CSIR-NET Statistical Thermodynamics — Partition Functions Worked in Full

By Aniket Bhardwaj · 22 September 2026 · CSIR-NET Chemistry

Statistical thermodynamics is the part of physical chemistry that connects a molecule's energy levels to the bulk quantities you already know — U, S, C_V and K. In CSIR-NET it usually appears as a short numerical: compute a partition function, find a population ratio, or decide which degrees of freedom are active at a given temperature. Every one of those is a two-line calculation once you have the right expression, so this article gives the expressions, then works four numbers all the way through with real constants.

The molecular partition function

q = Σ_i g_i e^−ε_i/k_BT

The sum runs over energy levels, with g_i the degeneracy of level i and ε_i measured from the lowest level. Read q as "the effective number of levels thermally accessible at temperature T". At T → 0 only the ground level is reachable and q → g₀; at high T, q becomes large.

Because translational, rotational, vibrational and electronic energies are (to a very good approximation) independent and additive, the partition function factorises:

q = q_trans × q_rot × q_vib × q_elec

For an assembly of N independent, indistinguishable molecules the canonical partition function is Q = q^N/N!. For localised, distinguishable particles — atoms fixed on lattice sites, for instance — the N! does not appear and Q = q^N. Choosing the wrong one is the single most common conceptual error in this topic.

The four building blocks

Mode	Expression	Notes
Translation	q_trans = V/Λ³ with Λ = h/√(2πmk_BT)	Λ is the thermal wavelength; m is the mass of one molecule in kg
Rotation (linear, T ≫ θ_rot)	q_rot = T/(σ θ_rot), θ_rot = hcB̃/k_B	σ is the symmetry number: 1 for a heteronuclear diatomic, 2 for a homonuclear one
Vibration (one mode)	q_vib = 1/(1 − e^−θ_vib/T), θ_vib = hcṽ/k_B	Energy zero taken at the vibrational ground state (see the caution below)
Electronic	q_elec ≈ g₀	Usually just the ground-state degeneracy, since excited electronic states lie far above k_BT

Constants used throughout: h = 6.62607 × 10⁻³⁴ J s, k_B = 1.380649 × 10⁻²³ J K⁻¹, N_A = 6.02214 × 10²³ mol⁻¹, R = 8.31446 J K⁻¹ mol⁻¹, c = 2.99792 × 10¹⁰ cm s⁻¹.

Worked example 1 — translational partition function of N₂

Find q_trans for N₂ at 298.15 K in a volume of 1.00 dm³ (= 1.00 × 10⁻³ m³).

Step 1 — mass of one molecule. m = 0.0280134 kg mol⁻¹ ÷ 6.02214 × 10²³ mol⁻¹ = 4.6517 × 10⁻²⁶ kg.

Step 2 — the quantity under the root.
m k_B = 4.6517 × 10⁻²⁶ × 1.380649 × 10⁻²³ = 6.4224 × 10⁻⁴⁹
× T: 6.4224 × 10⁻⁴⁹ × 298.15 = 1.91483 × 10⁻⁴⁶
× 2π: 1.91483 × 10⁻⁴⁶ × 6.28319 = 1.20320 × 10⁻⁴⁵

Step 3 — the thermal wavelength. √(1.20320 × 10⁻⁴⁵) = 3.46872 × 10⁻²³, so
Λ = 6.62607 × 10⁻³⁴ ÷ 3.46872 × 10⁻²³ = 1.9102 × 10⁻¹¹ m = 19.10 pm.

Step 4 — cube it and divide. Λ³ = (1.9102 × 10⁻¹¹)³ = 6.9707 × 10⁻³³ m³.
q_trans = 1.00 × 10⁻³ ÷ 6.9707 × 10⁻³³ = 1.43 × 10²⁹.

Sense check: Λ (19 pm) is far smaller than the average spacing between molecules, which is exactly the condition for classical behaviour — and it is why q_trans comes out enormous. If a calculation ever gives q_trans of order 1, you have made a unit error, almost always by leaving the volume in dm³ or the mass in g.

