Deep-Dive Guides · Chemical Thermodynamics
Classical Thermodynamics — the complete deep dive
From the three laws to Maxwell relations, spontaneity criteria, the ΔG–K link and its temperature dependence. The organising idea: thermodynamics is bookkeeping of state functions.
Free guide — the complete guide, open to everyone.
- How to use this guide — the state-function map
- Zeroth & First Law: energy, work, enthalpy
- Second & Third Law: entropy & spontaneity
- Thermodynamic potentials & Maxwell relations
- Chemical potential, ΔG–K & temperature dependence
- Worked problems (graded easy → hard)
- CSIR Part-C style problems
- Traps & look-alikes
- Quick-recall master tables (formulae)
0. How to use this guide — the state-function map
Everything below hangs on one distinction and one decision.
dS = dqrev/T holds as an equality.| Constraint | Potential minimised | Fundamental relation | Spontaneity test |
|---|---|---|---|
| isolated (U,V const) | — | — | dS ≥ 0 (2nd law, universe) |
| S, V const | Internal energy U | dU = T dS − P dV | dU ≤ 0 |
| S, P const | Enthalpy H = U+PV | dH = T dS + V dP | dH ≤ 0 |
| T, V const | Helmholtz A = U−TS | dA = −S dT − P dV | dA ≤ 0 |
| T, P const | Gibbs G = H−TS | dG = −S dT + V dP | dG ≤ 0 |
1. Zeroth & First Law — energy, work, enthalpy ▲ top
Zeroth law: if A is in thermal equilibrium with C and B is in thermal equilibrium with C, then A and B are in equilibrium with each other. This transitivity is what makes temperature a well-defined state function and lets a thermometer work.
1.1 State versus path functions
For a state function f(x,y), the differential is exact: df = (∂f/∂x)ydx + (∂f/∂y)xdy and the cross-derivatives are equal (Euler / reciprocity). This equality is the seed of every Maxwell relation (§3). For a path function such as work, ∮dw ≠ 0 in general, and you must specify the path to integrate it.
w = −∫PextdV is work done on the system. Work done on the system (compression) is positive. Heat absorbed by the system is positive. (Some Indian textbooks write ΔU = q − w with w = work done by the gas; the physics is identical — just track the sign of w.)1.2 Work, heat and reversibility
For expansion/compression, w = −∫V1V2 Pext dV. The external pressure — not the gas pressure — does the accounting.
- Free (Joule) expansion into vacuum: Pext = 0 ⇒ w = 0. For an ideal gas ΔU = 0 (U depends only on T), so q = 0 too — yet ΔS > 0 (irreversible).
- Irreversible against constant Pext: w = −Pext(V2−V1).
- Reversible isothermal (ideal gas): Pext = P = nRT/V at every instant, so
wrev = −nRT ln(V2/V1). This is the maximum magnitude of work — reversible work is the envelope of all paths.
1.3 First law and the definition of enthalpy
At constant volume, δw = 0, so qV = ΔU: the heat measured in a bomb calorimeter is ΔU directly. Most reactions run at constant pressure, where the system also does expansion work; to make heat a state-function-like quantity we define enthalpy.
Deriving qP = ΔH
- At constant P: qP = ΔU − w = ΔU + P ΔV = (U2−U1) + P(V2−V1).
- Group by state: qP = (U2+PV2) − (U1+PV1).
- Define H ≡ U + PV (a state function, since U, P, V all are). Then
qP = H2 − H1 = ΔH. ∎
Relation for reactions: ΔH = ΔU + Δ(PV); for ideal gases at constant T, ΔH = ΔU + ΔngasRT, where Δngas = (moles gaseous products − reactants).
1.4 Heat capacities and the Cp − Cv relation
CV = (∂U/∂T)V and CP = (∂H/∂T)P. CP exceeds CV because at constant P part of the added heat does expansion work rather than raising T.
General CP − CV (any substance), then ideal-gas limit
- Start from H = U + PV and differentiate at constant P: CP = (∂U/∂T)P + P(∂V/∂T)P.
- Expand (∂U/∂T)P using U(T,V): (∂U/∂T)P = CV + (∂U/∂V)T(∂V/∂T)P.
- Combine: CP − CV = [P + (∂U/∂V)T](∂V/∂T)P. The bracket is the "internal pressure + external pressure".
- Using the thermodynamic equation of state (∂U/∂V)T = T(∂P/∂T)V − P (derived in §3), this becomes the exact result CP − CV = T V α²/κT, with α = isobaric expansivity, κT = isothermal compressibility. This is always ≥ 0.
