magnitude and sign of cooling per atm) against temperature, revealing what the pass/fail comparison hides: the cooling coefficient is not constant — it grows as you pre-cool, is largest well below $T_{i}$ , and slides continuously through zero as you cross the red fence at $T_{i}$ . So pre-cooling hydrogen from 300 K to 77 K does not merely "switch cooling on"; it moves you from a negative coefficient (warming) up to a large positive one (strong cooling). The steepness of the curve at $T_{i}$ is exactly why the process is so sensitive to getting below the inversion temperature.

Figure — Liquefaction of gases — Linde, Claude processes (concept)

The single idea to carry away: $μ_{J T}$ passes smoothly through zero at $T_{i}$ , so the closer you push below $T_{i}$ , the more cooling you extract per atmosphere — the payoff of pre-cooling is more than a yes/no flip.

Worked example Ex 5 — Cell E (degenerate): the ideal gas,

$μ_{J T} = 0$ Statement. A hypothetical ideal gas ( $P V = n R T$ exactly) is throttled from 200 atm to 1 atm. Find $Δ T$ .

Forecast: Does an ideal gas cool, warm, or do nothing?

Use the identity we derived above. From the boxed result, $μ_{J T} = \frac{1}{C _{p}} [T (\frac{\partial V}{\partial T})_{P} - V]$ , where $C_{p}$ is the constant-pressure heat capacity defined earlier.
- Why this step? This form isolates exactly the real-gas effect inside the bracket; $C_{p} > 0$ only scales it, so the sign lives in the bracket.
Plug in ideal-gas $V$ . For one mole, $V = R T / P$ , so $(\frac{\partial V}{\partial T})_{P} = R / P$ . Then $T (\frac{\partial V}{\partial T})_{P} - V = T \cdot \frac{R}{P} - \frac{R T}{P} = 0.$
- Why this step? Substituting the ideal law shows the bracket cancels exactly — no leftover attraction term ( $a = 0$ ).
Conclusion: $μ_{J T} = 0$ , so $Δ T = 0$ regardless of the 199 atm drop.

Verify: Bracket $= R T / P - R T / P = 0$ . The ideal gas has no attractions, nothing to "stretch," so no self-cooling. ✔ This kills the common "spray can" misconception for throttling.

Worked example Ex 7 — Cell G: compute

$T_{i}$ from van der Waals $a, b$ Statement. For a gas with van der Waals equation constants $a = 1.35 L^{2} atm mol^{- 2}$ and $b = 0.0387 L mol^{- 1}$ , estimate the inversion temperature. Use $R = 0.0821 L atm K^{- 1} mol^{- 1}$ .

Forecast: Roughly how hot is the inversion fence — tens, hundreds, or thousands of K?

Recall the low-pressure estimate. $T_{i} \approx \frac{2 a}{R b}$ .
- Why this step? Setting $μ_{J T} = 0$ (the fence) for a van der Waals gas gives this — but only in the limit $P \to 0$ (low pressure). It is a single number, the maximum inversion temperature, not the whole story.
The full picture (range of validity). In reality $μ_{J T} = 0$ traces a whole inversion curve in the $T$ – $P$ plane: at each pressure there are generally two inversion temperatures (an upper and a lower one) that pinch together at high $P$ . The formula $2 a / (R b)$ is just the low-pressure upper branch — valid when pressure is small compared with $P_{c}$ . Above a certain pressure the gas never cools by throttling at any $T$ . Treat $2 a / (R b)$ as a "best-case ceiling," accurate near $P \to 0$ .
Plug in. $T_{i} \approx \frac{2 ( 1.35 )}{( 0.0821 ) ( 0.0387 )} = \frac{2.70}{0.003177} \approx 850 K .$
- Why this step? Units cancel: $\frac{L ^{2} atm mol ^{- 2}}{( L atm K ^{- 1} mol ^{- 1} ) ( L mol ^{- 1} )} = K$ . ✔
Interpret. $T_{i} \approx 850 K ≫ 300 K$ : at low pressure this gas cools on throttling at room temperature (like $N_{2}$ ).

