CSIR-NET Thermodynamics — Partial Molar Quantities Made Clear
Partial molar quantities are where solution thermodynamics stops being additive arithmetic and starts being real thermodynamics. Candidates lose marks here not because the mathematics is hard, but because they never quite settle what the symbol means. Once you can say in one sentence what a partial molar volume physically is, the Gibbs–Duhem equation, chemical potential, activity and excess functions all follow from the same idea.
The definition, and what it physically means
In words: the partial molar quantity of component i is the change in the total extensive property X when you add one mole of i to a solution so large that the composition does not change, at constant T, P and amount of everything else. That "so large" clause is the whole point — add a mole of water to a bucket of ethanol and the composition shifts; add it to a lake and it does not.
Partial molar quantities therefore depend on composition, not just on the substance: the partial molar volume of water is 18.0 cm³/mol in pure water but about 14 cm³/mol in ethanol-rich mixtures, where a water molecule slots into gaps in the hydrogen-bonded ethanol structure.
They can even be negative. The partial molar volume of MgSO₄ in dilute aqueous solution is about −1.4 cm³/mol: adding the salt makes the solution shrink, because the highly charged ions pull surrounding water into a tighter shell than it occupies in bulk. This electrostriction is impossible to explain if you think of partial molar volume as "the volume the substance occupies".
Euler's theorem — the property that makes them useful
Because X is a homogeneous function of degree one in the amounts, Euler's theorem gives
Worked example 1 — volume of a non-ideal mixture. At a certain composition the partial molar volumes are V̄(water) = 17.4 cm³/mol and V̄(ethanol) = 57.4 cm³/mol. What is the volume of a mixture of 4.000 mol water and 1.000 mol ethanol at that composition, and how does it compare with simple additivity?
By Euler: V = (4.000 × 17.4) + (1.000 × 57.4) = 69.6 + 57.4 = 127.0 cm³
Simple additivity using the pure molar volumes (water 18.0, ethanol 58.0 cm³/mol): Videal = (4.000 × 18.0) + (1.000 × 58.0) = 72.0 + 58.0 = 130.0 cm³
ΔVmix = 127.0 − 130.0 = −3.0 cm³, a contraction of 2.3%: ethanol and water hydrogen bond to each other more efficiently than either does to itself. Any question asking for "the volume of the solution" from partial molar volumes is testing Euler's theorem and nothing else.
Experimentally, partial molar volumes are obtained by the method of intercepts: plot the mean molar volume Vm against the mole fraction x₂, draw the tangent at the composition of interest, and read the intercepts at x₂ = 0 and x₂ = 1. Those two intercepts are V̄₁ and V̄₂ at that composition.
The Gibbs–Duhem equation
Differentiating Euler's relation and comparing with the total differential of X at constant T and P leaves a constraint: the partial molar quantities of a mixture cannot vary independently.
Binary form: x1 dX̄1 + x2 dX̄2 = 0
Physically: if one component's partial molar quantity rises with composition, the other's must fall. That is what lets you obtain a solute's activity coefficient from measurements on the solvent alone — a real experimental technique and a standard Part C derivation.
Worked example 2 — a classic Gibbs–Duhem derivation. For a binary solution the partial molar volume of A is found to obey V̄A = VA° + k xB². Find V̄B.
Differentiate: dV̄A = 2k xB dxB
Substitute into Gibbs–Duhem, xA dV̄A + xB dV̄B = 0:
xA (2k xB dxB) + xB dV̄B = 0 ⟹ dV̄B = −2k xA dxB = −2k(1 − xB) dxB
Integrate from pure B (xB = 1, where V̄B = VB°) to xB:
V̄B − VB° = −2k [x − x²/2]1xB = −2k[(xB − xB²/2) − (1 − ½)]
= −2k xB + k xB² + k = k(1 − 2xB + xB²) = k(1 − xB)²
V̄B = VB° + k xA² — a pleasingly symmetric result. Check the limits: at xB = 1 (pure B) we get VB° as required, and at xA = 1 the deviation is largest. That limit check is worth two lines in an answer and catches sign errors instantly.
