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Thermodynamics

  1. Zeroth Law: If AA is in thermal equilibrium with BB, and BB with CC, then AA is in thermal equilibrium with CC. This establishes temperature as a transitive property and justifies the use of thermometers.

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

    dU=δq+δwdU = \delta q + \delta w

    where UU is internal energy, qq is heat, and ww is work. The notation δ\delta indicates inexact differentials: qq and ww are path-dependent, but UU is a state function.

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

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

wrev=−∫ViVfP dV=−nRTln⁡VfViw_{\text{rev}} = -\int_{V_i}^{V_f} P\,dV = -nRT\ln\frac{V_f}{V_i}

Definition 1 (Enthalpy): The enthalpy HH is defined as:

H=U+PVH = U + PV

For a process at constant pressure:

dH=dU+P dV+V dP=δqp+V dP  ⟹  ΔH=qpdH = 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:

Cp=(∂H∂T)P,CV=(∂U∂T)VC_p = \left(\frac{\partial H}{\partial T}\right)_P, \quad C_V = \left(\frac{\partial U}{\partial T}\right)_V

For an ideal gas: Cp−CV=nRC_p - C_V = nR.

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

∮δqT≤0\oint \frac{\delta q}{T} \leq 0

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

dS≥δqTdS \geq \frac{\delta q}{T}

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

For a reversible process at temperature TT:

ΔS=∫T1T2CT dT\Delta S = \int_{T_1}^{T_2} \frac{C}{T}\,dT

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

ΔtrsS=ΔtrsHTtrs\Delta_{\text{trs}}S = \frac{\Delta_{\text{trs}}H}{T_{\text{trs}}}

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

ΔS=n ΔfusHT=2×6010273=44.0 J/K\Delta S = \frac{n\,\Delta_{\text{fus}}H}{T} = \frac{2 \times 6010}{273} = 44.0 \text{ J/K}

■\blacksquare

Theorem 2 (Boltzmann Entropy):

S=kBln⁡WS = k_B \ln W

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

For NN distinguishable particles with nin_i in each energy level εi\varepsilon_i:

W=N!n1! n2! ⋯W = \frac{N!}{n_1!\,n_2!\,\cdots}

The entropy of mixing two ideal gases:

ΔmixS=−nR(xAln⁡xA+xBln⁡xB)\Delta_{\text{mix}}S = -nR\left(x_A \ln x_A + x_B \ln x_B\right)

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

lim⁡T→0S=0\lim_{T \to 0} S = 0

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

3. Gibbs Free Energy and Chemical Potential

Section titled “3. Gibbs Free Energy and Chemical Potential”

Definition 2 (Helmholtz Free Energy):

A=U−TSA = U - TS

dA=−S dT−P dVdA = -S\,dT - P\,dV

Definition 3 (Gibbs Free Energy):

G=H−TS=U+PV−TSG = H - TS = U + PV - TS

dG=−S dT+V dPdG = -S\,dT + V\,dP

At constant TT and PP: ΔG=wnon-PV\Delta G = w_{\text{non-PV}}, so the Gibbs free energy change equals the maximum non-expansion work.

ConditionCriterion
Constant TT, VV (closed)dA<0dA < 0
Constant TT, PP (closed)dG<0dG < 0
Isolated systemdS>0dS > 0

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

dU=T dS−P dVdU = T\,dS - P\,dV dH=T dS+V dPdH = T\,dS + V\,dP dA=−S dT−P dVdA = -S\,dT - P\,dV dG=−S dT+V dPdG = -S\,dT + V\,dP

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

dG=−S dT+V dP+∑i=1kμi dnidG = -S\,dT + V\,dP + \sum_{i=1}^{k} \mu_i\,dn_i

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

For an ideal gas: μ=μ∘+RTln⁡PP∘\mu = \mu^\circ + RT\ln\frac{P}{P^\circ}.

Theorem 4 (Maxwell Relations): Since UU, HH, AA, GG are state functions, their mixed second partial derivatives are equal:

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

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

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

ΔS=∫V1V2nRV dV=nRln⁡V2V1\Delta S = \int_{V_1}^{V_2} \frac{nR}{V}\,dV = nR\ln\frac{V_2}{V_1}

Theorem 5 (Gibbs-Helmholtz Equation):

[∂(G/T)∂T]P=−HT2\left[\frac{\partial(G/T)}{\partial T}\right]_P = -\frac{H}{T^2}

Equivalently:

ΔG2T2−ΔG1T1=−ΔH(1T2−1T1)\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 ΔH\Delta H is constant over the temperature range).

