Thermodynamics and Statistical Mechanics

Zeroth Law: If system $A$ is in thermal equilibrium with system $B$And $B$ is in thermal Equilibrium with system $C$Then $A$ is in thermal equilibrium with $C$.

This law establishes the existence of temperature as an equivalence relation. Two systems are in Thermal equilibrium if and only if they are at the same temperature.

First Law (Conservation of Energy): The change in internal energy of a system equals the heat Added to the system minus the work done by the system:

$dU = \delta Q - \delta W$

Where $\delta Q$ and $\delta W$ are inexact differentials (path-dependent), while $dU$ is an Exact differential (state function).

For a quasi-static process with pressure-volume work:

$\delta W = P\,dV$

So the first law becomes:

$dU = \delta Q - P\,dV$

Definition (Heat capacity). The heat capacity at constant volume and constant pressure are:

$C_V = \left(\frac{\partial U}{\partial T}\right)_V, \quad C_P = \left(\frac{\partial H}{\partial T}\right)_P$

Where $H = U + PV$ is the enthalpy.

Second Law (Clausius Statement): Heat cannot spontaneously flow from a colder body to a hotter Body without external work.

Second Law (Kelvin-Planck Statement): No cyclic process can convert heat entirely into work.

Theorem 1.1 (Carnot”s Theorem). No engine operating between two heat reservoirs is more Efficient than a Carnot engine. All reversible engines operating between the same two reservoirs Have the same efficiency.

Proof. Suppose engine $A$ (claimed more efficient than Carnot) operates between reservoirs at $T_h$ and $T_c$. Let $A$ extract heat $Q_h$ from the hot reservoir, do work $W$And reject heat $Q_c = Q_h - W$ to the cold reservoir. Run a Carnot engine $C$ in reverse as a refrigerator using Work $W$ from $A$: it extracts $Q_c'$ from the cold reservoir and delivers $Q_h' = W + Q_c'$ to the Hot reservoir. If $\eta_A \gt \eta_C$Then $Q_c \lt Q_c'$So the combined system transfers net Heat from cold to hot with no external work, violating the Clausius statement. $\blacksquare$

Definition (Entropy). For a reversible process, the entropy change is:

dS = \frac{\delta Q_{\mathrm{rev}}{T}}

Clausius Inequality: For any cyclic process:

$\oint \frac{\delta Q}{T} \leq 0$

With equality if and only if the process is reversible.

Proof of entropy increase for irreversible processes. Consider a system undergoing an Irreversible process from state $1$ to state $2$Then returning via a reversible process. By the Clausius inequality:

\int_1^2 \frac{\delta Q_{\mathrm{irrev}}{T} + \int_2^1 \frac{\delta Q_{\mathrm{rev}}{T} \leq 0}}

\int_1^2 \frac{\delta Q_{\mathrm{irrev}}{T} - \int_1^2 \frac{\delta Q_{\mathrm{rev}}{T} \leq 0}}

Since $dS = \delta Q_{\mathrm{rev}/T}$:

$\Delta S \geq \int_1^2 \frac{\delta Q}{T}$

With equality for reversible processes. $\blacksquare$

Third Law (Nernst Heat Theorem): As $T \to 0$The entropy of a perfect crystal approaches zero:

$\lim_{T \to 0} S(T) = 0$

This sets an absolute reference for entropy and implies that it is impossible to reach absolute zero In a finite number of steps.

Several material-dependent quantities characterise how a system responds to changes in its state Variables.

Definition (Isothermal compressibility).

$\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$

Definition (Adiabatic compressibility).

$\kappa_S = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_S$

Definition (Coefficient of thermal expansion).

$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$

Theorem 1.2 (Relation between heat capacities).

$C_P - C_V = \frac{TV\alpha^2}{\kappa_T}$

Proof. From the identity $dS(T, V) = (\partial S/\partial T)_V\,dT + (\partial S/\partial V)_T\,dV$ And writing $dV$ in terms of $dT$ and $dP$ along a constant-$P$ path:

$\left(\frac{\partial S}{\partial T}\right)_P = \left(\frac{\partial S}{\partial T}\right)_V + \left(\frac{\partial S}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P$

Multiply by $T$ and use $C_V = T(\partial S/\partial T)_V$, $C_P = T(\partial S/\partial T)_P$And the Maxwell relation $(\partial S/\partial V)_T = (\partial P/\partial T)_V$:

$C_P - C_V = T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P$

Now use the cyclic relation $(\partial P/\partial T)_V = -(\partial V/\partial T)_P / (\partial V/\partial P)_T$ To obtain:

$C_P - C_V = -T\frac{\left(\frac{\partial V}{\partial T}\right)_P^2}{\left(\frac{\partial V}{\partial P}\right)_T} = \frac{TV\alpha^2}{\kappa_T}$

$\blacksquare$

Since $\kappa_T \gt 0$ for stable systems and $\alpha^2 \geq 0$We always have $C_P \geq C_V$.

