Thermodynamics

1. The Laws of Thermodynamics

1.1 The Zeroth and First Laws

1. Zeroth Law: If $A$ is in thermal equilibrium with $B$, and $B$ with $C$, then $A$ is in thermal equilibrium with $C$. This establishes temperature as a transitive property and justifies the use of thermometers.

2. First Law: Energy is conserved. For a closed system:

   $dU = \delta q + \delta w$

   where $U$ is internal energy, $q$ is heat, and $w$ is work. The notation $\delta$ indicates inexact differentials: $q$ and $w$ are path-dependent, but $U$ is a state function.

1.2 Work of Expansion

For a reversible expansion of an ideal gas against an external pressure:

$\delta w_{\text{rev}} = -P\,dV$

$w_{\text{rev}} = -\int_{V_i}^{V_f} P\,dV = -nRT\ln\frac{V_f}{V_i}$

1.3 Enthalpy

Definition 1 (Enthalpy): The enthalpy $H$ is defined as:

$H = U + PV$

For a process at constant pressure:

$dH = dU + P\,dV + V\,dP = \delta q_p + V\,dP \implies \Delta H = q_p$

The molar heat capacities relate to enthalpy and internal energy:

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

For an ideal gas: $C_p - C_V = nR$.

2. The Second and Third Laws

2.1 The Second Law

Theorem 1 (Clausius Inequality): For any cyclic process:

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

Equality holds only for reversible processes. This implies the existence of a state function $S$ (entropy) such that:

$dS \geq \frac{\delta q}{T}$

For a spontaneous (irreversible) process in an isolated system: $dS > 0$.

2.2 Entropy Changes

For a reversible process at temperature $T$:

$\Delta S = \int_{T_1}^{T_2} \frac{C}{T}\,dT$

Entropy of phase transition: At the transition temperature $T_{\text{trs}}$:

$\Delta_{\text{trs}}S = \frac{\Delta_{\text{trs}}H}{T_{\text{trs}}}$

Example 1: Calculate $\Delta S$ when 2 mol of ice melts at 273 K ($\Delta_{\text{fus}}H = 6.01$ kJ/mol).

$\Delta S = \frac{n\,\Delta_{\text{fus}}H}{T} = \frac{2 \times 6010}{273} = 44.0 \text{ J/K}$

$\blacksquare$

2.3 Statistical Interpretation of Entropy

Theorem 2 (Boltzmann Entropy):

$S = k_B \ln W$

where $W$ is the number of microstates and $k_B = 1.381 \times 10^{-23}$ J/K is Boltzmann”s constant.

For $N$ distinguishable particles with $n_i$ in each energy level $\varepsilon_i$:

$W = \frac{N!}{n_1!\,n_2!\,\cdots}$

The entropy of mixing two ideal gases:

$\Delta_{\text{mix}}S = -nR\left(x_A \ln x_A + x_B \ln x_B\right)$

2.4 The Third Law

Theorem 3 (Third Law of Thermodynamics): The entropy of a perfect crystal at absolute zero is zero:

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

This provides a reference point for absolute entropies (standard molar entropies $S^\circ$).

3. Gibbs Free Energy and Chemical Potential

3.1 Gibbs and Helmholtz Free Energy

Definition 2 (Helmholtz Free Energy):

$A = U - TS$

$dA = -S\,dT - P\,dV$

Definition 3 (Gibbs Free Energy):

$G = H - TS = U + PV - TS$

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

At constant $T$ and $P$: $\Delta G = w_{\text{non-PV}}$, so the Gibbs free energy change equals the maximum non-expansion work.

3.2 Spontaneity Criteria

3.3 Fundamental Equations

The four fundamental equations of thermodynamics (for closed systems of constant composition):

$dU = T\,dS - P\,dV$ $dH = T\,dS + V\,dP$ $dA = -S\,dT - P\,dV$ $dG = -S\,dT + V\,dP$

3.4 Chemical Potential

Definition 4 (Chemical Potential): For an open system with $k$ components:

$dG = -S\,dT + V\,dP + \sum_{i=1}^{k} \mu_i\,dn_i$

where $\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_{j\neq i}}$ is the chemical potential of component $i$.