- Ideal gas: (∂U/∂V)T = 0 and (∂V/∂T)P = nR/P, so
CP − CV = nR(Mayer's relation). Per mole, CP,m − CV,m = R.
| Ideal gas | CV,m | CP,m | γ = CP/CV |
|---|---|---|---|
| Monatomic (He, Ar) | (3/2)R ≈ 12.47 | (5/2)R ≈ 20.79 | 5/3 ≈ 1.67 |
| Diatomic (N₂, O₂), 298 K | (5/2)R ≈ 20.79 | (7/2)R ≈ 29.10 | 7/5 = 1.40 |
| Nonlinear polyatomic (translation+rotation) | 3R ≈ 24.94 | 4R ≈ 33.26 | 4/3 ≈ 1.33 |
1.5 Adiabatic reversible processes
Deriving PVγ = const for a reversible adiabat (ideal gas)
- Adiabatic: δq = 0 ⇒ dU = δw ⇒ CVdT = −P dV = −(nRT/V)dV.
- Separate: CV dT/T = −nR dV/V. Integrate: CV ln(T₂/T₁) = −nR ln(V₂/V₁).
- Use nR = CP−CV and divide by CV: ln(T₂/T₁) = −(γ−1) ln(V₂/V₁), i.e. T Vγ−1 = const.
- Substitute T ∝ PV: P Vγ = const, and Tγ P1−γ = const. Work:
w = CVΔT = (P₂V₂−P₁V₁)/(γ−1).
1.6 Thermochemistry & Kirchhoff's law
Hess's law is just "ΔH is a state function": the enthalpy of a reaction is the sum over any convenient path. ΔH°rxn = Σν ΔH°f(products) − Σν ΔH°f(reactants) (elements in standard states have ΔH°f = 0). The temperature dependence is Kirchhoff's law:
with ΔCP = Σν CP(products) − Σν CP(reactants). An identical relation holds for ΔS with ΔCP/T inside the integral.
Solved example — ΔH vs ΔU for a combustion
Q: For C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔU° = −2212.5 kJ/mol at 298 K. Find ΔH°.
Count gas moles: products gaseous = 3 (CO₂; water is liquid); reactants gaseous = 1 + 5 = 6. Δngas = 3 − 6 = −3.
Apply ΔH = ΔU + ΔngasRT = −2212.5 + (−3)(8.314×10⁻³)(298) = −2212.5 − 7.43 = −2219.9 kJ/mol.
Because gas moles decrease, the system is compressed by the surroundings (work done on it), so |ΔH| > |ΔU| here.
2. Second & Third Law — entropy & spontaneity ▲ top
The first law conserves energy but says nothing about direction. Ice melting and water freezing both conserve energy; only one is spontaneous at 25 °C. The second law supplies the arrow of time through entropy.
dS ≥ δq/T (equality for reversible paths); for an isolated system ΔS ≥ 0.2.1 The Carnot cycle and the working-substance independence of efficiency
Carnot's engine (two isotherms + two adiabats, all reversible) is the theoretical maximum. It fixes the thermodynamic temperature scale and forces the existence of entropy.
Carnot efficiency and the birth of entropy
- Along the two isotherms (ideal gas, ΔU=0): qh = nRThln(VB/VA) > 0; qc = nRTcln(VD/VC) < 0.
- The two adiabats give ThVBγ−1=TcVCγ−1 and ThVAγ−1=TcVDγ−1, so VB/VA = VC/VD.
- Hence qc/qh = −Tc/Th, giving η = 1 + qc/qh = 1 − Tc/Th — independent of the working substance.
- Rearranged: qh/Th + qc/Tc = 0. Summing over a reversible cycle built from Carnot strips, ∮ δqrev/T = 0 ⇒ dS = δqrev/T defines a state function S. For any irreversible engine η < ηCarnot, giving the Clausius inequality ∮δq/T ≤ 0. ∎
2.2 Computing entropy changes
Because S is a state function, always integrate along a convenient reversible path between the same endpoints.
| Process (ideal gas unless noted) | ΔS |
|---|---|
| Isothermal reversible volume change | nR ln(V₂/V₁) = −nR ln(P₂/P₁) |
| Isochoric heating | nCV,m ln(T₂/T₁) |
| Isobaric heating | nCP,m ln(T₂/T₁) |
| General (T and V both change) | nCV,mln(T₂/T₁) + nR ln(V₂/V₁) |
| Phase change at Ttrs | ΔHtrs/Ttrs |
| Mixing of ideal gases (const T,P) | −nR Σ xi ln xi > 0 |
Solved example — entropy of the universe for irreversible expansion
Q: 2.0 mol ideal gas at 300 K expands isothermally and irreversibly from 10.0 L to 20.0 L against a constant external pressure equal to the final pressure. Find ΔSsys, ΔSsurr, ΔSuniv.
