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

A quick reminder of the toolbox, so you never meet an undefined symbol below:

Here $R = 0.0821 L atm K^{- 1} mol^{- 1} = 8.314 J K^{- 1} mol^{- 1}$ (the gas constant).

Three gases have critical temperatures $T_{c} (CO_{2}) = 304 K$ , $T_{c} (N_{2}) = 126 K$ , $T_{c} (He) = 5.2 K$ . Which of these can be liquefied by pressure alone at room temperature ( $298 K$ )?

Recall Solution

What we do: Compare each $T_{c}$ with the room temperature $298 K$ . Why: Liquefaction by pressure only works when $T < T_{c}$ (below $T_{c}$ attractions can win; above it, no pressure helps).

$CO_{2}$ : $304 K > 298 K$ ✔ — room temperature is below its $T_{c}$ , so yes, compression alone liquefies it.
$N_{2}$ : $126 K < 298 K$ ✘ — must cool below $126 K$ first.
$He$ : $5.2 K < 298 K$ ✘ — must cool below $5.2 K$ first. Answer: Only $CO_{2}$ .

L1-Q2

For an ideal gas passing through a throttle (porous plug), what is $μ_{J T}$ , and does the gas cool, warm, or stay the same?

Recall Solution

Answer: $μ_{J T} = 0$ ; the temperature stays the same. Why: An ideal gas has no intermolecular attraction ( $a = 0$ ). Throttling cools a real gas because pulling attracting molecules apart steals kinetic energy. With no attraction, there is nothing to "pay for" — so $Δ T = 0$ .

Level 2 — Application

L2-Q1

For nitrogen, $a = 1.37 L^{2} atm mol^{- 2}$ and $b = 0.0387 L mol^{- 1}$ . Estimate its inversion temperature $T_{i}$ using $T_{i} \approx \frac{2 a}{R b}$ with $R = 0.0821 L atm K^{- 1} mol^{- 1}$ .

Recall Solution

What we do: Plug into $T_{i} = \frac{2 a}{R b}$ . Why this formula: Setting $μ_{J T} = 0$ (the boundary between cooling and warming) for a low-pressure van der Waals gas gives exactly this expression. $T_{i} = \frac{2 ( 1.37 )}{( 0.0821 ) ( 0.0387 )} = \frac{2.74}{0.003177} \approx 863 K$ Answer: $T_{i} \approx 863 K$ . Sanity check: Room temperature ( $298 K$ ) is far below $863 K$ , so $N_{2}$ cools on throttling at room temperature — consistent with the fact that the Linde process works on air.

L2-Q2

Helium has $a = 0.0341 L^{2} atm mol^{- 2}$ , $b = 0.0237 L mol^{- 1}$ . Estimate $T_{i}$ . Will helium cool or warm if throttled at $298 K$ ?

Recall Solution

$T_{i} = \frac{2 ( 0.0341 )}{( 0.0821 ) ( 0.0237 )} = \frac{0.0682}{0.001946} \approx 35 K$ Answer: $T_{i} \approx 35 K$ . Since $298 K > 35 K$ , helium is above its inversion temperature at room conditions. Therefore $μ_{J T} < 0$ and throttling warms it. To liquefy He we must pre-cool below $35 K$ first (this is exactly why the parent note says He and $H_{2}$ resist the simple trick).

Level 3 — Analysis

L3-Q1

Using the exact identity $μ_{J T} = \frac{1}{C _{p}} [T (\frac{\partial V}{\partial T})_{P} - V]$ , explain why the sign of $μ_{J T}$ flips as temperature rises past $T_{i}$ . Reference the figure.

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

Recall Solution

The bracket decides the sign (because $C_{p} > 0$ always). Two competing terms:

$V$ — the actual volume the gas occupies.
$T (\frac{\partial V}{\partial T})_{P}$ — how strongly the volume wants to expand on heating, scaled by $T$ .

Low $T$ (below $T_{i}$ ): attractions pull molecules inward, so on heating the volume expands eagerly — $T (\frac{\partial V}{\partial T})_{P}$ is large, larger than $V$ . Bracket $> 0$ , so $μ_{J T} > 0$ → cooling (green region in the figure).

High $T$ (above $T_{i}$ ): molecules move too fast for attractions to matter much; the gas behaves more ideally, and the repulsive $b$ -volume term dominates. Now $T (\frac{\partial V}{\partial T})_{P} < V$ , the bracket flips negative, $μ_{J T} < 0$ → warming (coral region).

At $T = T_{i}$ : the two terms exactly cancel, $μ_{J T} = 0$ — the crossover line.

L3-Q2

In the Linde process, a single throttling step produces only a small temperature drop. Explain why the process nonetheless reaches liquefaction, and what would happen if the counter-current heat exchanger were removed.

Recall Solution

Why it still works: The regenerative (counter-current) heat exchanger feeds the cold, already-expanded gas back around the incoming warm gas. Each fresh batch therefore starts colder than the previous one, and its small $Δ T$ stacks on top of an already-lowered baseline. This is a cascade: many small drops accumulate until the gas dips below $T_{c}$ and part of it condenses. Without the exchanger: every batch would start at the same warm temperature and drop by the same small amount — the temperature could never ratchet down to $T_{c}$ . Liquefaction would fail. The exchanger is the true engine of the process; the throttle alone is far too weak.

Level 4 — Synthesis

L4-Q1

Air at $298 K$ (mostly $N_{2}$ , $T_{i} \approx 863 K$ ) is to be liquefied by Linde, while $H_{2}$ ( $T_{i} \approx 200 K$ ) also needs liquefying. Design a correct ordering of steps for each, and justify why they differ.