Worked example 2 — rotation and vibration of N₂ at 298.15 K

Take B̃ = 1.9982 cm⁻¹ and ṽ = 2358.6 cm⁻¹ for N₂. First convert hc into convenient units: hc = 6.62607 × 10⁻³⁴ × 2.99792 × 10¹⁰ = 1.98645 × 10⁻²³ J cm.

Rotational temperature. θ_rot = hcB̃/k_B = (1.98645 × 10⁻²³ × 1.9982) ÷ 1.380649 × 10⁻²³ = 3.96955 × 10⁻²³ ÷ 1.380649 × 10⁻²³ = 2.875 K.

N₂ is homonuclear, so σ = 2:
q_rot = T/(σθ_rot) = 298.15 ÷ (2 × 2.875) = 298.15 ÷ 5.750 = 51.9.

Vibrational temperature. θ_vib = hcṽ/k_B = (1.98645 × 10⁻²³ × 2358.6) ÷ 1.380649 × 10⁻²³ = 4.68544 × 10⁻²⁰ ÷ 1.380649 × 10⁻²³ = 3394 K.

θ_vib/T = 3394 ÷ 298.15 = 11.383, and e^−11.383 = 1.139 × 10⁻⁵, so
q_vib = 1 ÷ (1 − 1.139 × 10⁻⁵) = 1.00001.

Read the physics off the numbers. q_rot ≈ 52 says about fifty rotational levels are populated at room temperature; q_vib ≈ 1 says essentially every molecule is in v = 0. The vibration is frozen out, and that immediately predicts the heat capacity.

From q to thermodynamics

U − U(0) = Nk_BT² (∂ ln q/∂T)_V · A − A(0) = −k_BT ln Q · S = [U − U(0)]/T + k_B ln Q

Applying the first of these to each mode reproduces equipartition in the high-temperature limit: every quadratic term in the energy contributes ½RT per mole to U and ½R to C_V,m. For a linear molecule that means 3/2 R from translation, R from rotation (two rotational axes) and R per fully active vibrational mode.

For N₂ at 298 K the vibration is frozen, so the prediction is C_V,m = 3/2 R + R = 5/2 R = 2.5 × 8.31446 = 20.79 J K⁻¹ mol⁻¹, which is what is measured. A non-linear molecule would take 3/2 R + 3/2 R = 3R from translation and rotation instead.

Worked example 3 — Sackur–Tetrode entropy of argon

This is the classic "does statistical mechanics really work?" calculation, and it is a favourite because the answer can be checked against a tabulated value.

S_m = R [ ln(V_m / N_AΛ³) + 5/2 ]

Argon at 298.15 K and 1 bar, M = 39.948 g mol⁻¹.

Step 1 — mass and thermal wavelength. m = 0.039948 ÷ 6.02214 × 10²³ = 6.6336 × 10⁻²⁶ kg.
m k_B T = 6.6336 × 10⁻²⁶ × 1.380649 × 10⁻²³ × 298.15 = 2.73042 × 10⁻⁴⁶
× 2π = 1.71549 × 10⁻⁴⁵; its square root is 4.14185 × 10⁻²³.
Λ = 6.62607 × 10⁻³⁴ ÷ 4.14185 × 10⁻²³ = 1.5998 × 10⁻¹¹ m, so Λ³ = 4.0945 × 10⁻³³ m³.

Step 2 — molar volume. V_m = RT/p = (8.31446 × 298.15) ÷ 1.00 × 10⁵ = 2478.96 ÷ 10⁵ = 0.0247896 m³ mol⁻¹.

Step 3 — the logarithm. N_AΛ³ = 6.02214 × 10²³ × 4.0945 × 10⁻³³ = 2.46574 × 10⁻⁹ m³ mol⁻¹.
V_m/(N_AΛ³) = 0.0247896 ÷ 2.46574 × 10⁻⁹ = 1.00537 × 10⁷.
ln(1.00537 × 10⁷) = 0.00536 + 16.11810 = 16.1235.

Step 4 — finish. S_m = 8.31446 × (16.1235 + 2.5) = 8.31446 × 18.6235 = 154.8 J K⁻¹ mol⁻¹.