Theorem 1.3 (Adiabatic index). The ratio $\gamma = C_P/C_V$ satisfies:

$\gamma = \frac{\kappa_T}{\kappa_S}$

Proof. Along an adiabat, $dS = 0$. Using the chain rule:

$dS = \left(\frac{\partial S}{\partial T}\right)_P\,dT + \left(\frac{\partial S}{\partial P}\right)_T\,dP = 0$

$\left(\frac{\partial P}{\partial T}\right)_S = -\frac{\left(\frac{\partial S}{\partial T}\right)_P}{\left(\frac{\partial S}{\partial P}\right)_T} = -\frac{C_P/T}{-(\partial V/\partial T)_P} = \frac{C_P}{T(\partial V/\partial T)_P}$

Similarly, along an isotherm:

$\left(\frac{\partial P}{\partial T}\right)_V = -\frac{(\partial V/\partial T)_P}{(\partial V/\partial P)_T} = \frac{\alpha}{\kappa_T}$

Using the identity $C_P/C_V = (\partial P/\partial T)_S / (\partial P/\partial T)_V$ and simplifying gives $\gamma = \kappa_T/\kappa_S$. $\blacksquare$

Theorem 1.4 (Statistical basis of the second law). For an isolated system, the entropy $S = k_B \ln \Omega$ can only increase or remain constant.

Proof. Consider an isolated system with fixed energy $E$Volume $V$And particle number $N$. The system evolves through accessible microstates. If the system starts in a non-equilibrium Macrostate $A$ with $\Omega_A$ microstates and evolves to macrostate $B$ with $\Omega_B$ microstates, The evolution is driven by the ergodic exploration of phase space.

Since the system explores all accessible microstates with equal probability over time, it will Overwhelmingly be found in the macrostate with the largest number of microstates. If $\Omega_B \geq \Omega_A$, then $S_B = k_B \ln \Omega_B \geq k_B \ln \Omega_A = S_A$.

More rigorously: suppose the system starts in a subset of $w_0$ microstates out of a total $\Omega$. The probability that the system remains in this subset after randomising over all Microstates is $w_0/\Omega \ll 1$. The system therefore evolves toward the macrostate that Maximises $\Omega$And hence maximises $S$. $\blacksquare$

This shows that the second law is not absolute but statistical: fluctuations can temporarily Decrease entropy, but the probability of a macroscopic fluctuation is exponentially small in $N$.

The internal energy $U$ is the fundamental thermodynamic potential. By performing Legendre Transformations on $U(S, V, N)$We obtain the other potentials:

The Legendre transforms are:

$H = U + PV, \quad F = U - TS, \quad G = H - TS = U - TS + PV$

Starting from the first law for a reversible process:

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

This tells us $T = (\partial U/\partial S)_{V,N}$, $P = -(\partial U/\partial V)_{S,N}$And $\mu = (\partial U/\partial N)_{S,V}$.

For enthalpy, $H = U + PV$So:

$dH = dU + P\,dV + V\,dP = T\,dS + V\,dP + \mu\,dN$

For Helmholtz free energy, $F = U - TS$So:

$dF = dU - T\,dS - S\,dT = -S\,dT - P\,dV + \mu\,dN$

For Gibbs free energy, $G = U - TS + PV$So:

$dG = -S\,dT + V\,dP + \mu\,dN$

• $U$: Total energy at constant entropy and volume.
• $H$: Total energy plus the work needed to make room for the system ($PV$). Useful for constant-pressure processes (e.g., chemical reactions at atmospheric pressure).
• $F$: The maximum work extractable from a system at constant temperature. Minimised at equilibrium for systems in contact with a heat bath at fixed $T, V$.
• $G$: The maximum non-expansion work extractable. Minimised at equilibrium for systems at fixed $T, P$.

Since each thermodynamic potential is a state function, its differential is exact. By Euler’s Reciprocity, mixed second partial derivatives are equal. This yields the four Maxwell relations Directly:

These relations allow us to express unmeasurable quantities (entropy changes) in terms of Measurable ones (equations of state).