For an ideal gas: $\mu = \mu^\circ + RT\ln\frac{P}{P^\circ}$.

4. Maxwell Relations

4.1 Derivation from Exact Differentials

Theorem 4 (Maxwell Relations): Since $U$, $H$, $A$, $G$ are state functions, their mixed second partial derivatives are equal:

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

4.2 Applications

Using the Maxwell relation $\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$:

For an ideal gas: $\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V}$, so:

$\Delta S = \int_{V_1}^{V_2} \frac{nR}{V}\,dV = nR\ln\frac{V_2}{V_1}$

5. Gibbs-Helmholtz Equation

5.1 Temperature Dependence of $G$

Theorem 5 (Gibbs-Helmholtz Equation):

$\left[\frac{\partial(G/T)}{\partial T}\right]_P = -\frac{H}{T^2}$

Equivalently:

$\frac{\Delta G_2}{T_2} - \frac{\Delta G_1}{T_1} = -\Delta H\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

(approximate form when $\Delta H$ is constant over the temperature range).

6. The Clausius-Clapeyron Equation

6.1 Derivation

At phase equilibrium between two phases $\alpha$ and $\beta$:

$\mu_\alpha(T, P) = \mu_\beta(T, P)$

Differentiating along the coexistence curve:

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

$\frac{dP}{dT} = \frac{\Delta_{\text{trs}}S}{\Delta_{\text{trs}}V} = \frac{\Delta_{\text{trs}}H}{T\,\Delta_{\text{trs}}V}$

6.2 The Integrated Form

For liquid-vapor equilibrium, assuming $\Delta_{\text{vap}}H$ is constant and $V_g \gg V_l$:

$\ln\frac{P_2}{P_1} = -\frac{\Delta_{\text{vap}}H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Example 2: The normal boiling point of benzene is 353 K with $\Delta_{\text{vap}}H = 30.8$ kJ/mol. Find the vapor pressure at 298 K.

$\ln\frac{P}{1.013 \times 10^5} = -\frac{30800}{8.314}\left(\frac{1}{298} - \frac{1}{353}\right) = -1.93$

$P = 1.013 \times 10^5 \times e^{-1.93} = 1.47 \times 10^4 \text{ Pa} \approx 14.7 \text{ kPa}$

$\blacksquare$

7. Phase Diagrams and Phase Equilibria

7.1 Phase Rule

Theorem 6 (Gibbs Phase Rule): For a system with $C$ components and $P$ phases at equilibrium:

$F = C - P + 2$

where $F$ is the number of degrees of freedom (intensive variables that can be independently varied).

For a single-component system ($C = 1$): $F = 3 - P$. At a triple point ($P = 3$), $F = 0$.

7.2 Phase Diagrams of One-Component Systems

• Triple point: All three phases coexist; $F = 0$.
• Critical point: Termination of the liquid-vapor coexistence curve; above this point the fluid is supercritical.
• Slope of solid-liquid boundary: Positive for most substances (liquid is denser); negative for water (ice is less dense).

7.3 Two-Component Systems

For binary mixtures, common diagrams include:

• Temperature-composition diagrams for liquid-vapor equilibrium (distillation).
• Eutectic diagrams for solid-liquid equilibrium.
• Lever rule: Determines the mass fractions of phases in a two-phase region.

Definition 5 (Lever Rule): For a two-phase region with phases $\alpha$ and $\beta$ at overall composition $x$:

$\frac{n_\alpha}{n_\beta} = \frac{x_\beta - x}{x - x_\alpha}$

8. Chemical Equilibrium

8.1 Equilibrium Constant

At equilibrium, $\Delta_r G = 0$, giving:

$\Delta_r G^\circ = -RT\ln K$

For the reaction $aA + bB \rightleftharpoons cC + dD$:

$K = \frac{a_C^c\,a_D^d}{a_A^a\,a_B^b}$

where $a_i$ are activities. For ideal gases: $a_i = P_i/P^\circ$, so:

$K_p = \frac{(P_C/P^\circ)^c\,(P_D/P^\circ)^d}{(P_A/P^\circ)^a\,(P_B/P^\circ)^b}$

8.2 van’t Hoff Equation

Theorem 7 (van’t Hoff Equation): The temperature dependence of the equilibrium constant:

$\frac{d\ln K}{dT} = \frac{\Delta_r H^\circ}{RT^2}$

Integrated form (assuming $\Delta_r H^\circ$ is constant):

$\ln\frac{K_2}{K_1} = -\frac{\Delta_r H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

8.3 Le Chatelier’s Principle

Definition 6 (Le Chatelier’s Principle): If a system at equilibrium is subjected to a disturbance, the system shifts to partially counteract the change.