At phase equilibrium between two phases α\alpha and β\beta:

μα(T,P)=μβ(T,P)\mu_\alpha(T, P) = \mu_\beta(T, P)

Differentiating along the coexistence curve:

dμα=dμβ  ⟹  −Sα dT+Vα dP=−Sβ dT+Vβ dPd\mu_\alpha = d\mu_\beta \implies -S_\alpha\,dT + V_\alpha\,dP = -S_\beta\,dT + V_\beta\,dP

dPdT=ΔtrsSΔtrsV=ΔtrsHT ΔtrsV\frac{dP}{dT} = \frac{\Delta_{\text{trs}}S}{\Delta_{\text{trs}}V} = \frac{\Delta_{\text{trs}}H}{T\,\Delta_{\text{trs}}V}

For liquid-vapor equilibrium, assuming ΔvapH\Delta_{\text{vap}}H is constant and Vg≫VlV_g \gg V_l:

ln⁡P2P1=−ΔvapHR(1T2−1T1)\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 ΔvapH=30.8\Delta_{\text{vap}}H = 30.8 kJ/mol. Find the vapor pressure at 298 K.

ln⁡P1.013×105=−308008.314(1298−1353)=−1.93\ln\frac{P}{1.013 \times 10^5} = -\frac{30800}{8.314}\left(\frac{1}{298} - \frac{1}{353}\right) = -1.93

P=1.013×105×e−1.93=1.47×104 Pa≈14.7 kPaP = 1.013 \times 10^5 \times e^{-1.93} = 1.47 \times 10^4 \text{ Pa} \approx 14.7 \text{ kPa}

■\blacksquare

Theorem 6 (Gibbs Phase Rule): For a system with CC components and PP phases at equilibrium:

F=C−P+2F = C - P + 2

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

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

7.2 Phase Diagrams of One-Component Systems

Section titled “7.2 Phase Diagrams of One-Component Systems”
  • Triple point: All three phases coexist; F=0F = 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).

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 α\alpha and β\beta at overall composition xx:

nαnβ=xβ−xx−xα\frac{n_\alpha}{n_\beta} = \frac{x_\beta - x}{x - x_\alpha}

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

ΔrG∘=−RTln⁡K\Delta_r G^\circ = -RT\ln K

For the reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD:

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

where aia_i are activities. For ideal gases: ai=Pi/P∘a_i = P_i/P^\circ, so:

Kp=(PC/P∘)c (PD/P∘)d(PA/P∘)a (PB/P∘)bK_p = \frac{(P_C/P^\circ)^c\,(P_D/P^\circ)^d}{(P_A/P^\circ)^a\,(P_B/P^\circ)^b}

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

dln⁡KdT=ΔrH∘RT2\frac{d\ln K}{dT} = \frac{\Delta_r H^\circ}{RT^2}

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

ln⁡K2K1=−ΔrH∘R(1T2−1T1)\ln\frac{K_2}{K_1} = -\frac{\Delta_r H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

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 TT favors the endothermic direction.
  • Increasing PP favors the direction with fewer moles of gas.
  • Adding a reactant shifts equilibrium toward products.

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.

ΔrH=∑ΔfH∘(products)−∑ΔfH∘(reactants)\Delta_r H = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})

  • Standard enthalpy of formation: ΔfH∘\Delta_f H^\circ — enthalpy change when 1 mol of compound forms from its elements in their standard states.
  • Standard enthalpy of combustion: ΔcH∘\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.

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

dΔrH∘dT=ΔrCp∘\frac{d\Delta_r H^\circ}{dT} = \Delta_r C_p^\circ

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

Vi=(∂V∂ni)T,P,nj≠iV_i = \left(\frac{\partial V}{\partial n_i}\right)_{T,P,n_{j\neq i}}

The total volume of a mixture:

V=∑iniViV = \sum_i n_i V_i

Theorem 10 (Gibbs-Duhem Equation): At constant TT and PP:

∑ini dμi=0\sum_i n_i\,d\mu_i = 0

For a binary mixture: nA dμA+nB dμB=0n_A\,d\mu_A + n_B\,d\mu_B = 0.