Theorem 2.1. At equilibrium:

• For an isolated system: $S$ is maximised (at fixed $U, V, N$).
• For a system in contact with a heat bath at temperature $T$ and pressure $P$: $G$ is minimised.
• For a system at constant $T, V$: $F$ is minimised.

Proof (for $G$). Consider a system in contact with a reservoir at $T_0, P_0$. The total entropy Of system plus reservoir is $S_{\mathrm{tot} = S + S_R}$. At equilibrium, $S_{\mathrm{tot}}$ is Maximised, so $\delta S_{\mathrm{tot} \leq 0}$ for any variation. Since $dS_R = \delta Q_R / T_0$ And by energy conservation $\delta Q_R = -\delta Q = -(dU + P_0\,dV)$:

$\delta S_{\mathrm{tot} = \delta S - \frac{1}{T_0}(dU + P_0\,dV) = -\frac{1}{T_0}\delta G \leq 0}$

Where $\delta G = \delta U + P_0\,\delta V - T_0\,\delta S$. Hence $\delta G \geq 0$So $G$ is Minimised. $\blacksquare$

Since $U, H, F, G$ are state functions, their differentials are exact. By the symmetry of second Derivatives (Euler’s reciprocity), if $dz = M\,dx + N\,dy$Then:

$\left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y$

Applying this to each thermodynamic potential:

From $dU = T\,dS - P\,dV + \mu\,dN$:

$\left(\frac{\partial T}{\partial V}\right)_{S,N} = -\left(\frac{\partial P}{\partial S}\right)_{V,N}$

$\left(\frac{\partial T}{\partial N}\right)_{S,V} = \left(\frac{\partial \mu}{\partial S}\right)_{V,N}$

$\left(\frac{\partial P}{\partial N}\right)_{S,V} = -\left(\frac{\partial \mu}{\partial V}\right)_{S,N}$

From $dH = T\,dS + V\,dP + \mu\,dN$:

$\left(\frac{\partial T}{\partial P}\right)_{S,N} = \left(\frac{\partial V}{\partial S}\right)_{P,N}$

From $dF = -S\,dT - P\,dV + \mu\,dN$:

$\left(\frac{\partial S}{\partial V}\right)_{T,N} = \left(\frac{\partial P}{\partial T}\right)_{V,N}$

From $dG = -S\,dT + V\,dP + \mu\,dN$:

$\left(\frac{\partial S}{\partial P}\right)_{T,N} = -\left(\frac{\partial V}{\partial T}\right)_{P,N}$

Derivation of the heat capacity relation. From $dU = T\,dS - P\,dV$:

$C_V = T\left(\frac{\partial S}{\partial T}\right)_V, \quad C_P = T\left(\frac{\partial S}{\partial T}\right)_P$

Using the chain rule and Maxwell relations:

$C_P - C_V = T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P$

Proof. Expand $S(T, V)$ as $S(T, P(T, V))$:

$\left(\frac{\partial S}{\partial T}\right)_V = \left(\frac{\partial S}{\partial T}\right)_P + \left(\frac{\partial S}{\partial P}\right)_T \left(\frac{\partial P}{\partial T}\right)_V$

Multiply by $T$:

$C_V = C_P + T\left(\frac{\partial S}{\partial P}\right)_T \left(\frac{\partial P}{\partial T}\right)_V$

Using the Maxwell relation $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$:

$C_V = C_P - T\left(\frac{\partial V}{\partial T}\right)_P \left(\frac{\partial P}{\partial T}\right)_V$

$\blacksquare$

Application: entropy change of an ideal gas. Using $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ And the ideal gas law $P = Nk_B T/V$:

$\left(\frac{\partial S}{\partial V}\right)_T = \frac{Nk_B}{V}$

Integrating: $\Delta S = Nk_B \ln(V_f/V_i)$ for an isothermal expansion.

Similarly, using $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$:

$\left(\frac{\partial S}{\partial P}\right)_T = -\frac{Nk_B}{P}$

So $\Delta S = -Nk_B \ln(P_f/P_i) = Nk_B \ln(V_f/V_i)$Consistent.

Application: internal energy of an ideal gas. Using $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ (a Maxwell relation consequence from $dU = T\,dS - P\,dV$):

$\left(\frac{\partial U}{\partial V}\right)_T = T \cdot \frac{Nk_B}{V} - \frac{Nk_B T}{V} = 0$

This confirms that the internal energy of an ideal gas depends only on $T$ (Joule’s law).

