IIT-JAM Basic Mathematical Concepts for Chemistry — The Maths You Actually Need

By Aniket Bhardwaj · 13 September 2026 · IIT-JAM Chemistry

Physical chemistry is where most IIT-JAM candidates gain or lose their rank, and the honest reason a student struggles with it is rarely the chemistry — it is the calculus and the logarithms underneath. The good news is that the mathematics needed is a short, closed list. This article covers exactly that list, in the form in which chemistry uses it, with every example computed step by step.

1. Logarithms — the tool used in half the syllabus

log(ab) = log a + log b · log(a/b) = log a − log b · log(aⁿ) = n log a
ln x = 2.302585 × log₁₀ x · if ln y = z then y = e^z

pH, pK_a, the Nernst equation, the Arrhenius equation, integrated rate laws and entropy all live on logarithms. Two habits save marks: know that log 2 = 0.3010 and log 3 = 0.4771 by heart (they generate most exam-friendly numbers), and always split a power of ten out first.

Example 1. Find the pH of a solution with [H⁺] = 4.5 × 10⁻⁵ M.

pH = −log(4.5 × 10⁻⁵) = −[log 4.5 + log 10⁻⁵] = −[0.6532 − 5]
= 5 − 0.6532 = 4.347 ≈ 4.35.

Sanity check: [H⁺] lies between 10⁻⁵ and 10⁻⁴, so the pH must lie between 4 and 5. It does.

2. Differentiation — finding maxima and rates

d(xⁿ)/dx = n xⁿ⁻¹ · d(e^ax)/dx = a e^ax · d(ln x)/dx = 1/x
Product rule: (uv)′ = u′v + uv′ · Chain rule: dy/dx = (dy/du)(du/dx)
A maximum or minimum needs dy/dx = 0; the sign of d²y/dx² tells you which

Chemistry uses differentiation in three recurring places: defining a rate as −d[A]/dt, locating the most probable value of a distribution, and deriving thermodynamic relations. The classic JAM-level derivative is the radial distribution function of the hydrogen 1s orbital.

Example 2 — the most probable radius of the 1s orbital. The radial distribution function is P(r) = (4/a₀³) r² e^−2r/a₀. Find the r at which it is maximum.

Differentiate using the product rule, with u = r² and v = e^−2r/a₀:
dP/dr = (4/a₀³) [ 2r·e^−2r/a₀ + r²·(−2/a₀)e^−2r/a₀ ]
Take out the common factor 2r e^−2r/a₀:
dP/dr = (4/a₀³) · 2r e^−2r/a₀ [ 1 − r/a₀ ]
Set dP/dr = 0. The factor e^−2r/a₀ is never zero, and r = 0 is a minimum (P = 0 there), so the maximum comes from
1 − r/a₀ = 0 → r = a₀, the Bohr radius.

This is a genuinely satisfying result: the most probable distance of the 1s electron from the nucleus is exactly the Bohr radius, even though the model is quantum mechanical rather than Bohr's.

3. Integration — from a rate law to a usable equation

∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1) · ∫(1/x) dx = ln x + C · ∫e^ax dx = e^ax/a + C

Every integrated rate law in the kinetics chapter is this operation, done once. Do the first-order derivation yourself at least twice; after that you will never have to memorise it.

Example 3 — deriving and using the first-order integrated rate law.

Start from −d[A]/dt = k[A]. Separate the variables:
−d[A]/[A] = k dt
Integrate from [A]₀ at t = 0 to [A] at time t:
−(ln[A] − ln[A]₀) = kt → ln([A]₀/[A]) = kt.

Now use it. If k = 2.0 × 10⁻³ s⁻¹, how long until the reaction is 75 % complete?
75 % complete means [A]/[A]₀ = 0.25, so [A]₀/[A] = 4.
ln 4 = 1.386294.
t = 1.386294 ÷ (2.0 × 10⁻³) = 693.1 s ≈ 11.6 minutes.

Cross-check by a second route: t½ = ln 2 ÷ k = 0.693147 ÷ (2.0 × 10⁻³) = 346.6 s, and 75 % completion is exactly two half-lives, 2 × 346.6 = 693.1 s ✓.

Example 4 — a definite integral doing thermodynamic work. Find the work for the reversible isothermal expansion of 1 mol of an ideal gas at 300 K from V₁ to V₂ = 2V₁.

w = −∫_V₁^V₂ p dV, and for an ideal gas p = nRT/V, so
w = −nRT ∫_V₁^V₂ dV/V = −nRT ln(V₂/V₁).
w = −(1)(8.314)(300) ln 2 = −2494.2 × 0.693147 = −1728.9 J ≈ −1.73 kJ.

Sign convention warning. This uses the IUPAC convention ΔU = q + w, in which w is the work done on the system, so an expansion gives a negative w. Some older textbooks write ΔU = q − w and define w as the work done by the system, which would make the same number +1.73 kJ. Both are correct within their own convention — state which one you are using and never mix them inside one problem.

