States, Processes, and the Laws of Thermodynamics

Thermodynamics begins with reproducible macroscopic states, not with atoms or probability. Its laws identify temperature and entropy operationally, keep track of heat and work, and rule out otherwise imaginable cyclic machines. The framework applies when a system has meaningful equilibrium states and controlled exchanges; it does not by itself predict how quickly equilibrium is reached.

Required background. Calculus and Linear Algebra supplies total differentials and multivariable derivatives. Helpful background. Classical and Variational Mechanics supplies mechanical work and energy conservation; Quantum Mechanics becomes useful when the low-temperature origin of entropy is discussed.

Equilibrium states and paths between them

An equilibrium state is a reproducible macroscopic condition with no unresolved change on the observation timescale. For a simple one-component compressible system, a complete independent set might be $(U,V,N)$ or $(T,V,N)$. An equation of state, such as $pV=NT$ for a dilute classical ideal gas, relates variables that would otherwise be redundant. Extensive quantities such as $U$, $V$, $N$, and $S$ scale when the system is replicated; intensive quantities such as $T$, $p$, and $\mu$ do not.

A state function assigns one value to each equilibrium state. Its change depends only on the endpoints. Internal energy $U$ and entropy $S$ are state functions. Heat and work instead describe transfers during a process. Their values depend on the path, which is why their infinitesimal symbols are written $\delta Q$ and $\delta W$ rather than $dQ$ and $dW$.

A quasistatic process is slow enough that a sequence of equilibrium states models its intermediate configurations. A reversible process is stronger: an infinitesimal reversal of the controls reverses the process without leaving a change in the system or its surroundings. Friction, heat flow across a finite temperature difference, diffusion, and hysteresis produce entropy even when they occur slowly. Quasistatic therefore does not mean reversible. The equilibrium and process distinctions are developed in Tong, current notes, §§4.1–4.3, printed pp. 109–127.

We set $k_{\mathrm B}=1$, count heat absorbed by the system as positive, and count pressure–volume work done by the system as positive. With these signs,

$$dU=\delta Q-\delta W, \qquad \delta W_{\mathrm{rev}}=p\,dV.$$

The second relation uses the system pressure only for a quasistatic mechanical step. In a rapid expansion the actual boundary work is controlled by the external stress; it need not equal $p\,dV$.

Zeroth and first laws: temperature and energy

The zeroth law states that thermal equilibrium is transitive. If systems $A$ and $B$ are each in thermal equilibrium with system $C$, then $A$ and $B$ are in equilibrium with one another. This makes it possible to label equivalence classes of equilibrium states by one empirical quantity, temperature. A thermometer works because its own equilibrium reading can be calibrated against that ordering.

The first law is energy conservation with the transfer channels exposed:

$$\Delta U=Q-W.$$

It applies to irreversible processes as well as reversible ones, provided the system boundary and every transfer—mechanical, electrical, chemical, radiative, and so on—are included consistently. It does not say that heat or work is separately conserved, and it does not determine whether an energy-conserving process can occur spontaneously.

Second law: entropy balance and direction

For a closed system receiving amounts of heat through boundary parts held at temperatures $T_b\gt 0$, the entropy balance can be written

$$\Delta S =\int\frac{\delta Q}{T_b}+S_{\mathrm{gen}}, \qquad S_{\mathrm{gen}}\geq0.$$

The first term is entropy transferred with heat; $S_{\mathrm{gen}}$ is entropy produced inside the system by irreversible processes. Equality in the Clausius inequality corresponds to $S_{\mathrm{gen}}=0$, not merely to a slowly executed path. For an isolated system, $Q=0$, so $\Delta S=S_{\mathrm{gen}}\geq0$. For a cycle, $\Delta S=0$ and therefore $\oint \delta Q/T_b\leq0$.