ΔSsys (state function, use reversible path) = nR ln(V₂/V₁) = 2.0×8.314×ln2 = 2.0×8.314×0.6931 = +11.53 J/K.
Heat to surroundings: ΔU=0 ⇒ q = −w. Pext = P₂ = nRT/V₂ = (2×8.314×300)/(20×10⁻³) = 2.494×10⁵ Pa. w = −PextΔV = −(2.494×10⁵)(10×10⁻³) = −2494 J. So qsys = +2494 J; qsurr = −2494 J.
ΔSsurr = qsurr/T = −2494/300 = −8.31 J/K. ΔSuniv = 11.53 − 8.31 = +3.22 J/K > 0 ✓ (irreversible). For a reversible expansion qsys would be nRT ln2 = 3458 J, ΔSsurr = −11.53 J/K and ΔSuniv = 0.
2.3 The Third Law and absolute entropies
- Residual entropy S(0) > 0 occurs for frozen-in disorder: CO (head-tail), ice (H-bond proton disorder, R ln(3/2) ≈ 3.4 J/K/mol), FClO₃. These deviate because the crystal is not "perfect".
- Unattainability of absolute zero in a finite number of steps is an equivalent statement of the third law.
3. Thermodynamic potentials & Maxwell relations ▲ top
Combining the first and second laws for a reversible process (δqrev=TdS, δw=−PdV) gives the fundamental equation, the single most useful line in the subject:
Everything else is a Legendre transform of this. Add or subtract PV and TS to slide between potentials so that the natural variables match your experiment:
The four potentials and their differentials
- U(S,V): dU = T dS − P dV.
- H = U + PV: dH = dU + P dV + V dP = T dS + V dP → natural variables (S,P).
- A = U − TS (Helmholtz): dA = dU − T dS − S dT = −S dT − P dV → (T,V).
- G = H − TS (Gibbs): dG = dH − T dS − S dT = −S dT + V dP → (T,P).
Read coefficients directly: e.g. from dG, (∂G/∂T)P = −S and (∂G/∂P)T = V. Since S > 0, G always decreases as T rises; since V > 0, G increases with P (steeper for gases).
3.1 The four Maxwell relations
Each potential's differential is exact, so its mixed second derivatives are equal. Applying (∂²/∂x∂y) to each fundamental equation gives:
| From | Maxwell relation |
|---|---|
| dU = T dS − P dV | (∂T/∂V)S = −(∂P/∂S)V |
| dH = T dS + V dP | (∂T/∂P)S = (∂V/∂S)P |
| dA = −S dT − P dV | (∂S/∂V)T = (∂P/∂T)V |
| dG = −S dT + V dP | (∂S/∂P)T = −(∂V/∂T)P |
Worked derivation — internal pressure of a gas
- Want (∂U/∂V)T. From dU = TdS − PdV at constant T: (∂U/∂V)T = T(∂S/∂V)T − P.
- Apply the Helmholtz Maxwell relation (∂S/∂V)T = (∂P/∂T)V: (∂U/∂V)T = T(∂P/∂T)V − P.
- Ideal gas: P = nRT/V ⇒ (∂P/∂T)V = nR/V ⇒ T(nR/V) − P = P − P = 0. So U(ideal) depends on T only. ✓
- van der Waals: (∂P/∂T)V = nR/(V−nb) gives (∂U/∂V)T = a n²/V² > 0 — the internal pressure from attractions, which is why real gases cool on Joule–Thomson expansion.
3.2 Gibbs–Helmholtz & the pressure dependence of G
Gibbs–Helmholtz equation
- Consider ∂(G/T)/∂T at constant P. Quotient rule: (1/T)(∂G/∂T)P − G/T².
- Insert (∂G/∂T)P = −S and G = H − TS: = (1/T)(−S) − (H−TS)/T² = −S/T − H/T² + S/T = −H/T².
- So [∂(G/T)/∂T]P = −H/T², and for a reaction [∂(ΔG/T)/∂T]P = −ΔH/T² — the route to the van't Hoff equation (§4).
Pressure dependence: from (∂G/∂P)T = V, integrating for an ideal gas gives G(P) = G° + nRT ln(P/P°). For a condensed phase V is nearly constant, so ΔG ≈ VΔP is small. This log term is the origin of the chemical potential of a gas and of the reaction quotient.