• Increasing $T$ favors the endothermic direction.
• Increasing $P$ favors the direction with fewer moles of gas.
• Adding a reactant shifts equilibrium toward products.

9. Thermochemistry

9.1 Hess’s Law

Theorem 8 (Hess’s Law): The enthalpy change for a reaction is independent of the pathway; it equals the sum of enthalpy changes for any series of steps into which the reaction can be divided.

$\Delta_r H = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$

9.2 Standard Enthalpies

• Standard enthalpy of formation: $\Delta_f H^\circ$ — enthalpy change when 1 mol of compound forms from its elements in their standard states.
• Standard enthalpy of combustion: $\Delta_c H^\circ$ — enthalpy change for complete combustion of 1 mol of substance.
• Bond enthalpies: Average energy required to break a bond in the gas phase.

9.3 Kirchhoff’s Law

Theorem 9 (Kirchhoff’s Law): Temperature dependence of reaction enthalpy:

$\frac{d\Delta_r H^\circ}{dT} = \Delta_r C_p^\circ$

10. Partial Molar Quantities and Mixing

10.1 Partial Molar Quantities

Definition 7 (Partial Molar Volume): The partial molar volume of component $i$:

$V_i = \left(\frac{\partial V}{\partial n_i}\right)_{T,P,n_{j\neq i}}$

The total volume of a mixture:

$V = \sum_i n_i V_i$

10.2 Gibbs-Duhem Equation

Theorem 10 (Gibbs-Duhem Equation): At constant $T$ and $P$:

$\sum_i n_i\,d\mu_i = 0$

For a binary mixture: $n_A\,d\mu_A + n_B\,d\mu_B = 0$.

10.3 Chemical Potential of Real Solutions

For a real solution, the chemical potential is:

$\mu_i = \mu_i^\circ + RT\ln a_i = \mu_i^\circ + RT\ln(\gamma_i\,x_i)$

where $\gamma_i$ is the activity coefficient and $x_i$ is the mole fraction. For ideal solutions ($\gamma_i = 1$):

$\mu_i = \mu_i^\circ + RT\ln x_i$

11. Carnot Cycle for Chemical Systems

11.1 Efficiency

Theorem 11 (Carnot Efficiency): A heat engine operating between hot reservoir $T_h$ and cold reservoir $T_c$:

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

This is the maximum possible efficiency for any engine operating between these temperatures.

11.2 Refrigerators and Heat Pumps

• Coefficient of Performance (refrigerator): $\text{COP}_{\text{ref}} = \frac{T_c}{T_h - T_c}$
• Coefficient of Performance (heat pump): $\text{COP}_{\text{hp}} = \frac{T_h}{T_h - T_c}$

12. Thermodynamic Properties of Ideal Gases

12.1 Joule-Thomson Effect

For a real gas undergoing throttling (isenthalpic expansion):

$\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H = \frac{1}{C_p}\left[\frac{2a}{RT} - b\right]$

For an ideal gas: $\mu_{JT} = 0$ (no temperature change on throttling).

12.2 Adiabatic Processes

For a reversible adiabatic process with an ideal gas ($\gamma = C_p/C_V$):

$TV^{\gamma-1} = \text{const}, \quad PV^\gamma = \text{const}$

Work done:

$w = \frac{nR(T_2 - T_1)}{\gamma - 1}$

13. Fugacity and Activity

13.1 Fugacity

Definition 8 (Fugacity): For a real gas:

$\mu = \mu^\circ + RT\ln\left(\frac{f}{P^\circ}\right)$

where $f = \phi P$ and $\phi$ is the fugacity coefficient. As $P \to 0$, $f \to P$ and $\phi \to 1$.

