Thermodynamic processes

The puzzle: two ways to double a volume, neither of them heated¶

A cylinder of helium sits inside a perfect insulator. No heat can enter it and none can leave. Whatever we do to this gas, $Q = 0$ .

First way. The cylinder is closed by a piston carrying a pile of sand. Remove the grains one at a time. The gas pushes the piston out grain by grain, and by the time the volume has doubled, the helium has cooled from 300 K to 189 K.

Second way. The cylinder is one half of the insulated box; the other half is empty, and a thin partition separates them. Pull the partition out. The gas rushes into the empty half. Its volume has doubled. Its temperature is still 300 K, to the last decimal place.

Same gas, same insulation, same start, same finish volume. In the first case the temperature fell by 37%; in the second it did not move at all.

The answer is not a subtlety about insulation. It is that “adiabatic” and “quasistatic” are two independent conditions, and the famous law needs both. This module separates them, and the separation turns out to be worth roughly half the work an engine could have produced.

Predict before you calculate¶

Commit to an answer for each before reading on. Write them down.

An insulated gas is compressed slowly to half its volume. Does its temperature rise, fall or stay put? Now the same gas, insulated, is allowed to expand slowly to twice its volume. By what factor does its temperature change?
A gas expands into a vacuum inside an insulated box. What are $Q$ , $W_{\mathrm{on}}$ , $\Delta U$ and $\Delta T$ ? Be careful with the third and fourth: they are not the same question.
You want to warm one mole of a gas by exactly one kelvin. Does it take more heat at constant volume or at constant pressure? Where does the difference go?
A gas is compressed by dropping a heavy weight onto its piston, so that it slams down to a new position and stops. Compare the work done on the gas with the work done in compressing it slowly between the same two volumes. Which is larger?

Explore the model¶

The laboratory for this module runs a gas along each of the four quasistatic families out of a common starting state, and then races the three adiabatic routes of the puzzle against each other: slow expansion, expansion against a fixed load, and free expansion into a vacuum.

First, the four families. Every one of them leaves the same point in the $P$ – $V$ plane, and each is a different constraint on what is held fixed:

Left, the four quasistatic families leaving one common state in the pressure–volume plane; right, three stacked panels tracking the work done on the gas, the heat into it, and its change of internal energy as each route is traversed. Blue is isochoric — a vertical line, since the volume never changes — orange isobaric (horizontal), green isothermal, red adiabatic. Watch which bars stay flat at zero: blue in the work panel, because a boundary that does not move does no work; green in the energy panel, because an isotherm ends where it began in temperature; and red in the heat panel, which is what “adiabatic” means. The red curve is the steep one, falling below the green isotherm, because the energy it spends on the piston comes out of the gas itself.

Now the puzzle. Three routes, all adiabatic, all doubling the volume, and nothing in common after that:

Three adiabatic expansions from one state to one final volume, in three panels. Left, the pressure–volume plane: the slow route (red) traces a curve and shades the largest area, the work it delivers; the route against a fixed load (orange) shades the rectangle under its constant external pressure, a third smaller; the free expansion (green) shades nothing, because zero external pressure means zero work. Middle, the work delivered as a bar per route — there are only two visible bars, and the missing green one is the point. Right, temperature against volume: the slow route has a curve because it passes through equilibrium states and has a temperature at every instant, while the other two have only a start dot and an end dot with nothing drawn between them, because in between they have no temperature at all. The three final temperatures are 189 K, 225 K and 300 K.