4. Chemical potential, ΔG–K & temperature dependence ▲ top
For a mixture that can change composition, add composition terms to dG: dG = −S dT + V dP + Σ μi dni, where μi = (∂G/∂ni)T,P,nj is the chemical potential (partial molar Gibbs energy). Matter flows from high μ to low μ, exactly as heat flows from high T to low T.
where activity ai = Pi/P° (ideal gas), = γixi (real solution — here γi is the activity coefficient, not the heat-capacity ratio γ = CP/CV of §1.5), = 1 (pure solid/liquid). Summing νμ over a reaction defines the reaction Gibbs energy ΔGrxn = (∂G/∂ξ)T,P, the slope of G against extent of reaction ξ.
Deriving ΔG = ΔG° + RT ln Q and ΔG° = −RT ln K
- ΔGrxn = Σνiμi = Σνi(μi° + RT ln ai) = ΔG° + RT ln Q, where Q = Π aiνi (products over reactants).
- At equilibrium the slope of G vs ξ is zero: ΔGrxn = 0 and Q = K (the equilibrium constant). So 0 = ΔG° + RT ln K ⇒ ΔG° = −RT ln K.
- Substituting back: ΔG = RT ln(Q/K). Sign test: Q<K ⇒ ΔG<0, reaction runs forward; Q>K ⇒ ΔG>0, runs backward; Q=K ⇒ at rest.
4.1 Temperature dependence — the van't Hoff equation
From Gibbs–Helmholtz to van't Hoff
- ln K = −ΔG°/RT. Differentiate w.r.t. T: d(ln K)/dT = −(1/R) d(ΔG°/T)/dT.
- By Gibbs–Helmholtz, d(ΔG°/T)/dT = −ΔH°/T². Hence d(ln K)/dT = ΔH°/RT² (differential van't Hoff).
- Assuming ΔH° roughly constant over the range, integrate: ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁). A plot of ln K vs 1/T is linear with slope −ΔH°/R and intercept ΔS°/R.
4.2 Clausius–Clapeyron & phase boundaries
Clapeyron → Clausius–Clapeyron
- Along a two-phase coexistence line the molar Gibbs energies are equal: μα=μβ. Moving along the line, dμα=dμβ ⇒ −SαdT+VαdP = −SβdT+VβdP.
- Rearrange to the exact Clapeyron equation: dP/dT = ΔStrs/ΔVtrs = ΔHtrs/(T ΔVtrs).
- For vaporisation/sublimation, ΔV ≈ Vgas = RT/P and assume ideal: dP/dT = ΔHvapP/(RT²) ⇒ d ln P/dT = ΔHvap/RT², i.e. ln(P₂/P₁) = −(ΔHvap/R)(1/T₂ − 1/T₁).
Solved example — vapour pressure via Clausius–Clapeyron
Q: Water boils (P = 1.00 atm) at 373 K with ΔHvap = 40.7 kJ/mol. Estimate its vapour pressure at 350 K.
ln(P₂/1.00) = −(40700/8.314)(1/350 − 1/373) = −4895.4 × (0.0028571 − 0.0026810) = −4895.4 × 1.7617×10⁻⁴ = −0.8624.
P₂ = e−0.8624 = 0.422 atm (≈ 321 mmHg). Reasonable — real value near 0.41 atm; the small gap is from assuming constant ΔHvap.
4.3 ΔG°, K, and electrochemistry — one unifying triangle
| Bridge | Relation | Use |
|---|---|---|
| Thermodynamics ↔ equilibrium | ΔG° = −RT ln K | get K from tabulated ΔG°f |
| Thermodynamics ↔ cell EMF | ΔG° = −nFE°cell | spontaneity of redox; E°>0 ⇒ ΔG°<0 |
| Equilibrium ↔ EMF | E° = (RT/nF) ln K; Nernst: E = E° − (RT/nF)ln Q | concentration cells, pH, Ksp |
| ΔG split | ΔG° = ΔH° − TΔS° | sign analysis / crossover temperature T = ΔH°/ΔS° |
Solved example — K from ΔG°, and its temperature shift
Q: A reaction has ΔH° = +52.0 kJ/mol and ΔS° = +120 J/K/mol. (a) Find ΔG° and K at 298 K. (b) Above what T does it become spontaneous? (c) Find K at 500 K.
(a) ΔG° = 52000 − 298×120 = 52000 − 35760 = +16240 J/mol. ln K = −ΔG°/RT = −16240/(8.314×298) = −6.556 ⇒ K = 1.42×10⁻³.
(b) Spontaneous when ΔG°<0 ⇒ T > ΔH°/ΔS° = 52000/120 = 433 K.
(c) ΔG°(500) = 52000 − 500×120 = −8000 J/mol ⇒ ln K = 8000/(8.314×500) = 1.925 ⇒ K = 6.85. (Cross-check via van't Hoff from 298 K gives the same order, confirming the ΔH°/ΔS° assumption.)