13.2 Activity

For condensed phases:

$a_i = \frac{f_i}{f_i^\circ} \approx \gamma_i\,x_i$

The equilibrium constant in terms of activities:

$K = \prod_i a_i^{\nu_i}$

Common Pitfalls

1. Confusing heat ($q$) and temperature ($T$). Heat is energy in transit due to a temperature difference; temperature is a state property. Fix: $q$ is path-dependent; $T$ is a state function. Use $dU = \delta q + \delta w$, not $dU = T\,dS + \ldots$ for irreversible processes.
2. Using $\Delta G < 0$ as the sole spontaneity criterion. This only applies at constant $T$ and $P$. Fix: Use $dA < 0$ at constant $T$, $V$, or $dS > 0$ for isolated systems.
3. Ignoring the standard state. $\Delta G^\circ$ and $K$ are related, but $\Delta G = \Delta G^\circ + RT\ln Q$, where $Q$ is the reaction quotient. Fix: Only at equilibrium does $\Delta G = 0$ and $Q = K$.
4. Assuming $\Delta H$ and $\Delta S$ are temperature-independent. This is an approximation valid only over small temperature ranges. Fix: Use Kirchhoff’s law or integrate $C_p$ data when precision is needed.
5. Confusing intensive and extensive properties. $G$ is extensive; $\mu = G/n$ is intensive. Fix: Always use molar quantities when comparing substances with different amounts.
6. Wrong sign in the Clausius-Clapeyron equation. The negative sign appears because $\ln(P)$ decreases as $1/T$ increases for exothermic vaporization. Fix: Write it as $\ln(P_2/P_1) = -(\Delta H/R)(1/T_2 - 1/T_1)$ and check units.
7. Applying the ideal gas law to phase equilibrium without correction. The integrated Clausius-Clapeyron equation assumes $V_g \gg V_l$ and ideal gas behavior. Fix: Use fugacity corrections for high-pressure systems.

Summary

• First Law: $dU = \delta q + \delta w$; energy conservation.
• Second Law: $dS \geq \delta q/T$; entropy always increases in isolated systems.
• Third Law: $S \to 0$ as $T \to 0$ for a perfect crystal.
• Gibbs free energy: $\Delta G = \Delta H - T\Delta S$; spontaneity criterion at constant $T$, $P$.
• Chemical potential: $\mu_i = (\partial G/\partial n_i)_{T,P}$; drives mass transfer and chemical equilibrium.
• Maxwell relations: Connect measurable quantities derived from exact differentials of state functions.
• Phase rule: $F = C - P + 2$; determines degrees of freedom at equilibrium.
• Clausius-Clapeyron: $\ln(P_2/P_1) = -(\Delta H/R)(1/T_2 - 1/T_1)$; describes vapor pressure vs temperature.
• Equilibrium: $\Delta_r G^\circ = -RT\ln K$; van’t Hoff equation for temperature dependence.

Worked Examples

Example 1: Clausius-Clapeyron Calculation

Problem: The boiling point of water is 100 degrees C at 1 atm. The enthalpy of vaporization is 40.7 kJ/mol. Calculate the boiling point at 0.8 atm. Solution: ln(P2/P1) = -(Delta H_vap/R)(1/T2 - 1/T1). ln(0.8/1.0) = -(40700/8.314)(1/T2 - 1/373). -0.2231 = -4893(1/T2 - 0.00268). 1/T2 = 0.00268 + 0.2231/4893 = 0.00268 + 4.56e-5 = 0.002726. T2 = 366.8 K = 93.7 degrees C.

Example 2: Calculating Gibbs Free Energy of Reaction

Problem: For the reaction N2(g) + 3H2(g) -> 2NH3(g), Delta H = -92.4 kJ/mol, Delta S = -198.8 J K^-1 mol^-1. At 298 K, determine if the reaction is spontaneous. Solution: Delta G = Delta H - T Delta S = -92,400 - 298(-198.8) = -92,400 + 59,200 = -33,200 J/mol = -33.2 kJ/mol. Delta G < 0, so the reaction is spontaneous at 298 K. At what T does it become non-spontaneous? Delta G = 0 when T = Delta H/Delta S = 92,400/198.8 = 464.8 K.

Cross-References