Model specification

System: a fixed amount of ideal gas, $N$ particles with $f$ quadratic degrees of freedom each, in a cylinder closed by a piston — or in an insulated box divided by a removable partition.
Dynamics: two regimes. Quasistatic, where the gas holds one $(P, V, T)$ throughout and passes through equilibrium states only; and irreversible, where only the endpoints are equilibrium states and the work follows the external pressure.
Boundary: a frictionless piston set in a wall that is either diathermal (heat may cross) or adiabatic (it may not). The partition of the free expansion is adiabatic and does no work.
Ensemble: not applicable to the thermodynamics; the microscopic free expansion in the laboratory is microcanonical, conserving energy exactly.
Ignored: friction, the piston’s own mass and inertia, gas non-ideality, heat leaking through an adiabatic wall, and the time the gas takes to re-equilibrate after each step.
Valid when: quasistatic results — the process is slow compared with the gas’s relaxation time; irreversible results — the external pressure is known and uniform at the moving boundary.
Failure modes: real gases, whose internal energy depends on volume, so their free expansion does change the temperature; compressions fast enough that even the external pressure is not uniform; a wall that leaks, which makes “adiabatic” false rather than approximate.

Three things worth doing before you read the derivation:

Slide the number of degrees of freedom from 3 to 5 to 6 and watch the adiabat flatten towards the isotherm. Ask yourself what $f \to \infty$ would mean physically.
Set the external pressure of the irreversible expansion to zero and check that it becomes the free expansion exactly, not approximately.
Run the microscopic free expansion, where a box of particles is suddenly widened. Nothing in that code touches a single velocity — watch what the measured temperature does.

Derive the result¶

System and boundary. The gas in the cylinder; the piston is the moving boundary.

Independent variables. $N$ , $V$ , $T$ and the number of quadratic degrees of freedom $f$ per particle, with $P$ fixed by the equation of state. Every result below is macroscopic thermodynamics — no microstate is counted anywhere in this section — except the final check in the laboratory, which is kinetic.

Sign convention. As everywhere in this course,

dU = \delta Q + \delta W_{\mathrm{on}}, \qquad \delta W_{\mathrm{on}} = -P_{\mathrm{ext}}\, dV .

(1)

Note which pressure appears there. Work is done at the boundary, by whatever the piston pushes against, so the honest statement of the second equation carries $P_{\mathrm{ext}}$ . Module 5 wrote it with the gas’s own $P$ , which is correct only when the two are equal — and that equality is the definition we are about to make precise.

Heat capacities, and why there are two of them¶

The heat needed to raise a gas’s temperature depends on what you hold fixed while you do it. At constant volume the gas cannot expand, so no work is done and all the heat goes into internal energy:

C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{f}{2} N k_B .

(2)

At constant pressure the gas also pushes its piston out, and that expansion has to be paid for. Writing $\delta Q = dU + P\,dV$ and using $P\,dV = N k_B\, dT$ along an isobar:

C_P = C_V + N k_B .

(3)

The adiabat¶

Now shut the heat off and move slowly. Slowly enough that the gas is in equilibrium at every instant, so it has a single pressure and $P_{\mathrm{ext}} = P$ . Then $\delta Q = 0$ gives

C_V\, dT = -P\, dV .

(5)

Differentiate the equation of state $PV = N k_B T$ to eliminate $dT$ :

P\, dV + V\, dP = N k_B\, dT .

(6)

Substituting, and using $C_V = N k_B/(\gamma - 1)$ :

\frac{P\, dV + V\, dP}{\gamma - 1} = -P\, dV \qquad\Longrightarrow\qquad V\, dP = -\gamma P\, dV .

(7)

Separate and integrate:

\frac{dP}{P} = -\gamma \frac{dV}{V} \qquad\Longrightarrow\qquad \boxed{\; P V^{\gamma} = \text{constant} \;}

(8)

and, folding the equation of state back in, the same law in the variables that matter for the puzzle:

T V^{\gamma - 1} = \text{constant} .

(9)

For helium ( $\gamma = 5/3$ ) doubling the volume multiplies the temperature by $2^{-2/3} = 0.630$ : from 300 K to 189 K, exactly as the first cylinder did.

The four families, side by side¶

Each family is one constraint, and the first law does the rest.