The Carnot cycle consists of four reversible stages operating between temperatures $T_h$ (hot) and $T_c$ (cold):

1. Isothermal expansion at $T_h$: absorb heat $Q_h$ from hot reservoir.
2. Adiabatic expansion: temperature drops from $T_h$ to $T_c$.
3. Isothermal compression at $T_c$: reject heat $Q_c$ to cold reservoir.
4. Adiabatic compression: temperature rises from $T_c$ to $T_h$.

The efficiency is:

$\eta = 1 - \frac{Q_c}{Q_h} = 1 - \frac{T_c}{T_h}$

Derivation. For the isothermal steps, $\Delta S_{\mathrm{hot} = Q_h/T_h}$ and $\Delta S_{\mathrm{cold} = -Q_c/T_c}$. Since entropy is a state function and the cycle returns to The initial state, $\Delta S_{\mathrm{total} = 0}$So $Q_h/T_h = Q_c/T_c$. $\blacksquare$

A refrigerator is a Carnot engine run in reverse. The coefficient of performance (COP):

\mathrm{COP_}{\mathrm{ref} = \frac{Q_c}{W} = \frac{T_c}{T_h - T_c}}

A heat pump heats the hot reservoir:

\mathrm{COP_}{\mathrm{hp} = \frac{Q_h}{W} = \frac{T_h}{T_h - T_c}}

Otto cycle (idealised petrol engine): two isochoric and two adiabatic processes.

$\eta_{\mathrm{Otto} = 1 - \frac{1}{r^{\gamma - 1}}}$

Where $r = V_{\mathrm{max}/V_{\mathrm{min}}}$ is the compression ratio and $\gamma = C_P/C_V$.

Diesel cycle: one isobaric, two adiabatic, and one isochoric process:

$\eta_{\mathrm{Diesel} = 1 - \frac{1}{r^{\gamma - 1}} \cdot \frac{\alpha^\gamma - 1}{\gamma(\alpha - 1)}}$

Where $\alpha = V_{\mathrm{max}/V_{\mathrm{cutoff}}}$ is the cutoff ratio.

Definition (Boltzmann Entropy). For a macrostate with $\Omega$ accessible microstates:

$S = k_B \ln \Omega$

Where $k_B = 1.381 \times 10^{-23}$ J/K is Boltzmann’s constant.

Justification. Consider two independent systems $A$ and $B$. The total number of microstates is $\Omega_{AB} = \Omega_A \cdot \Omega_B$. We require $S_{AB} = S_A + S_B$ (additivity). The logarithm Is the unique function satisfying $f(xy) = f(x) + f(y)$. $\blacksquare$

For a system with probability $p_i$ of being in microstate $i$:

$S = -k_B \sum_i p_i \ln p_i$

This reduces to the Boltzmann formula when all accessible microstates are equally probable: $p_i = 1/\Omega$.

Theorem 5.1 (Concavity of entropy). The Gibbs entropy is maximised when all accessible Microstates are equally probable.

Proof. Maximise $S = -k_B \sum_i p_i \ln p_i$ subject to $\sum_i p_i = 1$ using a Lagrange Multiplier $\lambda$:

$\frac{\partial}{\partial p_j}\left[-\sum_i p_i \ln p_i - \lambda\left(\sum_i p_i - 1\right)\right] = -\ln p_j - 1 - \lambda = 0$

This gives $p_j = e^{-1-\lambda} = \mathrm{const}$ for all $j$. The constraint $\sum_i p_i = 1$ Then gives $p_i = 1/\Omega$. $\blacksquare$

Derivation from Boltzmann. For $N$ identical systems distributed among $\Omega$ equally probable Microstates, the most probable macrostate has $n_i = N/\Omega$ systems in each microstate. The Number of ways to arrange this is:

$W = \frac{N!}{\prod_i n_i!}$

Using Stirling’s approximation $\ln N! \approx N \ln N - N$:

$\ln W = N \ln N - N - \sum_i (n_i \ln n_i - n_i) = -N \sum_i p_i \ln p_i$

Where $p_i = n_i/N$. Multiplying by $k_B$ gives the Gibbs entropy. $\blacksquare$

The Helmholtz free energy connects thermodynamics to statistical mechanics:

$F = -k_B T \ln Z$

Where $Z = \sum_i e^{-\beta E_i}$ is the canonical partition function and $\beta = 1/(k_B T)$.