For a real solution, the chemical potential is:

μi=μi∘+RTln⁡ai=μi∘+RTln⁡(γi xi)\mu_i = \mu_i^\circ + RT\ln a_i = \mu_i^\circ + RT\ln(\gamma_i\,x_i)

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

μi=μi∘+RTln⁡xi\mu_i = \mu_i^\circ + RT\ln x_i

Theorem 11 (Carnot Efficiency): A heat engine operating between hot reservoir ThT_h and cold reservoir TcT_c:

η=1−TcTh\eta = 1 - \frac{T_c}{T_h}

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

  • Coefficient of Performance (refrigerator): COPref=TcTh−Tc\text{COP}_{\text{ref}} = \frac{T_c}{T_h - T_c}
  • Coefficient of Performance (heat pump): COPhp=ThTh−Tc\text{COP}_{\text{hp}} = \frac{T_h}{T_h - T_c}

12. Thermodynamic Properties of Ideal Gases

Section titled “12. Thermodynamic Properties of Ideal Gases”

For a real gas undergoing throttling (isenthalpic expansion):

μJT=(∂T∂P)H=1Cp[2aRT−b]\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H = \frac{1}{C_p}\left[\frac{2a}{RT} - b\right]

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

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

TVγ−1=const,PVγ=constTV^{\gamma-1} = \text{const}, \quad PV^\gamma = \text{const}

Work done:

w=nR(T2−T1)γ−1w = \frac{nR(T_2 - T_1)}{\gamma - 1}

Definition 8 (Fugacity): For a real gas:

μ=μ∘+RTln⁡(fP∘)\mu = \mu^\circ + RT\ln\left(\frac{f}{P^\circ}\right)

where f=ϕPf = \phi P and ϕ\phi is the fugacity coefficient. As P→0P \to 0, f→Pf \to P and ϕ→1\phi \to 1.

For condensed phases:

ai=fifi∘≈γi xia_i = \frac{f_i}{f_i^\circ} \approx \gamma_i\,x_i

The equilibrium constant in terms of activities:

K=∏iaiνiK = \prod_i a_i^{\nu_i}

  1. Confusing heat (qq) and temperature (TT). Heat is energy in transit due to a temperature difference; temperature is a state property. Fix: qq is path-dependent; TT is a state function. Use dU=δq+δwdU = \delta q + \delta w, not dU=T dS+…dU = T\,dS + \ldots for irreversible processes.
  2. Using ΔG<0\Delta G < 0 as the sole spontaneity criterion. This only applies at constant TT and PP. Fix: Use dA<0dA < 0 at constant TT, VV, or dS>0dS > 0 for isolated systems.
  3. Ignoring the standard state. ΔG∘\Delta G^\circ and KK are related, but ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT\ln Q, where QQ is the reaction quotient. Fix: Only at equilibrium does ΔG=0\Delta G = 0 and Q=KQ = K.
  4. Assuming ΔH\Delta H and ΔS\Delta S are temperature-independent. This is an approximation valid only over small temperature ranges. Fix: Use Kirchhoff’s law or integrate CpC_p data when precision is needed.
  5. Confusing intensive and extensive properties. GG is extensive; μ=G/n\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)\ln(P) decreases as 1/T1/T increases for exothermic vaporization. Fix: Write it as ln⁡(P2/P1)=−(ΔH/R)(1/T2−1/T1)\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 Vg≫VlV_g \gg V_l and ideal gas behavior. Fix: Use fugacity corrections for high-pressure systems.
  • First Law: dU=δq+δwdU = \delta q + \delta w; energy conservation.
  • Second Law: dS≥δq/TdS \geq \delta q/T; entropy always increases in isolated systems.
  • Third Law: S→0S \to 0 as T→0T \to 0 for a perfect crystal.
  • Gibbs free energy: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S; spontaneity criterion at constant TT, PP.
  • Chemical potential: μi=(∂G/∂ni)T,P\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+2F = C - P + 2; determines degrees of freedom at equilibrium.
  • Clausius-Clapeyron: ln⁡(P2/P1)=−(ΔH/R)(1/T2−1/T1)\ln(P_2/P_1) = -(\Delta H/R)(1/T_2 - 1/T_1); describes vapor pressure vs temperature.
  • Equilibrium: ΔrG∘=−RTln⁡K\Delta_r G^\circ = -RT\ln K; van’t Hoff equation for temperature dependence.

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

Section titled “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.

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