Process	Held fixed	$W_{\mathrm{on}}$	$Q$	$\Delta U$
Isochoric	$V$	0	$C_V \Delta T$	$C_V \Delta T$
Isobaric	$P$	$-P \Delta V$	$C_P \Delta T$	$C_V \Delta T$
Isothermal	$T$	$-N k_B T \ln(V_2/V_1)$	$+N k_B T \ln(V_2/V_1)$	0
Adiabatic	$Q = 0$	$\dfrac{P_2V_2 - P_1V_1}{\gamma - 1}$	0	$C_V \Delta T$

The last column is one expression, $C_V \Delta T$ , in every row — the isotherm’s zero is that same expression with $\Delta T = 0$ — because $U$ is a state function and does not care which row it is in. Everything that distinguishes the rows lives in the two path functions beside it. Note in particular that $C_V$ appears in the isobaric row’s $\Delta U$ : the heat capacity at constant volume is what converts a temperature change into an energy change, whatever the process was.

Irreversible processes: the work follows the external pressure¶

Now drop the quasistatic assumption. The gas is compressed by a weight slammed onto the piston, or it expands into a vacuum. During the process the gas near the piston is at a different pressure from the gas at the far end; there is no single $P$ , no point in the $P$ – $V$ plane, and therefore no curve. What survives is the definition:

W_{\mathrm{on}} = -\int P_{\mathrm{ext}}\, dV .

(10)

For a constant external pressure this is just $-P_{\mathrm{ext}} (V_2 - V_1)$ , whatever the gas is doing internally.

The free expansion. Set $P_{\mathrm{ext}} = 0$ : the gas expands into a vacuum, so nothing pushes back on the piston — or, in the partition version, there is no piston at all. Then

W_{\mathrm{on}} = 0, \qquad Q = 0, \qquad \Delta U = 0 .

(11)

And for an ideal gas, whose internal energy depends on temperature alone, $\Delta U = 0$ forces $\Delta T = 0$ . The gas does not cool. It cannot: cooling would mean losing energy, and there is no channel open for energy to leave through.

Expansion against a fixed load. Between the two extremes sits the case where the piston carries a fixed weight, so $P_{\mathrm{ext}}$ is constant and non-zero. The gas expands until its own pressure has fallen to meet the load, and

T_2 = \frac{C_V T_1 + P_{\mathrm{ext}} V_1}{C_P} .

(12)

Set the load so that the gas stops at exactly twice its volume, and this gives 225 K — between the 189 K of the slow route and the 300 K of the free one.

The three routes, in numbers

All three are adiabatic. All three start at 300 K and finish at twice the volume.

Route	Final $T$	$-W_{\mathrm{on}}$ in units of $P_1V_1$
Slow (quasistatic adiabat)	189.0 K	0.555
Against a fixed load	225.0 K	0.375
Free expansion	300.0 K	0

The slow route delivers 48% more work than the loaded one and infinitely more than the free one, for an identical change of volume. The ordering is not an accident of these numbers, and the reason it can never reverse is the subject of module 7.

Now run it backwards, which is prediction 4. Compress the same gas to half its volume, insulated, two ways. Slowly, grain by grain: it ends at 476 K, having absorbed $0.881\,P_1V_1$ of work. Or drop a block heavy enough to hold $P_{\mathrm{ext}} = 6P_1$ until the piston stops — which is what “slamming” it means quantitatively — and it ends at 900 K, having absorbed $3.000\,P_1V_1$ .

The sudden compression costs 3.4 times the work and leaves the gas 424 K hotter, for the identical change of volume. Both facts are the same fact: $Q = 0$ , so $\Delta U = W_{\mathrm{on}}$ , and work that went in had nowhere else to go. If you predicted that slamming the piston was cheaper because it was quicker, this is the paragraph to reread.

So the slow route is the best of both worlds and it is not a coincidence: it extracts the most work on the way out and costs the least on the way in. Every departure from quasistatic is paid for, in the same direction, whichever way the piston is travelling.

Verify computationally¶

Six checks, each of which is also an automated test in the project’s suite, so nothing on this page can quietly rot.

1. The closed forms. Numerically integrating $-\int P\,dV$ along each family reproduces the corresponding row of the table above to a relative error below 10^-6, and the trapezoid rule’s error falls at second order as the sampling is refined.