Derivation. From the Gibbs entropy with the Boltzmann distribution $p_i = e^{-\beta E_i}/Z$:

$S = -k_B \sum_i \frac{e^{-\beta E_i}}{Z} \left(-\beta E_i - \ln Z\right) = k_B \beta \langle E \rangle + k_B \ln Z$

Since $\langle E \rangle = U$ and $k_B \beta = 1/T$:

$S = \frac{U}{T} + k_B \ln Z$

$F = U - TS = U - T\left(\frac{U}{T} + k_B \ln Z\right) = -k_B T \ln Z$

$\blacksquare$

First-order transition: Discontinuity in first derivatives of $G$ (e.g., $S$, $V$). There is a Latent heat $L = T \Delta S$.

Second-order (continuous) transition: First derivatives are continuous, but second derivatives (e.g., $C_P$, $\kappa_T$, $\alpha$) diverge or are discontinuous.

Ehrenfest classification: An $n$-th order transition has discontinuities in the $n$-th Derivatives of $G$With all lower derivatives continuous.

For a first-order phase transition between phases $\alpha$ and $\beta$ in equilibrium ($G_\alpha = G_\beta$):

$\frac{dP}{dT} = \frac{S_\beta - S_\alpha}{V_\beta - V_\alpha} = \frac{L}{T \Delta V}$

Where $L$ is the latent heat and $\Delta V = V_\beta - V_\alpha$.

Derivation. Along the coexistence curve, $dG_\alpha = dG_\beta$. Since $dG = -S\,dT + V\,dP$:

$-S_\alpha\,dT + V_\alpha\,dP = -S_\beta\,dT + V_\beta\,dP$

$\frac{dP}{dT} = \frac{S_\beta - S_\alpha}{V_\beta - V_\alpha} = \frac{L}{T \Delta V}$

$\blacksquare$

Application: liquid-gas coexistence. Assuming the vapour is an ideal gas and $V_{\mathrm{gas} \gg V_{\mathrm{liquid}}}$:

$\frac{dP}{dT} \approx \frac{L}{T \cdot nRT/P} = \frac{PL}{nRT^2}$

Integrating (assuming $L$ is constant) gives the Clausius equation:

$\ln P = -\frac{L}{nRT} + \mathrm{const}$

Consider a system $S$ in thermal contact with a large heat reservoir $R$ at temperature $T$. The Total energy $E_{\mathrm{tot} = E_S + E_R}$ is conserved.

The probability that $S$ is in state $i$ with energy $E_i$ is proportional to the number of Microstates of the reservoir:

$P_i \propto \Omega_R(E_{\mathrm{tot} - E_i)}$

Since the reservoir is large, expand $\ln \Omega_R$ to first order:

$\ln \Omega_R(E_{\mathrm{tot} - E_i) \approx \ln \Omega_R(E_{\mathrm{tot}) - E_i \left(\frac{\partial \ln \Omega_R}{\partial E}\right)_V}}$

$= \ln \Omega_R(E_{\mathrm{tot}) - \frac{E_i}{k_B T}}$

Where we used $\partial \ln \Omega_R / \partial E = 1/(k_B T)$ (the thermodynamic definition of Temperature). Therefore:

$P_i \propto e^{-E_i/(k_B T)} = e^{-\beta E_i}$

Normalising:

$P_i = \frac{e^{-\beta E_i}}{Z}, \quad Z = \sum_i e^{-\beta E_i}$

This is the Boltzmann distribution (canonical ensemble).

From the partition function, all thermodynamic quantities follow:

• Internal energy: $U = -\frac{\partial \ln Z}{\partial \beta}$
• Entropy: $S = k_B(\ln Z + \beta U)$
• Helmholtz free energy: $F = -k_B T \ln Z$
• Pressure: $P = \frac{1}{\beta}\frac{\partial \ln Z}{\partial V}$
• Heat capacity: $C_V = k_B \beta^2 \left(\langle E^2 \rangle - \langle E \rangle^2\right)$

A system has two energy levels: $E_0 = 0$ and $E_1 = \varepsilon$.

$Z = 1 + e^{-\beta\varepsilon}$

$U = -\frac{\partial \ln Z}{\partial \beta} = \frac{\varepsilon e^{-\beta\varepsilon}}{1 + e^{-\beta\varepsilon}} = \frac{\varepsilon}{e^{\beta\varepsilon} + 1}$

$C = \frac{\partial U}{\partial T} = k_B \beta^2 \varepsilon^2 \frac{e^{\beta\varepsilon}}{(1 + e^{\beta\varepsilon})^2}$

At high $T$ ($\beta \to 0$): $U \to \varepsilon/2$ and $C \to 0$ (equipartition). At low $T$ ($\beta \to \infty$): $U \to 0$ and $C \to 0$ (Schottky anomaly).