Chemical potential — the partial molar Gibbs energy
Of all the partial molar quantities, one has its own name and symbol:
Ideal solution: μi = μi° + RT ln xi
Real solution: μi = μi° + RT ln ai, with ai = γixi
Note that μ also equals (∂A/∂ni) at constant T, V and (∂U/∂ni) at constant S, V — but only the Gibbs-energy derivative, at constant T and P, is a genuine partial molar quantity in the strict sense. Examiners test exactly this distinction.
Ideal mixing — the four standard results
ΔSmix = −nR Σ xi ln xi (positive)
ΔHmix = 0 · ΔVmix = 0
Worked example 3 — mixing one mole of each. Mix 1.000 mol A with 1.000 mol B to form an ideal solution at 298 K. Find ΔGmix and ΔSmix.
xA = xB = 0.500, n = 2.000 mol.
Σ x ln x = 0.500 ln 0.500 + 0.500 ln 0.500 = ln 0.500 = −0.6931
ΔGmix = 2.000 × 8.314 × 298 × (−0.6931)
8.314 × 298 = 2477.6 ; × 2.000 = 4955.2 ; × 0.6931 = 3434.6
ΔGmix = −3.43 kJ
ΔSmix = −ΔGmix ÷ T (because ΔHmix = 0, so ΔG = −TΔS) = 3434.6 ÷ 298 = +11.53 J K⁻¹
Check against the direct formula: −nR Σ x ln x = 2.000 × 8.314 × 0.6931 = 11.53 J K⁻¹. In an ideal solution mixing is driven entirely by entropy — there is no energetic reward at all.
Worked example 4 — chemical potential of a dilute component. In an ideal solution at 298 K a component has x = 0.250, so μ − μ° = RT ln x = 8.314 × 298 × ln 0.250 = 2477.6 × (−1.3863) = −3435 J/mol. Since ln 0.250 = 2 ln 0.500, this happens to match the magnitude of ΔGmix above — a coincidence of the numbers chosen, but a useful check on your logarithms.
Excess functions — how real solutions are described
An excess function is the real value minus the ideal value at the same T, P and composition:
GE = RT Σ ni ln γi · HE = ΔHmix · VE = ΔVmix
Because ΔHmix and ΔVmix vanish for an ideal solution, HE and VE are simply the measured enthalpy and volume of mixing. Positive GE means the components dislike each other relative to ideal behaviour and gives a positive azeotrope; negative GE gives a negative one. A regular solution has SE = 0 with HE ≠ 0.
Mistakes to avoid
- Treating X̄i as a constant of the substance. It is a function of composition and must be quoted with one.
- Forgetting that partial molar volume can be negative. Electrostriction by multiply-charged ions is the standard example.
- Writing V = n₁V₁° + n₂V₂° for a real mixture. That is the ideal approximation; Euler's theorem requires the partial molar volumes.
- Applying Gibbs–Duhem without fixing T and P. The equation in that form holds only at constant temperature and pressure.
- Using mole fraction where activity is required. In a non-ideal solution it is ai, not xi, that enters the logarithm.
- Assuming ΔHmix = 0 for a real solution. Only ideal solutions mix athermally.
Summary table
| Concept | Relation | Exam use |
|---|---|---|
| Definition | X̄i = (∂X/∂ni)T,P,nj | Interpreting negative V̄ |
| Euler's theorem | X = Σ niX̄i | Volume of a mixture from V̄ values |
| Gibbs–Duhem | x₁dX̄₁ + x₂dX̄₂ = 0 | Get V̄₂ or γ₂ from component 1 data |
| Chemical potential | μi = μi° + RT ln ai | Equilibrium and phase conditions |
| Ideal mixing | ΔG = nRTΣx ln x, ΔH = ΔV = 0 | Numericals on ΔG and ΔS of mixing |
| Excess function | GE = RTΣni ln γi | Azeotrope sign, regular solutions |
Check the Gibbs-energy arithmetic instantly. The Gibbs Free Energy calculator handles ΔG = ΔH − TΔS and the ΔG° = −RT ln K relation — the same algebra that sits underneath chemical potential, mixing and equilibrium questions, so you can verify a step without breaking your train of thought.
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