4. Partial derivatives — because chemistry has more than one variable

(∂z/∂x)_y means: differentiate with respect to x while holding y constant
Total differential: dz = (∂z/∂x)_y dx + (∂z/∂y)_x dy

Thermodynamic quantities depend on several variables at once, which is why every partial derivative in that chapter carries a subscript telling you what is held fixed. Losing that subscript changes the physical meaning entirely.

Example 5. For one mole of an ideal gas, V = RT/P. Find (∂V/∂T)_P at P = 1.00 bar.

Hold P constant, so R/P is just a constant multiplying T:
(∂V/∂T)_P = R/P = 0.083145 ÷ 1.00 = 0.0831 L K⁻¹ (using R = 0.083145 L bar K⁻¹ mol⁻¹).
For comparison, (∂V/∂P)_T = −RT/P², which is negative — raising the pressure at fixed temperature shrinks the gas, exactly as expected.

5. Straight lines, slopes and least squares

An enormous fraction of physical chemistry is "rearrange until it is y = mx + c, then read the constant off the slope". Learn to spot the pattern:

Relationship	Plot y against x	Slope gives
Arrhenius: ln k = ln A − E_a/RT	ln k vs 1/T	−E_a/R
First-order kinetics	ln[A] vs t	−k
Second-order kinetics	1/[A] vs t	+k
Beer–Lambert: A = εcl	A vs c	εl
Freundlich isotherm	log(x/m) vs log p	1/n
Clausius–Clapeyron	ln p vs 1/T	−ΔH_vap/R

Example 6. An Arrhenius plot of ln k against 1/T has slope −1.20 × 10⁴ K. Find the activation energy.

Slope = −E_a/R → E_a = −slope × R = (1.20 × 10⁴) × 8.314
= 99 768 J/mol = 99.8 kJ/mol.

Note the unit check: the slope is in kelvin and R is in J K⁻¹ mol⁻¹, so the product is in J mol⁻¹. If your answer comes out in the wrong unit, the plot axes were wrong.

6. Approximations, quadratics and significant figures

For small x: e^x ≈ 1 + x · ln(1 + x) ≈ x · (1 + x)ⁿ ≈ 1 + nx
Quadratic: ax² + bx + c = 0 → x = [−b ± √(b² − 4ac)] ÷ 2a

Example 7 — when is the "x is small" approximation safe? For a weak acid with K_a = 1.8 × 10⁻⁵ at concentration C = 0.10 M, find the pH both ways.

Approximate route: x = √(K_aC) = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.3416 × 10⁻³ M.
pH = −log(1.3416 × 10⁻³) = 3 − 0.1276 = 2.87.
Validity check: x/C = 1.3416 × 10⁻³ ÷ 0.10 = 0.0134 = 1.34 %, comfortably under the usual 5 % limit, so the approximation is acceptable.

Exact route: x² + K_ax − K_aC = 0.
b² − 4ac = (1.8 × 10⁻⁵)² + 4(1.8 × 10⁻⁶) = 3.24 × 10⁻¹⁰ + 7.2 × 10⁻⁶ = 7.200324 × 10⁻⁶.
√(7.200324 × 10⁻⁶) = 2.683342 × 10⁻³.
x = (−1.8 × 10⁻⁵ + 2.683342 × 10⁻³) ÷ 2 = 1.33267 × 10⁻³ M.
pH = 3 − 0.1247 = 2.88.

The two answers differ by 0.01 pH unit. That is the size of the error you accept when you use the approximation — negligible here, but not if the acid were stronger or the solution far more dilute, where x/C would exceed 5 % and the quadratic becomes compulsory.

Common mistakes that cost marks

Mixing ln and log. The factor 2.303 belongs between them. Writing pH = −ln[H⁺] is an instant loss of the whole question.
Dropping the subscript on a partial derivative. (∂V/∂T)_P and (∂V/∂T)_S are different physical quantities.
Using degrees instead of kelvin inside any exponential or logarithm.
Forgetting the constant of integration, or forgetting to apply the limits in a definite integral. Chemistry almost always wants the definite form.
Rounding intermediate results. Carry at least four significant figures through the working and round only at the end; in Example 7 early rounding would hide the difference between the two routes entirely.
Using the small-x approximation without checking it. Always compute x/C and state that it is below 5 %.
Not checking units at the end. A slope in K times R in J K⁻¹ mol⁻¹ must give J mol⁻¹. If it does not, something upstream is wrong.

Where this maths is examined

Chemistry topic	Mathematics it needs
Chemical kinetics	Separation of variables, integration, semilog plots
Thermodynamics	Definite integrals, partial derivatives, exact differentials
Ionic equilibrium	Logarithms, quadratic equations, validity of approximations
Quantum chemistry	Differentiation for maxima, normalisation integrals, operators
Spectroscopy and analysis	Straight-line fitting, least squares, error handling
Gaseous state	Distribution functions, square roots, partial derivatives

Check the exact scope of the mathematics expected in the current official IIT-JAM notification before you spend time on anything beyond this list.

Practise the calculus, then verify it. Derive every result by hand first — that is what earns marks — and then confirm your derivative or your maximum with the free Derivative Calculator, which returns f′(a) and f″(a) at a chosen point so you can also confirm whether a turning point is a maximum or a minimum.

Open the Derivative Calculator →

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