2. Mayer’s relation, as a consistency check rather than an assumption. Along an isobar the work is computed from $-P\Delta V$ and the heat from $C_P \Delta T$ , by two independent formulae. The first law then has to close, and it does — which is a live test of $C_P - C_V = N k_B$ , since any other value would break it.

3. The adiabatic invariants, and what that check is worth. Sampling the adiabat and evaluating $P V^{\gamma}$ and $T V^{\gamma-1}$ at every point gives two constants, flat to machine precision along the whole curve. Be clear about how much of that is evidence: the curve is drawn from $P V^{\gamma} = \text{constant}$ , so the first invariant is true by construction and only the second — that the temperature form follows through the equation of state — is a result. A test that can only pass is worth naming as such rather than counting.

4. The adiabat rebuilt from below, which is the check that is not circular. Divide the expansion into $n$ equal steps and let the gas expand irreversibly against a constant load at each one, matched to its pressure at the start of that step. No step uses $P V^{\gamma}$ anywhere; each is nothing but the first law and the equation of state. One step lands at 100 K, sixteen at 185.6 K, and 256 within 0.11% of the quasistatic 189.0 K, with the error falling as $1/n$ .

That is the sand-grain argument made numerically, and it is the module’s strongest single verification: it shows the quasistatic adiabat is the limit of a sequence of irreversible processes rather than a separate law, and it reaches the same 189 K by a route that never assumes the answer.

5. The ordering of the three routes. The final temperatures come out 189.0 K, 225.0 K and 300.0 K, in that order, while the works run the other way. For the free expansion both $W_{\mathrm{on}}$ and $Q$ are exactly zero — they are not computed, they are put in — and its $\Delta U$ comes out within one part in 10¹⁵ of zero, which is as close to zero as a float can get. Note that this is checked fractionally: these energies are around $10^{-18}\,\mathrm{J}$ , so an absolute tolerance would report every route’s work as equal to every other’s.

6. The free expansion from underneath. This is the check worth running yourself.

Transfer the idea¶

The quasistatic adiabat and its irreversible cousins are everywhere once you know the shape.

The atmosphere. A parcel of air pushed up a mountainside expands as the pressure around it drops, does work on the air it displaces, and has no time to exchange heat — a quasistatic adiabat in the sky. The resulting lapse rate, near $10\ \mathrm{K/km}$ for dry air, follows from $T V^{\gamma-1}$ being constant and nothing else. Clouds form where the cooling reaches the dew point.
The diesel engine. No spark plug. Air is compressed adiabatically by a factor of about 20, which by $T V^{\gamma-1} = \text{constant}$ raises it past 700 K, and the fuel injected at that moment ignites on contact. The engine is a working demonstration that adiabatic compression heats.
A bicycle pump gets hot at the barrel for the same reason, and a spray can gets cold for its mirror image. Both are the work channel, not the heat channel.
Throttling. Push a gas slowly through a porous plug from high pressure to low. It is irreversible and adiabatic, and what is conserved is not $U$ but the enthalpy $H = U + PV$ . For an ideal gas that still gives $\Delta T = 0$ ; for a real one it gives the cooling that liquefies air industrially. Enthalpy earns its own module (10).
Sound. A sound wave compresses air far too fast for heat to diffuse between compression and rarefaction, so the compressions are adiabatic rather than isothermal. Newton assumed isothermal and got the speed of sound wrong by the factor $\sqrt{\gamma}$ ; Laplace supplied the $\gamma$ a century later.

