IIT-JAM Chemical Thermodynamics — The Complete Unit

By Aniket Bhardwaj · 17 September 2026 · IIT-JAM Chemistry

Chemical thermodynamics is the largest single block of physical chemistry in the IIT-JAM syllabus, and it is also the most connected: the first law feeds thermochemistry, thermochemistry feeds free energy, free energy feeds equilibrium, and equilibrium feeds electrochemistry in the next unit. Students who learn it as a list of disconnected formulas keep re-learning it. This article puts the whole unit in one place — the laws, the state functions, the process types, the Maxwell relations and the link to K — with every worked number computed line by line and cross-checked by a second route wherever one exists.

First law, and the sign convention problem

ΔU = q + w (IUPAC convention: w is work done on the system)
For expansion against a constant external pressure: w = −P_ext ΔV

Two conventions are in genuine use and you must know which one your paper is using. The modern IUPAC form, used by NCERT and by most current textbooks, is ΔU = q + w with w counted positive when work is done on the system. Many older texts, and some engineering books, write ΔU = q − W where W is work done by the system. The two are the same physics: w = −W. Whichever you use, apply the physical test — a gas that expands does work on the surroundings and loses internal energy, so in the IUPAC form w must come out negative. Never pick a sign from memory; derive it from that sentence.

The core equations of the unit

Enthalpy: H = U + PV · ΔH = ΔU + Δ(PV) = ΔU + Δn_gRT (ideal gases)
Heat capacities: C_P − C_V = R per mole (ideal gas) · γ = C_P/C_V

Isothermal reversible (ideal gas): w = −nRT ln(V₂/V₁), ΔU = 0, q = −w
Adiabatic reversible (ideal gas): q = 0, w = ΔU = nC_VΔT, T₁V₁^γ−1 = T₂V₂^γ−1

Second law: dS = δq_rev/T · ΔS_total > 0 for a spontaneous change
Entropy of an ideal gas: ΔS = nC_V ln(T₂/T₁) + nR ln(V₂/V₁)

Free energies: G = H − TS, A = U − TS · ΔG = ΔH − TΔS at constant T
Reaction free energy: ΔG = ΔG° + RT ln Q · at equilibrium ΔG = 0 so ΔG° = −RT ln K
Gibbs–Helmholtz: [∂(ΔG/T)/∂T]_P = −ΔH/T²

The four Maxwell relations

Each comes from one of the four fundamental equations by using the fact that mixed second partial derivatives of a state function are equal. Learn the fundamental equations and you never have to memorise the relations themselves.

Fundamental equation	Natural variables	Maxwell relation
dU = T dS − P dV	S, V	(∂T/∂V)_S = −(∂P/∂S)_V
dH = T dS + V dP	S, P	(∂T/∂P)_S = (∂V/∂S)_P
dA = −S dT − P dV	T, V	(∂S/∂V)_T = (∂P/∂T)_V
dG = −S dT + V dP	T, P	−(∂S/∂P)_T = (∂V/∂T)_P

The last two are the ones JAM uses most, because they convert an entropy derivative — which you cannot measure directly — into a P–V–T derivative, which you can.

Worked example 1 — isothermal reversible expansion, checked two ways

2.00 mol of an ideal gas expands reversibly and isothermally at 300 K from 5.00 L to 20.0 L. Find w, q, ΔU and ΔS_system.

V₂/V₁ = 20.0/5.00 = 4.00, and ln 4 = 1.38629

nRT = 2.00 × 8.314 × 300 = 4988.4 J

w = −nRT ln(V₂/V₁) = −4988.4 × 1.38629 = −6915 J = −6.92 kJ

Isothermal, ideal gas ⇒ ΔU = 0, therefore q = −w = +6.92 kJ

ΔS = nR ln(V₂/V₁) = 2.00 × 8.314 × 1.38629 = 16.628 × 1.38629 = +23.05 J K⁻¹

Cross-check by a second route: for a reversible isothermal step ΔS = q_rev/T = 6915 / 300 = 23.05 J K⁻¹. The two routes agree, which is the quickest way to catch a slip in an exam.

Note the signs make sense: the gas expands, so it does work on the surroundings (w negative) and must absorb an equal amount of heat to keep U constant.

Worked example 2 — adiabatic reversible expansion

1.00 mol of a monatomic ideal gas (C_V = 3R/2, γ = 5/3) at 300 K expands reversibly and adiabatically from 10.0 L to 20.0 L. Find T₂, ΔU, w, ΔH and ΔS.