Along a reversible path of a closed, fixed-$N$ simple system, $\delta Q_{\mathrm{rev}}=T\,dS$. Combining this identity with reversible pressure work gives $dU=T\,dS-p\,dV$. The more general equilibrium-state differential, allowing neighboring states with different particle number, is

$$dU=T\,dS-p\,dV+\mu\,dN.$$

It identifies the conjugate variables in the energy representation:

$$T=\left(\frac{\partial U}{\partial S}\right)_{V,N}, \qquad p=-\left(\frac{\partial U}{\partial V}\right)_{S,N}, \qquad \mu=\left(\frac{\partial U}{\partial N}\right)_{S,V}.$$

This is a differential relation between neighboring equilibrium states, not an open-system entropy balance and not an assertion that every real process is reversible. Matter crossing an actual boundary can carry both energy and entropy and must be included separately. The Kelvin–Planck statement forbids a cyclic device whose only effect is to extract heat from one reservoir and convert it completely into work. The Clausius statement forbids a cyclic device whose only effect is to transfer heat from a colder body to a hotter one. Under the standard reservoir and cyclic-device assumptions, either forbidden machine could be combined with an allowed engine or refrigerator to construct the other, so the statements are equivalent forms of the second law Tong, current notes, §4.3, printed pp. 113–127.

Third law: what happens near absolute zero

The third law is a low-temperature equilibrium statement. In its Nernst form, entropy differences between equilibrium states connected by changing ordinary external parameters approach zero as $T\to0$, subject to regularity assumptions. A commonly used stronger normalization says that a system with a unique ground state has $S\to0$. In quantum statistical mechanics, a finite ground-state degeneracy $g_0$ instead gives the canonical limit $S\to\ln g_0$ unless the degeneracy is lifted.

These formulations should not be collapsed into “every cold object has zero entropy.” A glass can fall out of equilibrium; a macroscopic degeneracy can make the order of limits consequential; and reaching $T=0$ is not licensed by taking an idealized formula literally. Tong’s discussion connects the third law to the quantum suppression of excitations in Tong, current notes, §4.5, printed pp. 133–134.

Exercise: free expansion

A dilute classical ideal gas with fixed particle number initially occupies volume $V_i$ and expands into vacuum to $V_f$ inside an isolated container. Find the changes in internal energy, temperature, and entropy between the initial and final equilibrium states.

Solution

The external pressure vanishes, so the gas does no boundary work: $W=0$. Isolation gives $Q=0$. The first law therefore yields $\Delta U=0$. At fixed $N$, the internal energy of a dilute classical ideal gas depends only on temperature, $U=U(T,N)$, so the final equilibrium temperature equals the initial one.

The actual expansion is irreversible, so one cannot compute its entropy change by integrating $\delta Q/T$ along the real nonequilibrium history. Entropy is nevertheless a state function. Connect the same endpoints by a fictitious reversible isothermal expansion, for which $pV=NT$:

$$\Delta S =\int_{V_i}^{V_f}\frac{p\,dV}{T} =N\int_{V_i}^{V_f}\frac{dV}{V} =N\ln\frac{V_f}{V_i}\gt 0.$$

For the actual isolated process this increase is entirely produced entropy: $S_{\mathrm{gen}}=\Delta S$. Energy conservation does not imply reversibility, and the reversible path is only a device for computing the endpoint difference.

Exercise: two bodies reach a common temperature

Two identical bodies have constant heat capacity $C\gt 0$ and initial temperatures $T_h\gt T_c\gt 0$. They are placed in thermal contact inside an isolated enclosure. Neglect all other energy storage. Find the final temperature and show that total entropy increases.

Solution

Energy conservation gives

$$C(T_f-T_h)+C(T_f-T_c)=0,$$

so $T_f=(T_h+T_c)/2$. The total entropy change is

$$\Delta S_{\mathrm{tot}} =C\ln\frac{T_f}{T_h}+C\ln\frac{T_f}{T_c} =C\ln\frac{(T_h+T_c)^2}{4T_hT_c}.$$

Because $(T_h-T_c)^2\gt 0$, one has $(T_h+T_c)^2\gt 4T_hT_c$, and hence $\Delta S_{\mathrm{tot}}\gt 0$. The first law fixes the final temperature; the second law selects the spontaneous direction of heat flow.

Scope and limitations

Thermodynamics does not identify microstates, probabilities, or relaxation mechanisms. At phase transitions, differentiability and local-response assumptions can fail. Finite, nonadditive, or long-range self-gravitating systems can additionally violate familiar extensivity assumptions. Thermodynamic entropy is also not automatically Gibbs, Shannon, von Neumann, entanglement, or black-hole entropy. Thermodynamic Potentials and Stability now asks which state function supplies the equilibrium criterion when the surroundings hold particular variables fixed.

References

• Tong, David. Statistical Physics. University of Cambridge Part II Mathematical Tripos lecture notes, current official undated-cover PDF, §§4.1–4.3, printed pp. 109–127, and §4.5, printed pp. 133–134. Open PDF.