5. Worked problems — graded easy → hard ▲ top
Constants used throughout: R = 8.314 J K⁻¹ mol⁻¹ = 0.08206 L atm K⁻¹ mol⁻¹; F = 96485 C mol⁻¹. Click "Show solution".
P1. Isothermal reversible expansion.
2.00 mol ideal gas expands reversibly and isothermally at 300 K from 10.0 L to 20.0 L. Find w, q, ΔU, ΔH, ΔSsys.
Show solution
Isothermal ideal gas ⇒ ΔU = 0 and ΔH = 0. w = −nRT ln(V₂/V₁) = −2.00×8.314×300×ln2 = −2.00×8.314×300×0.6931 = −3458 J. q = −w = +3458 J. ΔSsys = nR ln(V₂/V₁) = 2.00×8.314×0.6931 = +11.53 J/K. (Reversible ⇒ ΔSuniv=0.)
P2. Choose the potential.
A reaction is run in an open flask (constant T and P). Which thermodynamic potential's minimisation determines equilibrium, and what is the spontaneity criterion?
Show solution
Constant T,P ⇒ natural variables of Gibbs energy G. Equilibrium is the minimum of G; a change is spontaneous iff ΔG < 0 (equivalently ΔSuniv > 0). At constant T,V you would instead minimise Helmholtz A.
P3. ΔH − ΔU for ammonia synthesis.
N₂(g) + 3H₂(g) → 2NH₃(g). By how much do ΔH and ΔU differ at 298 K?
Show solution
Δngas = 2 − (1+3) = −2. ΔH − ΔU = ΔngasRT = (−2)(8.314)(298) = −4955 J ≈ −4.96 kJ. So ΔH is about 5 kJ more negative than ΔU (gas moles fall, surroundings do work on the system).
P4. Carnot efficiency.
A reversible heat engine works between 500 K and 300 K. Maximum efficiency? If it absorbs 1000 J from the hot reservoir, how much work and how much waste heat?
Show solution
η = 1 − Tc/Th = 1 − 300/500 = 0.40 (40%). Work = 0.40×1000 = 400 J; heat rejected |qc| = 1000 − 400 = 600 J (check: qc/qh = −Tc/Th = −0.6 ✓).
P5. Reversible adiabatic compression.
1.00 mol monatomic ideal gas (γ = 5/3) at 300 K is compressed reversibly and adiabatically to half its volume. Find the final temperature and the work done.
Show solution
Adiabat: T Vγ−1 = const ⇒ T₂ = T₁(V₁/V₂)γ−1 = 300×(2)2/3 = 300×1.5874 = 476.2 K. Since q = 0, w = ΔU = nCV,mΔT = 1.00×(3/2×8.314)×(476.2−300) = 12.471×176.2 = +2198 J (positive: work done on the gas). Temperature rises on compression.
P6. Entropy of mixing.
0.50 mol N₂ and 0.50 mol O₂, each at 1 bar and 298 K, are mixed at constant T and P to 1 bar total. Find ΔSmix and ΔGmix.
Show solution
Mole fractions x = 0.5 each; ntotal = 1.0. ΔSmix = −nR Σxiln xi = −(1.0)(8.314)(0.5 ln0.5 + 0.5 ln0.5) = −8.314×ln0.5 = +5.76 J/K. ΔHmix = 0 (ideal), so ΔGmix = −TΔSmix = −298×5.76 = −1717 J (spontaneous, entropy-driven).
P7. van't Hoff — ΔH° from K(T).
An equilibrium constant doubles when T rises from 300 K to 310 K. Estimate ΔH°.
Show solution
ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁). ln2 = 0.6931 = −(ΔH°/8.314)(1/310 − 1/300) = −(ΔH°/8.314)(−1.0753×10⁻⁴). Solve: ΔH° = 0.6931×8.314/1.0753×10⁻⁴ = +53.6 kJ/mol (endothermic — K rises with T ✓).
P8. Clausius–Clapeyron — ΔHvap from two vapour pressures.
A liquid has vapour pressure 100 mmHg at 300 K and 400 mmHg at 340 K. Find ΔHvap.
Show solution
ln(P₂/P₁) = −(ΔHvap/R)(1/T₂ − 1/T₁). ln(400/100) = ln4 = 1.3863 = −(ΔHvap/8.314)(1/340 − 1/300) = −(ΔHvap/8.314)(−3.9216×10⁻⁴). ΔHvap = 1.3863×8.314/3.9216×10⁻⁴ = +29.4 kJ/mol.
P9. Electrochemistry ↔ thermodynamics ↔ K.
The Daniell cell (Zn|Zn²⁺||Cu²⁺|Cu) has E° = 1.10 V, n = 2 at 298 K. Find ΔG° and the equilibrium constant K.