The tabulated standard molar entropy of argon at 298.15 K is 154.8 J K⁻¹ mol⁻¹. Two pages of statistics reproduce a calorimetric measurement to four figures, and that agreement is the reason the subject is trusted.

Worked example 4 — a population ratio

N_upper/N_lower = (g_upper/g_lower) e^−Δε/k_BT

What is the population of the J = 1 rotational level of N₂ relative to J = 0 at 298.15 K?

Rotational energies are ε(J) = hcB̃ J(J+1), so Δε(1 ← 0) = 2hcB̃ and Δε/k_BT = 2θ_rot/T = 5.750 ÷ 298.15 = 0.01929.

Degeneracy g_J = 2J + 1, so g₁/g₀ = 3/1 = 3.

N₁/N₀ = 3 × e^−0.01929 = 3 × 0.98090 = 2.94.

The upper level is more populated than the lower one — not because the Boltzmann factor favours it (it does not, being 0.98) but because it is three-fold degenerate. Forgetting g is the classic way to get this backwards.

Two conventions you must state, not assume

Vibrational energy zero. Measured from the vibrational ground state, q_vib = 1/(1 − e^−θ_vib/T). Measured from the bottom of the potential well, q_vib = e^−θ_vib/2T / (1 − e^−θ_vib/T), the extra factor being the zero-point energy. Both appear in standard textbooks; they differ, and a question is only well posed once the convention is fixed. Say which one you are using.
The high-temperature rotational formula. q_rot = T/(σθ_rot) is valid only when T ≫ θ_rot. For N₂ (θ_rot = 2.9 K) at room temperature that is safe. For H₂, θ_rot ≈ 88 K, so at 298 K the approximation is poor and the sum over J must be done explicitly. H₂ is exactly the molecule examiners choose to test this.

Other mistakes that cost marks

Dropping the symmetry number. σ = 2 for N₂, O₂, H₂ and CO₂; σ = 1 for CO, NO and HCl; σ = 3 for NH₃, σ = 12 for benzene and CH₄. Using σ = 1 for a homonuclear diatomic doubles q_rot and shifts the entropy by R ln 2 = 5.76 J K⁻¹ mol⁻¹.
Using molar mass instead of molecular mass. Λ needs the mass of one molecule in kilograms. Divide by N_A first, every time.
Mixing q and Q. The molecular q has no N! in it; the system Q for indistinguishable molecules does. Entropy expressions come from Q.
Forgetting the electronic degeneracy. Most closed-shell molecules have g₀ = 1, but radicals and many atoms do not — for a ²P_3/2 ground state g₀ = 4, and halogen atoms have a low-lying spin–orbit partner that must be included.
Leaving wavenumbers unconverted. B̃ and ṽ are in cm⁻¹; you must multiply by hc with c in cm s⁻¹, not m s⁻¹. A factor of 100 here changes every subsequent number.

Where this appears in the paper

Exam	Typical statistical-thermodynamics task
CSIR-NET Chemical Sciences	Compute q for a mode, population ratios, which modes contribute to C_V, entropy from Sackur–Tetrode
GATE Chemistry	Partition function expressions, equipartition heat capacities, Boltzmann distribution numericals
IIT-JAM / CUET-PG	Boltzmann distribution and degeneracy; the full partition-function machinery is usually beyond scope
MSc coursework	Deriving K from partition functions, ensembles, fluctuations, quantum statistics

The CSIR-NET paper is organised into Part A, Part B and Part C; for the current number of questions, marks and negative-marking scheme in each part, read the official notification for your session rather than relying on any secondary summary.

Everything above is exponentials, logarithms and careful powers of ten. That is precisely where marks are lost — a mis-typed exponent turns 10²⁹ into 10⁻²⁹ and the physics disappears. Do each step on paper, then repeat it on a calculator as an independent check. There is no dedicated partition-function tool in the suite, so this button honestly opens the suite itself rather than pretending one exists; the scientific calculator handles every step here.

Open the ABC Chemistry Calculator Suite →

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