Check your understanding¶

An ideal gas is heated at constant volume, then a second identical sample is heated at constant pressure through the same temperature rise. Which statement is correct?
A. Both gain the same internal energy, but the second absorbs more heat. B. Both absorb the same heat, but the second gains more internal energy. C. Both absorb the same heat and gain the same internal energy, since the temperature rise is the same. D. The first absorbs more heat, because at constant volume none of it is wasted on work.
Check your answer
Correct answer: A — Both gain the same internal energy, but the second absorbs more heat.
- A. Correct. Internal energy depends only on temperature for an ideal gas, so equal temperature rises mean equal changes in U. The constant-pressure sample also pushes its piston out, and that work has to be paid for by extra heat — exactly N k_B per kelvin, which is Mayer’s relation.
- B. This has the two quantities the wrong way round. Internal energy is fixed by the temperature alone, so it cannot differ between two samples with the same temperature rise. Heat is not fixed by the temperature, because it is a path function.
- C. The internal energies do match, for exactly that reason. The heats do not: the constant-pressure sample spends part of its heat on expansion work, so it needs more of it to reach the same temperature.
- D. The sign is inverted. Doing no work means the constant-volume sample needs LESS heat to achieve a given temperature rise, which is why C_V is the smaller of the two heat capacities.
An ideal gas fills half of a rigid, perfectly insulated box; the other half is evacuated. The partition is removed. What happens to the temperature of the gas?
A. It does not change, because no heat entered or left and the gas did no work on anything. B. It falls, because expanding gases cool. C. It falls, by the factor 2^(-2/3) that the adiabatic law gives for a monatomic gas. D. It rises, because the gas has more room and the molecules speed up.
Check your answer
Correct answer: A — It does not change, because no heat entered or left and the gas did no work on anything.
- A. Correct. There is nothing to push against, so W_on = 0; the box is insulated, so Q = 0; hence the internal energy is unchanged, and for an ideal gas that fixes the temperature exactly. The final pressure is halved, but nothing else moved.
- B. Expanding gases cool when they do work on something while insulated — a spray can, a rising air parcel. Here there is nothing outside the gas to push on, so no energy leaves by either channel. The cooling was never caused by the expansion; it was caused by the paying, and here nothing is paid.
- C. That factor comes from P V^gamma = constant, whose derivation needs the gas to have a single well-defined pressure at every instant. During a free expansion it does not, so the law does not apply here at all.
- D. Nothing accelerates the molecules. Their speeds are untouched by the removal of a partition, which is precisely why the temperature does not move in either direction.
The free expansion of the previous question has Q = 0, so it is adiabatic. Does it obey P V^gamma = constant?
A. No. That law needs the process to be quasistatic as well as adiabatic, and a free expansion is not. B. Yes. Adiabatic means Q = 0, and P V^gamma = constant is what Q = 0 implies. C. Yes, but only approximately, because the expansion is fast. D. No, because a free expansion is not adiabatic — no walls means no insulation.
Check your answer
Correct answer: A — No. That law needs the process to be quasistatic as well as adiabatic, and a free expansion is not.
- A. Correct. The derivation writes C_V dT = -P dV, and the P in there is the gas’s own pressure — which only exists if the gas is in equilibrium throughout. A free expansion is adiabatic and violently non-quasistatic, so it obeys P V = constant instead, which is just the isotherm.
- B. Q = 0 is only one of the two assumptions the derivation uses. The other is that the gas has a single pressure at every instant, and without it there is no P to put inside P dV — the differential equation cannot even be written down.
- C. It is not an approximation that degrades with speed. The free expansion obeys P V = constant exactly, and P V^gamma = constant would predict a final pressure smaller by the factor 2^(-2/3) — about 37% wrong, not slightly wrong.
- D. The box is insulated, so no heat crosses its boundary and the process is adiabatic by any definition. That is what makes it the interesting counterexample: it satisfies the heat condition and still fails the law.
Why is C_P larger than C_V for an ideal gas, and by how much?
A. Because heating at constant pressure also pays for expansion work; the excess is exactly N k_B. B. Because a gas at constant pressure is hotter, and hot gases have larger heat capacities. C. Because at constant pressure the molecules have more room to move, adding degrees of freedom. D. Because the excess N k_B accounts for the heat lost to the container walls.
Check your answer
Correct answer: A — Because heating at constant pressure also pays for expansion work; the excess is exactly N k_B.
- A. Correct, and the excess is remarkable for what it does not depend on: not the number of degrees of freedom, not the temperature, not the substance. It is the expansion work per kelvin, P dV = N k_B dT along an isobar, and nothing else.
- B. The two heat capacities are compared at the same temperature. And for an ideal gas both are independent of temperature, so “hot gases have larger heat capacities” is not true within this model at all.