Show solution
ΔG° = −nFE° = −2×96485×1.10 = −212.3 kJ/mol (strongly spontaneous). ln K = nFE°/RT = 212267/(8.314×298) = 85.67 ⇒ K = e85.67 ≈ 1.6×10³⁷. The huge K means the reaction goes essentially to completion — consistent with a large positive cell potential.
P10. Kirchhoff's law — ΔH at a new temperature.
For N₂ + 3H₂ → 2NH₃, ΔH°(298 K) = −92.2 kJ. Using CP,m (J K⁻¹mol⁻¹): NH₃ 35.1, N₂ 29.1, H₂ 28.8, estimate ΔH° at 398 K.
Show solution
ΔCP = 2(35.1) − [29.1 + 3(28.8)] = 70.2 − 115.5 = −45.3 J/K. Assuming ΔCP constant: ΔH°(398) = ΔH°(298) + ΔCPΔT = −92200 + (−45.3)(100) = −92200 − 4530 = −96.7 kJ. The reaction becomes slightly more exothermic on heating because the products have lower total heat capacity.
Note on the data: the CP values are standard 298 K figures treated as constants. Real heat capacities drift with temperature, but over a 100 K interval that drift moves ΔH° by well under 1 kJ, so the constant-ΔCP form of Kirchhoff’s law is the right tool here.
6. CSIR Part-C style problems ▲ top
Part-C rewards sign reasoning, derivations, and joining sub-topics. Each mirrors a typical stem.
C1. Four-quadrant spontaneity.
Classify the temperature dependence of spontaneity for each sign combination of (ΔH°, ΔS°), and give the crossover temperature where relevant.
Show solution
ΔG° = ΔH° − TΔS°.
(−, +): ΔG° < 0 at all T — always spontaneous (e.g. 2H₂O₂→2H₂O+O₂).
(+, −): ΔG° > 0 at all T — never spontaneous.
(−, −): spontaneous only below T = ΔH°/ΔS° (enthalpy-driven; e.g. condensation, most exothermic associations).
(+, +): spontaneous only above T = ΔH°/ΔS° (entropy-driven; e.g. vaporisation, CaCO₃ decomposition). The crossover T = ΔH°/ΔS° is where ΔG° = 0 and K = 1.
C2. Derive (∂CV/∂V)T.
Show that (∂CV/∂V)T = T(∂²P/∂T²)V, and hence that CV of an ideal gas is volume-independent.
Show solution
CV = (∂U/∂T)V. Then (∂CV/∂V)T = ∂/∂V(∂U/∂T)V = ∂/∂T(∂U/∂V)T (equality of mixed partials). Insert the thermodynamic equation of state (∂U/∂V)T = T(∂P/∂T)V − P: ∂/∂T[T(∂P/∂T)V − P]V = (∂P/∂T)V + T(∂²P/∂T²)V − (∂P/∂T)V = T(∂²P/∂T²)V. For an ideal gas P = nRT/V is linear in T, so (∂²P/∂T²)V = 0 ⇒ CV is independent of V. ∎
C3. Joule–Thomson coefficient.
Derive μJT = (∂T/∂P)H in terms of measurable quantities, show it is zero for an ideal gas, and explain the inversion temperature.
Show solution
Using the triple product rule on H(T,P): μJT = (∂T/∂P)H = −(∂H/∂P)T/(∂H/∂T)P = −(1/CP)(∂H/∂P)T. Now (∂H/∂P)T = V − T(∂V/∂T)P (from dH = TdS + VdP and a Maxwell relation). Hence μJT = (1/CP)[T(∂V/∂T)P − V]. Ideal gas: (∂V/∂T)P = V/T ⇒ bracket = 0 ⇒ μJT = 0 (no temperature change on throttling). Real gas: below the inversion temperature μJT > 0 (attractions dominate → gas cools on expansion, basis of Linde liquefaction); above it μJT < 0 (repulsions dominate → gas warms). H₂ and He have very low inversion temperatures, so they warm on ordinary throttling.
C4. Entropy generated by heat flow.
Two large reservoirs at 400 K and 300 K are connected by a conducting bar; 1000 J flows from hot to cold. Compute ΔSuniv and comment.
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Reservoirs are large ⇒ each stays at its T. ΔShot = −1000/400 = −2.50 J/K; ΔScold = +1000/300 = +3.33 J/K. ΔSuniv = +0.83 J/K > 0 — spontaneous, as required for heat flowing hot→cold. Reverse flow would give ΔSuniv < 0 (forbidden — the Clausius statement). The lost work (exergy destroyed) = T0ΔSuniv.
