Thermodynamics

Put a cup of hot tea on a table and leave it. Its temperature falls, the room’s rises by an imperceptible amount, and after a while nothing changes any further. Two systems in contact reach a common state called thermal equilibrium, in which no net heat flows between them even though they remain in contact. This is the state that lets you assign a single number, temperature, to describe “hotness” at all.

The trouble is that checking whether two systems are in equilibrium by placing them in direct contact is often inconvenient, you want to know if a patient and a healthy person are at the same temperature without pressing them together. The zeroth law rescues this:

If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

Let C be a thermometer. The zeroth law says you never need to touch the patient and the healthy person together, touching each in turn to the same thermometer and reading the same value is enough to guarantee they’d be in equilibrium if they were in contact. This is precisely what makes a thermometer, and a temperature scale, a logically sound idea rather than a convenient fiction. It sounds obvious once stated, which is exactly why it went unnoticed and unnamed until after the first and second laws were already established, hence “zeroth”.

Internal energy $U$ is the sum of all the microscopic kinetic and potential energy inside a system, the random translational, rotational and vibrational motion of its molecules, and the energy stored in the forces between them. The one property that matters most about $U$ is that it is a state variable: its value is fixed entirely by the current state of the system (for an ideal gas, by its temperature alone), with no memory of how that state was reached.

Heat $Q$ and work $W$ are different in kind. Neither is something a system has, both are modes of energy in transit, across the boundary of the system, and both can depend on the exact path taken between two states even when the start and end states are identical. Saying “this gas contains so much heat” is not loose language, it is simply false; the gas contains internal energy, and heat is one of the two ways that energy gets transferred in or out.

The first law of thermodynamics is nothing more than energy conservation, written so as to keep heat and work separately visible:

with the sign convention used throughout this chapter: $Q$ is positive when heat is absorbed by the system, and $W$ is positive when work is done by the system on its surroundings (such as a gas pushing a piston outward). Get either sign backwards and every later result in this chapter, isothermal work, adiabatic cooling, engine efficiency, comes out wrong, so it is worth fixing firmly: heat in is positive, work out is positive, and both appear on the side of the equation that increases $\Delta U$.

For a gas doing work only by changing its volume against a pressure $P$, the work done by the gas as it expands from $V_1$ to $V_2$ is the area under its P-V curve:

which depends on the exact shape of that curve, the path, not just its endpoints. $\Delta U$ depends only on the endpoints.$Q$ then has to absorb whatever path-dependence is left over, which is exactly why heat, like work, cannot be a state variable.

A thermodynamic process is any change from one equilibrium state of a gas to another. Two idealized processes matter more than any others in this chapter, because they sit at opposite extremes of how a gas exchanges heat with its surroundings while its volume changes.

How much heat raises a gas’s temperature by a fixed amount depends on how it is heated, specifically, whether its volume or its pressure is held fixed while heating. This gives two distinct molar specific heats: $C_v$, the heat needed per mole per kelvin at constant volume, and $C_p$, the same thing at constant pressure.

At constant volume, the gas cannot expand, so it can do no work at all ($W=0$), and the first law says every joule of heat supplied goes directly into $\Delta U$:

At constant pressure, raising the temperature by the same $\Delta T$ still requires that same $\Delta U$ (internal energy of an ideal gas depends on $T$ alone, regardless of path), but now the gas also expands and does positive work $P\Delta V$ on its surroundings, and that work has to be paid for with additional heat on top of what raised $\Delta U$:

so $C_p$ must exceed $C_v$, for exactly the same $\Delta T$, heating at constant pressure costs strictly more heat because some of it leaks out as expansion work rather than staying behind as internal energy. Substituting $P\Delta V = nR\Delta T$ (from the ideal gas law at constant $P$) gives the exact gap between them:

independent of the gas, a genuinely universal relation for any ideal gas, monatomic or diatomic.

A heat engine is any device that absorbs heat $Q_1$ from a hot reservoir, converts part of it into work $W$, rejects the rest, $Q_2$, to a cold reservoir, and returns to its starting state so it can repeat the cycle indefinitely. Returning to the starting state matters, it means $\Delta U = 0$ over one full cycle, so the first law forces:

Its efficiency is the fraction of the absorbed heat that actually becomes useful work:

It is tempting to imagine a sufficiently clever engine with $Q_2=0$, converting every joule of absorbed heat straight into work, $\eta = 1$. The second law of thermodynamics forbids this absolutely, not as an engineering shortfall but as a structural limit on what any cyclic process can do:

No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work. (Kelvin-Planck statement)

Every real heat engine must reject some heat to a cold sink; a refrigerator or heat pump is the mirror image, using external work to force heat to flow from cold to hot, which is the direction it never flows on its own, the companion Clausius statement of the same law. Both are experimentally equivalent statements of the same underlying fact: heat engines have a hard ceiling on efficiency, well short of 100%.

If every real engine falls short of $\eta=1$, the natural next question is: what is the best possible efficiency between two given temperatures? Sadi Carnot answered this with an idealized engine built from four reversible steps, working between a hot reservoir at $T_1$ and a cold one at $T_2$:

1. Isothermal expansion at $T_1$, absorbing heat $Q_1$ from the hot reservoir.
2. Adiabatic expansion, cooling the gas from $T_1$ down to $T_2$ with no heat exchange.
3. Isothermal compression at $T_2$, rejecting heat $Q_2$ to the cold reservoir.
4. Adiabatic compression, returning the gas from $T_2$ back up to $T_1$, closing the cycle.

Every one of these four steps is a process already derived in §11.3, the Carnot cycle adds nothing new physically, it simply chains two isothermal and two adiabatic steps together in the most efficient possible sequence. Working through the heat exchanged in the two isothermal legs gives $Q_2/Q_1 = T_2/T_1$, so the efficiency depends on nothing but the two reservoir temperatures:

This is the theoretical ceiling: no engine, however cleverly designed, can exceed $\eta_{Carnot}$ while operating between the same two temperatures, and a Carnot engine is the only kind that actually reaches it, because every one of its four steps is reversible. Real engines fall short of even this ceiling because real processes involve friction and irreversible heat loss that the idealized Carnot cycle has no room for. Notice, too, that $\eta_{Carnot}$ reaches 1 only if $T_2 = 0\text{ K}$, absolute zero, which is itself unreachable, a different, related limit this chapter does not derive but is worth knowing by name (the third law of thermodynamics).

A couple of the chapter’s own classic exercise questions, with original worked solutions.