- C. Degrees of freedom are set by the molecule’s structure, not by how much room it has. A monatomic gas has three translational degrees of freedom in any volume.
- D. Nothing is lost. The excess energy goes into the surroundings as work done by the expanding gas, which is an entirely different channel from heat leaking through a wall — and Mayer’s relation holds exactly for a perfectly insulated container too.
A heavy block is dropped onto a piston, compressing an insulated gas until the piston comes to rest. Which pressure belongs in the work integral W_on = -integral of P dV?
A. The external pressure applied at the boundary, which here is constant and larger than the gas’s own pressure. B. The gas’s own pressure, since it is the gas we are doing work on. C. The average of the gas’s initial and final pressures. D. Neither — work is undefined for an irreversible process.
Check your answer
Correct answer: A — The external pressure applied at the boundary, which here is constant and larger than the gas’s own pressure.
- A. Correct. Work is done at the boundary, against whatever the piston pushes on, so the external pressure is what appears. Only in a quasistatic process are the two equal — and that is the whole content of the word “quasistatic”.
- B. The gas is what receives the energy, but the force at the boundary is set by the block, not by the gas. Worse, during the compression the gas has no single pressure at all — it is denser near the piston than at the far wall.
- C. That average would be the right answer for a straight line in the P-V plane traversed quasistatically. Here there is no line to average along, and the external pressure is a genuinely different number from either endpoint.
- D. Work is perfectly well defined; it is a force acting through a distance at the boundary, and the block’s weight and the piston’s travel are both measurable. What is undefined is the gas’s own pressure during the process, and hence the area under a curve that does not exist.
One mole of a monatomic ideal gas at 400 K expands quasistatically and adiabatically from 5.0 L to 15.0 L. Find the final temperature in kelvin.
Check your answer
Correct answer: 192.3 (± 2.0)
Use T V^(gamma-1) = constant with gamma = 5/3, so T2 = T1 (V1/V2)^(2/3) = 400 x (1/3)^(2/3) = 192.3 K. If you obtained 400 K you applied the isothermal condition instead; 64 K means you put gamma in the exponent where gamma - 1 belongs; 133 K means you used an exponent of 1, which is the isobaric relation V/T = constant.
Two moles of an ideal gas in an insulated cylinder expand from 10.0 L to 25.0 L against a constant external pressure of 1.0e5 Pa. Compute the work done ON the gas, in joules.
Check your answer
Correct answer: -1500.0 (± 20.0)
W_on = -P_ext (V2 - V1) = -(1.0e5)(0.025 - 0.010) = -1500 J. Negative, because the gas expanded and did work on its surroundings. Note what did not appear anywhere: the amount of gas, its temperature, and its own pressure. Only the external pressure and the volume change matter.
An isotherm and an adiabat pass through the same point in the P-V plane. Which falls more steeply as the volume increases, and why?
A. The adiabat, by the factor gamma, because along it the gas thins out and cools at the same time. B. The isotherm, because a reservoir keeps feeding heat in and that supports the pressure. C. They have the same slope at the shared point; the curves separate only further along. D. The adiabat, because gamma is greater than 1 and the adiabatic exponent is applied to the temperature.
Check your answer
Correct answer: A — The adiabat, by the factor gamma, because along it the gas thins out and cools at the same time.
- A. Correct. Differentiating gives -P/V for the isotherm and -gamma P/V for the adiabat. Along the isotherm the pressure falls only because the density does; along the adiabat the temperature falls as well, and both effects push the pressure down.
- B. Heat flowing in does support the pressure — which makes the isotherm the SHALLOWER curve, not the steeper one. It is the adiabat, with no heat coming in, that falls away faster.
- C. The slopes differ at the shared point by exactly the factor gamma, which for a monatomic gas is 5/3. Curves through a common point with different slopes separate immediately.
- D. The conclusion is right and the reason is not. The exponent gamma appears on the volume in P V^gamma = constant; the temperature form carries gamma - 1 instead. Getting the right answer from the wrong exponent is worth catching now.
Before calculating: an insulated gas doubles its volume three ways — slowly against a matched load, in one step against a fixed smaller load, and freely into a vacuum. Rank the three by the work the gas delivers and by the final temperature, and predict how the two rankings are related.
Check your answer
Work delivered: slow > fixed load > free (0.555, 0.375 and 0 in units of P1 V1 for a monatomic gas). Final temperature: free (300 K) > fixed load (225 K) > slow (189 K). The rankings are exactly opposite, and one line explains it: all three have Q = 0, so the change in internal energy equals the work done on the gas. Every joule handed to the surroundings is a joule taken out of the gas’s own energy, and for an ideal gas internal energy is temperature. A common wrong prediction is that the free expansion should deliver the most work because it is the most violent; violence is not work, and with nothing to push against it delivers none at all.
In a few sentences, explain the difference between “adiabatic” and “quasistatic” to a classmate, using one process that is adiabatic but not quasistatic and one that is quasistatic but not adiabatic.