C5. Coupled equilibria (thermodynamic coupling).
Reaction A has ΔG° = +30 kJ (non-spontaneous). It is coupled to ATP hydrolysis with ΔG° = −31 kJ. Is the coupled process spontaneous, and what is the overall K at 298 K?
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Gibbs energies of coupled reactions add: ΔG°total = +30 + (−31) = −1 kJ ⇒ spontaneous (just barely). K = exp(−ΔG°/RT) = exp(1000/(8.314×298)) = exp(0.4037) = 1.50. This is the thermodynamic basis of biochemical coupling: an exergonic driver pulls an endergonic reaction forward by sharing a common intermediate.
C6. Maxwell relation to a measurement.
Using a Maxwell relation, find how the entropy of an ideal gas changes with pressure at constant T, and evaluate ΔS when 1 mol is compressed isothermally from 1 bar to 10 bar.
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From dG: (∂S/∂P)T = −(∂V/∂T)P. Ideal gas (∂V/∂T)P = nR/P, so (∂S/∂P)T = −nR/P. Integrate: ΔS = −nR ln(P₂/P₁) = −1×8.314×ln10 = −8.314×2.3026 = −19.14 J/K. Entropy falls on compression (fewer accessible positions), consistent with ΔS = nR ln(V₂/V₁) since V₂/V₁ = P₁/P₂.
7. Traps & look-alikes ▲ top
⚠ The classic confusions — memorise the discriminators
- Spontaneous ≠ fast. Thermodynamics (ΔG) predicts direction and extent, not rate. Diamond→graphite has ΔG<0 but is infinitely slow (kinetics).
- ΔSsys can be negative for a spontaneous process — freezing, protein folding. The test is always ΔSuniv ≥ 0 (or ΔGsys ≤ 0 at const T,P), never ΔSsys alone.
- q and w are path functions; ΔU, ΔH, ΔS, ΔG are state functions. "Heat of a state" and "work content" do not exist.
- Reversible ≠ ΔS = 0. Reversible means ΔSuniv = 0; ΔSsys can be large (e.g. reversible isothermal expansion).
- ΔG° vs ΔG. ΔG° = −RT ln K is a constant at fixed T (standard states); ΔG = ΔG° + RT ln Q is the actual driving force at the current Q. At equilibrium ΔG = 0 but ΔG° ≠ 0.
- ΔG° = 0 means K = 1, not "no reaction." Reaction proceeds to a 50:50-ish mixture.
- Le Chatelier for T uses ΔH. Heating shifts endothermic direction; adding a catalyst shifts nothing (changes rate, not K).
- Clapeyron (exact, any transition) vs Clausius–Clapeyron (approximation for vapour, assumes ideal gas and Vgas≫Vcond). Don't use C–C for solid↔liquid.
- CP,m − CV,m = R only for an ideal gas. (The extensive form is CP − CV = nR.) General result is TVα²/κT; for liquids/solids the difference is small but nonzero.
- Third law gives absolute S, not S = 0 for everything. Only a perfect crystal at 0 K has S = 0; residual entropy (CO, ice) is a genuine exception.
- Internal energy of ideal gas depends only on T — a consequence of (∂U/∂V)T = 0, not an axiom. It fails for real gases (a n²/V²).
- Maxwell relations: watch the sign. Only the U- and G-relations carry a minus sign on one side; deriving from the correct fundamental equation prevents errors.