Verify: $2 \times 1.35 = 2.70$ ; $0.0821 \times 0.0387 = 0.003177$ ; $2.70/0.003177 \approx 850 K$ . ✔

Worked example Ex 8 — Cell H: Claude — work-doing expansion cools even an ideal gas

Statement. One mole of an ideal gas ( $C_{v} = 20.8 J K^{- 1} mol^{- 1}$ ) expands adiabatically doing 2000 J of work in a Claude turbine. Find $Δ T$ .

Forecast: Ideal gas + turbine — does it cool, and by how much?

First law, adiabatic. From the First law of thermodynamics, $Δ U = q + w$ with $q = 0$ , so $Δ U = w$ .
- Why this step? Adiabatic means no heat crosses the wall; energy for the work comes from internal energy.
Work sign. The gas does 2000 J on the surroundings, so $w = - 2000 J$ (system loses energy).
- Why this step? We use the physicist's sign convention where $w$ is the work done on the system. When the gas pushes the turbine outward, energy leaves the system, so $w$ must carry a minus sign — otherwise the first law would let the gas gain energy while doing work, which is impossible.
Link $Δ U$ to $Δ T$ . $Δ U = C_{v} Δ T$ (this is what $C_{v}$ , defined at the top, is for), so $Δ T = \frac{Δ U}{C _{v}} = \frac{- 2000 J}{20.8 J K ^{- 1}} \approx - 96.2 K .$
- Why this step? Internal energy of an ideal gas depends only on $T$ ; spending it forces $T$ down.

Verify: $- 2000/20.8 = - 96.15 K$ — strong cooling, and it happened for an ideal gas (no van der Waals $a$ needed). This is precisely why Claude beats Linde: the work term cools where JT cannot. Units: $J / (J K^{- 1}) = K$ . ✔

Worked example Ex 9 — Cell I: real-world word problem

Statement. An air-liquefaction plant takes air at $300 K$ , $200 atm$ with $μ_{J T} \approx 0.20 K/atm$ and throttles it to $1 atm$ in one pass. (a) Estimate the temperature drop. (b) Air must reach $\approx 80 K$ to liquefy — does one pass suffice, and what fixes it?

Forecast: Guess: does one throttle liquefy air?

(a) One-pass drop. $Δ T \approx μ_{J T} Δ P = (0.20 K/atm) (1 - 200) atm = (0.20) (- 199) = - 39.8 K$ .
- Why this step? Rate $\times$ pressure lost. $Δ P = P_{final} - P_{initial} < 0$ .
Resulting temperature. $300 - 39.8 = 260.2 K$ , still far above 80 K.
(b) Fix — regenerative feedback. One pass fails; the counter-current heat exchanger recycles the cold output to pre-cool the incoming gas, stacking many $\approx 40 K$ drops until air reaches $\approx 80 K$ and condenses.
- Why this step? A single JT drop is small; the Linde loop ratchets temperature down cycle after cycle.

Verify: $0.20 \times 199 = 39.8 K$ per pass; $300 - 39.8 = 260.2 K > 80 K$ ⇒ needs many passes. ✔ Roughly $(300 - 80) /39.8 \approx 5.5$ ideal passes — realistically more, hence the exchanger.

Worked example Ex 10 — Cell J: exam twist — two thresholds at once

Statement. Helium sits at $300 K$ . Given $T_{c} (He) = 5.2 K$ and $T_{i} (He) \approx 40 K$ , list — in order — every step needed to liquefy it, and say which single number rules out throttling at room temperature.

Forecast: Which threshold, $T_{c}$ or $T_{i}$ , blocks you first at 300 K?

Check $T_{i}$ first. $300 K > 40 K = T_{i}$ , so at 300 K $μ_{J T} < 0$ : throttling would warm He. This is the number that rules out throttling right away.
- Why this step? Before you can even use JT cooling you must be below $T_{i}$ .
Pre-cool below $T_{i}$ . Cool He with cold hydrogen/nitrogen stages until $T < 40 K$ . Now $μ_{J T} > 0$ .
Check $T_{c}$ . To actually get liquid you must reach $T < 5.2 K$ . So keep throttling/pre-cooling below the critical ceiling.
Then compress to condense. Order: below $T_{i}$ → cool below $T_{c}$ → compress.