Exam-style problems for this module are collected in the problem set.

Explain it in your own words¶

Answer in a few sentences each.

A classmate says: “The free expansion doesn’t cool the gas because it happens too fast for the temperature to change.” Explain what is wrong with this, and give the correct reason.
Why is $C_P$ larger than $C_V$ ? Answer without writing an equation, in terms of what the heat is spent on.
Both the isothermal and the free expansion take an ideal gas from $V$ to $2V$ with $\Delta T = 0$ . Describe every way in which they nonetheless differ.
A gas is compressed to half its volume, once slowly and once by slamming the piston. Both are insulated. Which ends hotter, and why does that not violate energy conservation?

Advanced: one family, four members¶

Polytropes. The four families are not four separate objects. They are four values of one parameter in

P V^{n} = \text{constant},

(13)

with $n = 0$ isobaric, $n = 1$ isothermal, $n = \gamma$ adiabatic, and $n \to \infty$ isochoric (the curve becomes vertical). The work along any of them is

W_{\mathrm{on}} = \frac{P_2 V_2 - P_1 V_1}{n - 1},

(14)

which reproduces every row of the table above except the isotherm — and the isotherm is not an oversight but a genuine logarithm, since $\int dV/V$ is not a power. Evaluating the rational form at $n = 1 + 10^{-9}$ to “get around” the singularity is how a plausible wrong number gets produced; the course’s implementation branches on $n = 1$ instead.

A real process with heat leaking through an imperfect insulator sits at some $n$ between 1 and $\gamma$ , which is exactly how engineers fit measured compressor data.

Why the free expansion has no path. During it the gas is not in equilibrium, so it has no pressure and no temperature — not “a rapidly changing one”, but none at all, because those quantities are defined for equilibrium states. The $P$ – $V$ plane is a plane of states, and a process that leaves the set of states cannot be a curve in it. Drawing a dashed line between the endpoints is a common textbook habit and it is a lie; the honest picture has two dots and nothing between them.

Where the missing work went. The three routes started identically and ended at the same volume, yet delivered 0.555, 0.375 and 0 of $P_1V_1$ . Energy is conserved in all three — the gas that did less work simply kept more internal energy. So no conservation law forbids the free expansion; something else does. What distinguishes them is that only the slow route can be run backwards without leaving a trace on the universe, and the quantity that measures the trace is the entropy. All three have the same $\Delta U$ ; they have different $\Delta S$ , with the quasistatic adiabat alone at $\Delta S = 0$ . That is where module 8’s counting argument and module 7’s engines both begin.

The Joule coefficient. The statement “free expansion does not change $T$ ” is the ideal-gas value of

\left(\frac{\partial T}{\partial V}\right)_U ,

(15)

which vanishes precisely when $U$ is independent of volume. For a van der Waals gas it is negative and small, so a real free expansion cools a little. Joule’s original 1845 experiment could not resolve it — his water bath was too large — and the effect was found only when he and Thomson switched to throttling, which amplifies it.