T₂ = T₁ (V₁/V₂)^γ−1 = 300 × (0.500)^2/3

(0.500)^2/3 = e^{(2/3) ln 0.5} = e^−0.46210 = 0.62996

T₂ = 300 × 0.62996 = 189.0 K, so ΔT = −111.0 K

C_V = 1.5 × 8.314 = 12.471 J K⁻¹ mol⁻¹

ΔU = nC_VΔT = 1.00 × 12.471 × (−111.0) = −1384 J = −1.38 kJ

q = 0 (adiabatic), so w = ΔU = −1.38 kJ

C_P = C_V + R = 12.471 + 8.314 = 20.785 J K⁻¹ mol⁻¹
ΔH = nC_PΔT = 20.785 × (−111.0) = −2307 J = −2.31 kJ

Cross-check: ΔH = ΔU + nRΔT = −1384 + (1.00 × 8.314 × (−111.0)) = −1384 − 923 = −2307 J. Agreed.

ΔS = 0, because the process is both adiabatic and reversible — an isentropic change. Students often compute a non-zero entropy here out of habit; there is nothing to compute.

Worked example 3 — from ΔH° and ΔS° to K, and to the crossover temperature

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92.22 kJ mol⁻¹ and ΔS° = −198.75 J K⁻¹ mol⁻¹. Find ΔG° and K at 298.15 K, and the temperature above which the reaction is no longer spontaneous under standard conditions.

TΔS° = 298.15 × (−198.75) = −59257 J mol⁻¹

ΔG° = ΔH° − TΔS° = −92220 − (−59257) = −32963 J mol⁻¹ ≈ −33.0 kJ mol⁻¹

ΔG° = −RT ln K ⇒ ln K = 32963 / (8.314 × 298.15) = 32963 / 2478.8 = 13.298

K = e^13.298 ≈ 6.0 × 10⁵ (dimensionless, in terms of activities)

ΔG° = 0 when T = ΔH°/ΔS° = (−92220) / (−198.75) = 464 K

What the numbers are telling you. Both ΔH° and ΔS° are negative, so the reaction is enthalpy-driven and entropy-opposed. Below about 464 K the enthalpy term wins and ΔG° is negative; above it the −TΔS° term dominates and ΔG° turns positive. That single calculation is the thermodynamic reason ammonia synthesis is run at high pressure with a catalyst rather than simply at high temperature — heat helps the rate but hurts the equilibrium. JAM asks this as a reasoning question at least as often as it asks for the number.

Chemical potential in one paragraph

The chemical potential μ_i = (∂G/∂n_i)_{T,P,n_j} is the free energy carried by one mole of component i in a mixture. For an ideal gas μ = μ° + RT ln(P/P°), and for an ideal solution μ = μ° + RT ln x. Matter flows spontaneously from high chemical potential to low, which is why μ, not concentration, is the correct driving force for diffusion, for phase equilibrium and for osmosis. Equating μ across two phases is how the Clapeyron equation and Raoult's law are derived, so this definition connects three separate syllabus units.

Common mistakes that cost marks

Sign convention drift. Decide at the top of your answer whether w is work on or by the system, write it down, and stay with it.
Using ΔG° to judge a real mixture. ΔG° tells you about the standard state only. Spontaneity at the actual composition is decided by ΔG = ΔG° + RT ln Q. A reaction with positive ΔG° can still run forward if Q is small enough.
Forgetting the surroundings. The second law criterion is ΔS_total > 0, not ΔS_system > 0. Water freezing has negative ΔS_system and is perfectly spontaneous below 273 K.
Mixing J and kJ. ΔH is usually quoted in kJ mol⁻¹ and ΔS in J K⁻¹ mol⁻¹. Convert before subtracting, every time.
Computing entropy for an irreversible path. S is a state function, so you may — and must — invent a reversible path between the same two states and integrate δq_rev/T along that.
Using Δn_g wrongly. In ΔH = ΔU + Δn_gRT only gaseous moles count. For the ammonia reaction Δn_g = 2 − 4 = −2.
Giving K a unit. A thermodynamic equilibrium constant is defined through activities and is dimensionless. Only then does ln K make sense.

How to prepare the unit

Sub-topic	What you must be able to do without hesitation
First law and work	Compute w, q, ΔU, ΔH for isothermal, isobaric, isochoric and adiabatic ideal-gas paths
Thermochemistry	Hess's law, bond enthalpies, Kirchhoff's equation for ΔH at another temperature
Second and third laws	ΔS for expansion, heating, phase change; absolute entropies; ΔS_total as the criterion
Free energy	ΔG = ΔH − TΔS; ΔG = ΔG° + RT ln Q; crossover temperature
Maxwell and Gibbs–Helmholtz	Derive a relation on demand from the four fundamental equations
Chemical potential	Define it, use it for phase and solution equilibria
Equilibrium link	ΔG° = −RT ln K, and the van't Hoff equation for K at another temperature

Use this as a checklist for revision, not as a claim about what will be asked. The official IIT-JAM notification for your year is the only authority on syllabus and paper pattern.

Verify every ΔG line instantly. The Gibbs free energy calculator takes ΔH, ΔS and T and returns ΔG — so you can check a full problem set, and find the crossover temperature by solving for the T that makes ΔG zero.

Open the Gibbs Free Energy Calculator →

Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches — classroom at the Gurugram coaching centre and live online classes for students across India. Details at abcchemistry.in.