Verify: $300 > 40$ (throttle warms) and $300 > 5.2$ (can't condense) — both thresholds violated at 300 K, with $T_{i} = 40 K$ the first hurdle (it is the higher of the two thresholds, so you meet it first coming down from 300 K). Helium is the hardest gas exactly because both thresholds are so low. ✔

Worked example Ex 11 — Cell K: the exact boundaries

$T = T_{c}$ and $T = T_{i}$ Statement. Two knife-edge questions. (a) $C O_{2}$ at exactly $T = T_{c} = 304.2 K$ — can pressure liquefy it, and what does the meniscus look like? (b) A gas sitting at exactly $T = T_{i}$ is throttled — what is $Δ T$ ?

Forecast: At the exact boundary, do we get the "yes" side or the "no" side — or something new?

(a) At $T = T_{c}$ exactly. The gas–liquid boundary shrinks to a single point (the critical point). At $T = T_{c}$ the vapour pressure equals $P_{c}$ and there is no distinct liquid–gas interface: compressing to $P_{c}$ gives the critical opalescent state, not a normal liquid with a meniscus. For any $T > T_{c}$ liquefaction is impossible (Cell A); it becomes possible only for $T < T_{c}$ (Cell B). So $T = T_{c}$ is the exact "last chance" edge — the density of gas and liquid become equal there.
- Why this step? The boundary case tells you Cell A and Cell B meet at a single point where the two phases become indistinguishable — no contradiction, just a smooth handover.
(b) At $T = T_{i}$ exactly. By definition the inversion temperature is where $μ_{J T} = 0$ . Then $Δ T = μ_{J T} Δ P = 0 \times Δ P = 0,$ for any pressure drop. The gas neither cools nor warms on throttling — it behaves momentarily like an ideal gas.
- Why this step? This is exactly what makes $T_{i}$ the fence: it is the single temperature where the Cell C (cools) and Cell D (warms) behaviours cancel, so throttling does nothing.

Verify: (a) At $T = T_{c}$ , $P_{vap} = P_{c}$ and the interface vanishes — consistent with Ex 2's remark that $P_{vap} \to P_{c}$ as $T \to T_{c}$ . (b) $0 \times Δ P = 0$ for every $Δ P$ , so $Δ T = 0$ identically at $T_{i}$ . ✔

Recall Rapid self-test (cover the answers)

Which threshold decides if pressure alone can liquefy a gas? ::: $T_{c}$ (critical temperature) — need $T < T_{c}$ . Once $T < T_{c}$ , what pressure actually liquefies it, and what tool finds it? ::: $P \geq P_{vap} (T)$ , the saturation vapour pressure; estimate it with the Clausius–Clapeyron equation. What happens exactly at $T = T_{c}$ ? ::: Gas and liquid densities become equal, no meniscus — the critical point; for $T > T_{c}$ liquefaction is impossible. Which threshold decides if throttling cools or warms? ::: $T_{i}$ (inversion temperature) — cools when $T < T_{i}$ . What is $Δ T$ on throttling exactly at $T = T_{i}$ ? ::: Zero — $μ_{J T} = 0$ there, so $Δ T = μ_{J T} Δ P = 0$ for any $Δ P$ . $μ_{J T}$ for an ideal gas throttled from 200 to 1 atm? ::: Exactly $0$ — no attractions, no self-cooling. What is $C_{v}$ and where did we use it? ::: Constant-volume heat capacity; it links $Δ U = C_{v} Δ T$ in the Claude work-doing example. Formula for $Δ T$ across a small throttle? ::: $Δ T \approx μ_{J T} Δ P$ . Low-pressure van der Waals estimate of $T_{i}$ ? ::: $T_{i} \approx 2 a / (R b)$ , valid only as $P \to 0$ (upper branch of the full inversion curve). Why does Claude cool an ideal gas but Linde cannot? ::: Claude's gas does external work ( $Δ U = w < 0 \Rightarrow Δ T < 0$ ); Linde relies on real-gas attractions only.