Recall Solution

Air ( $N_{2}$ ): $298 K < T_{i} = 863 K$ , so it already cools on throttling. Sequence:

Compress (~200 atm), cool back the compression heat with water.
Pass down counter-current exchanger.
Throttle → JT cooling (works immediately, since below $T_{i}$ ).
Cold gas returns, pre-cools next batch → cascade → liquid air.

Hydrogen: $298 K > T_{i} = 200 K$ , so throttling warms it. We must pre-cool below $200 K$ first:

Pre-cool $H_{2}$ using liquid nitrogen (~ $77 K$ , safely below $200 K$ ).
Now $H_{2}$ is below $T_{i}$ → throttling cools it.
Proceed with the Linde cascade.

Why they differ: The whole difference is one comparison — is the starting temperature below $T_{i}$ ? For $N_{2}$ yes; for $H_{2}$ no, hence the extra pre-cooling stage.

L4-Q2

One mole of an ideal monatomic gas ( $C_{v} = \frac{3}{2} R$ , $R = 8.314 J K^{- 1} mol^{- 1}$ ) expands adiabatically and reversibly (Claude-style, doing work), losing $1500 J$ of internal energy. Find its temperature drop $Δ T$ . Then contrast with what throttling the same ideal gas would do.

Recall Solution

Claude expansion (does work): Adiabatic ( $q = 0$ ) ⇒ $Δ U = w$ . The gas does work on the surroundings, so $w < 0$ and $Δ U = - 1500 J$ . Using $Δ U = C_{v} Δ T$ : $Δ T = \frac{Δ U}{C _{v}} = \frac{- 1500}{\frac{3}{2} ( 8.314 )} = \frac{- 1500}{12.471} \approx - 120.3 K$ Answer: the gas cools by about $120 K$ . Contrast with throttling the same ideal gas: throttling keeps $H$ constant and does no work; for an ideal gas $μ_{J T} = 0$ , so $Δ T = 0$ — no cooling at all. This is precisely why Claude beats Linde: the work-doing term cools even an ideal gas, whereas JT cooling needs real-gas attractions.

Level 5 — Mastery

L5-Q1

For a van der Waals gas, the low-pressure Joule–Thomson coefficient is approximately $μ_{J T} \approx \frac{1}{C _{p}} (\frac{2 a}{R T} - b) .$ (a) Derive the inversion temperature from this. (b) For $CO_{2}$ ( $a = 3.59 L^{2} atm mol^{- 2}$ , $b = 0.0427 L mol^{- 1}$ ), compute $T_{i}$ . (c) Is $CO_{2}$ cooled or warmed on throttling at $298 K$ ? Tie the answer back to why $CO_{2}$ is easy to liquefy.

Recall Solution

(a) The inversion temperature is where $μ_{J T} = 0$ . Since $C_{p} > 0$ , set the bracket to zero: $\frac{2 a}{R T _{i}} - b = 0 ⟹ \frac{2 a}{R T _{i}} = b ⟹ T_{i} = \frac{2 a}{R b} .$ This recovers exactly the van der Waals estimate from the parent note. ✔

(b) With $R = 0.0821$ : $T_{i} = \frac{2 ( 3.59 )}{( 0.0821 ) ( 0.0427 )} = \frac{7.18}{0.003506} \approx 2048 K .$

(c) $298 K ≪ 2048 K$ , so $CO_{2}$ is far below its inversion temperature ⇒ $μ_{J T} > 0$ ⇒ cools on throttling. Its huge $a$ (strong attractions) both raises $T_{i}$ (easy JT cooling over a wide range) and raises $T_{c}$ (= $304 K$ , above room temperature). Both facts flow from the same cause: strong intermolecular attraction (see Intermolecular forces). That is why $CO_{2}$ is one of the easiest gases to liquefy.

L5-Q2

Rank $He$ , $H_{2}$ , $N_{2}$ , $CO_{2}$ by "ease of liquefaction by the JT route at $298 K$ ," using their inversion temperatures ( $T_{i} \approx 35, 200, 863, 2048 K$ respectively). State which need pre-cooling and explain the single underlying molecular property that orders them.

Recall Solution

Rank (easiest → hardest): $CO_{2}$ > $N_{2}$ > $H_{2}$ > $He$ .

$CO_{2}$ ( $T_{i} = 2048$ ) and $N_{2}$ ( $T_{i} = 863$ ): both above room temperature is false — room ( $298 K$ ) is below their $T_{i}$ , so both cool directly on throttling. Easiest.
$H_{2}$ ( $T_{i} = 200$ ) and $He$ ( $T_{i} = 35$ ): $298 K$ is above their $T_{i}$ , so they warm on throttling — both need pre-cooling (H $_{2}$ with liquid N $_{2}$ ; He needs even colder liquid H $_{2}$ ).

The single ordering property: the strength of intermolecular attraction, captured by the van der Waals constant $a$ . Larger $a$ → larger $T_{i} = 2 a / (R b)$ → the gas cools by JT over a wider temperature range → easier to liquefy. The order of $a$ (CO $_{2}$ ≫ N $_{2}$ ≫ H $_{2}$ ≫ He) is exactly the order of ease.

Recall One-line summary of the whole ladder

Compare $T$ with $T_{c}$ (can it liquefy at all?) and with $T_{i}$ (does throttling cool it?) — everything on this page is a variation on those two comparisons, and both are governed by the attraction constant $a$ .

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

Level 1 — Recognition

L1-Q1

L1-Q2

Level 2 — Application

L2-Q1

L2-Q2

Level 3 — Analysis

L3-Q1

L3-Q2

Level 4 — Synthesis

L4-Q1

L4-Q2

Level 5 — Mastery

L5-Q1

L5-Q2