8. Quick-recall master tables ▲ top
8.1 Fundamental equations & potentials
| Potential | Definition | Differential | Coefficients (from dΦ) |
|---|---|---|---|
| U(S,V) | — | dU = TdS − PdV | (∂U/∂S)V=T, (∂U/∂V)S=−P |
| H(S,P) | U + PV | dH = TdS + VdP | (∂H/∂S)P=T, (∂H/∂P)S=V |
| A(T,V) | U − TS | dA = −SdT − PdV | (∂A/∂T)V=−S, (∂A/∂V)T=−P |
| G(T,P) | H − TS | dG = −SdT + VdP | (∂G/∂T)P=−S, (∂G/∂P)T=V |
8.2 Key working formulae
| Quantity | Formula |
|---|---|
| First law | ΔU = q + w; w = −∫PextdV |
| Enthalpy link | ΔH = ΔU + ΔngasRT (ideal gas, const T) |
| Isothermal rev. work (ideal) | w = −nRT ln(V₂/V₁) |
| Adiabatic rev. (ideal) | PVγ=const; TVγ−1=const; w=CVΔT |
| Mayer | CP,m − CV,m = R |
| Entropy (heating) | ΔS = nC ln(T₂/T₁) (C = CV,m or CP,m — the molar heat capacity, since the n already sits outside) |
| Entropy (isothermal, ideal) | ΔS = nR ln(V₂/V₁) = −nR ln(P₂/P₁) |
| Carnot | η = 1 − Tc/Th |
| Clausius inequality | dS ≥ δq/T; ΔSuniv ≥ 0 |
| Gibbs–Helmholtz | [∂(ΔG/T)/∂T]P = −ΔH/T² |
| Reaction Gibbs energy | ΔG = ΔG° + RT ln Q; ΔG° = −RT ln K |
| van't Hoff | ln(K₂/K₁) = −(ΔH°/R)(1/T₂−1/T₁) |
| Clausius–Clapeyron | ln(P₂/P₁) = −(ΔHvap/R)(1/T₂−1/T₁) |
| Clapeyron (exact) | dP/dT = ΔHtrs/(TΔVtrs) |
| Kirchhoff | ΔH(T₂)=ΔH(T₁)+∫ΔCPdT |
| Electro-thermo | ΔG°=−nFE°=−RT ln K |
| Joule–Thomson | μJT=(1/CP)[T(∂V/∂T)P−V] |
8.3 The four Maxwell relations (sign map)
| Source | Relation | Mnemonic |
|---|---|---|
| U | (∂T/∂V)S = −(∂P/∂S)V | no cross-sign shortcut — derive from dU |
| H | (∂T/∂P)S = +(∂V/∂S)P | both "natural pair" positive |
| A | (∂S/∂V)T = +(∂P/∂T)V | the workhorse (→ internal pressure) |
| G | (∂S/∂P)T = −(∂V/∂T)P | the workhorse (→ S vs P, J–T) |
9. Revision summary — the results worth memorising & the four diagrams ▲ top
Use this as your last pass before the exam. Every line below is a result derived earlier in the guide, written in the form you should be able to reproduce from a blank page.
- Sign convention — this guide uses ΔU = q + w with w = work done on the system (§1.2 note). If your course writes w = work done by the gas, every isothermal-work sign flips, so settle the convention before touching a first-law problem.
- CP−CV = TVα²/κT is the general result for any substance; it collapses to R per mole for an ideal gas (§1.4).
- Ice residual entropy is R ln(3/2) ≈ 3.37 J K⁻¹mol⁻¹ on Pauling’s counting (§2.3), which reproduces the measured ~3.4.
- Water’s melting line slopes backwards (§4.2): dP/dT = ΔHfus/(TΔVfus) is negative because ΔVfus < 0 — ice is less dense than water.
- P5 adiabat: T₂ = 300×22/3 = 476.2 K and w = +2198 J. Keep 22/3 = 1.5874 in memory; it appears constantly for monatomic gases.
- P7 van’t Hoff: doubling K between 300 K and 310 K requires ΔH° = +53.6 kJ/mol — a useful feel for how little enthalpy a big change in K needs.
- P8 Clausius–Clapeyron: a fourfold rise in vapour pressure from 300 K to 340 K gives ΔHvap = +29.4 kJ/mol.
- P9 Daniell: ΔG° = −212.3 kJ/mol, ln K = 85.67, so K ≈ 1.6×10³⁷ — proof of how steeply K responds to a cell potential of only 1.10 V.
- P10 Kirchhoff: ΔCP = −45.3 J/K gives ΔH°(398) = −96.7 kJ; the CP values are 298 K standards treated as constant.
- The ΔG°(500 K) example in §4: the direct route and the van’t Hoff route both land on K ≈ 6.9, as they must whenever ΔH° and ΔS° are taken temperature-independent.
- Joule–Thomson (C3): μJT = (1/CP)[T(∂V/∂T)P − V] is zero for an ideal gas, positive (cooling on expansion) below the inversion temperature and negative above it. H₂ and He have inversion temperatures far below room temperature, so they warm on ordinary throttling.
The four diagrams, and the one idea each of them carries:
- Fig T-1 — reversible vs irreversible isothermal work: the work is the area under the path on a P–V plot, and the reversible path encloses the larger area.
- Fig T-2 — the Carnot cycle: two isotherms joined by two adiabats, η = 1 − Tc/Th, and ΔS = 0 once round the closed loop.
- Fig T-3 — G against extent of reaction: equilibrium is the minimum of the curve, where ΔGrxn = (∂G/∂ξ)T,P = 0 — not where G itself is zero.
- Fig T-4 — the van’t Hoff plot of ln K against 1/T: the slope is −ΔH°/R, so an endothermic reaction falls and an exothermic reaction rises as you move to the right.
All four diagrams are drawn inline, so they print and display with no external images. Redraw each one freehand from memory: if you can place the axes, the curve shapes and the slope signs unaided, you own the section.
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