GATE Statistical Thermodynamics — Partition Functions and the Numericals They Generate

By Aniket Bhardwaj · 13 September 2026 · GATE Chemistry

Statistical thermodynamics frightens people because it is taught as theory and examined as arithmetic. In a GATE paper you are almost never asked to derive the canonical ensemble. You are asked to evaluate a rotational partition function, decide whether a vibration is excited at 300 K, or compute a standard entropy from molecular data. All of those are four-line calculations once you know which formula to use and, crucially, which conventions the numbers were tabulated in. This guide works through the standard question types with every arithmetic step shown.

The Boltzmann distribution and the molecular partition function

Everything starts from the population of energy level i relative to the total:

n_i/N = g_i e^−ε_i/kT / q where q = Σ_i g_i e^−ε_i/kT

Here g_i is the degeneracy of level i and q is the molecular partition function. The physical reading of q is simple and worth carrying into the exam: q is roughly the number of energy levels thermally accessible at temperature T. If q ≈ 1, only the ground level is occupied. If q is large, many levels are populated.

Because the different kinds of motion are approximately independent, their energies add and the partition functions multiply:

q = q_trans × q_rot × q_vib × q_elec

For N indistinguishable independent molecules, the canonical partition function is Q = q^N/N!. The N! is the correction for indistinguishability; leaving it out is what produced the Gibbs paradox, and leaving it out is also what turns a correct entropy into a wrong one in an exam.

The four building blocks

Motion	Partition function	Typical size at 300 K
Translation	q_trans = V/Λ³, with Λ = h/√(2πmkT)	10²⁸–10³¹ for a molar volume
Rotation (linear, high T)	q_rot = kT/(σ hcB̃) = T/(σ θ_R)	10–10³
Vibration (harmonic, from v = 0)	q_vib = 1/(1 − e^−hcν̃/kT)	Usually very close to 1
Electronic	q_elec = g₀ (+ excited terms if low-lying)	1, 2 or 3 for most molecules

Λ is the thermal de Broglie wavelength, and θ_R = hcB̃/k is the rotational temperature. σ is the symmetry number — the number of indistinguishable orientations produced by pure rotation of the molecule. Getting σ wrong is the single most common error in this topic, so here is the short table:

Molecule	σ	Molecule	σ
CO, HCl, NO (heteronuclear diatomic)	1	H₂, N₂, O₂ (homonuclear diatomic)	2
CO₂ (linear, symmetric)	2	H₂O	2
NH₃	3	CH₄	12
N₂O (linear, unsymmetric N–N–O)	1	Benzene	12

A useful conversion constant to memorise, because it turns spectroscopic wavenumbers directly into temperatures: hc/k = 1.4388 cm K. Multiply any wavenumber in cm⁻¹ by 1.4388 and you get the characteristic temperature in kelvin.

Worked example 1 — rotational partition function of CO at 298 K

For ¹²C¹⁶O the rotational constant is B̃ = 1.9313 cm⁻¹. Find q_rot at 298 K.

Step 1 — rotational temperature.
θ_R = (hc/k) × B̃ = 1.4388 × 1.9313 = 2.779 K

Step 2 — check the high-temperature approximation is allowed.
T/θ_R = 298 ÷ 2.779 = 107, which is far greater than 1, so the classical formula is safe. (If this ratio were near 1 you would have to sum the levels term by term.)

Step 3 — symmetry number. CO is heteronuclear, so σ = 1.

Step 4 — evaluate.
q_rot = T/(σ θ_R) = 298 ÷ (1 × 2.779) = 107.2

Interpretation: about a hundred rotational states are thermally accessible at room temperature. If the molecule had been N₂ instead, σ = 2 would have halved the answer — and that factor of two is exactly what examiners test.

Worked example 2 — is the vibration excited?

The CO stretching wavenumber is about 2170 cm⁻¹. Find q_vib at 298 K.

Step 1 — the dimensionless ratio.
x = hcν̃/kT = (1.4388 × 2170) ÷ 298 = 3122.2 ÷ 298 = 10.477

Step 2 — the Boltzmann factor.
e^−10.477 = 2.82 × 10⁻⁵

Step 3 — the partition function.
q_vib = 1 ÷ (1 − 2.82 × 10⁻⁵) = 1.0000282 ≈ 1

What this means. Only about 28 molecules in a million are vibrationally excited. The vibration is effectively frozen at room temperature, contributes essentially nothing to the heat capacity, and can be dropped from q entirely. This is why C–H and C≡O stretches never matter thermodynamically at 298 K — but low-frequency bends and torsions do.

Contrast — the CO₂ bending mode at about 667 cm⁻¹:
x = (1.4388 × 667) ÷ 298 = 959.7 ÷ 298 = 3.220
e^−3.220 = 0.0399, and the bend is doubly degenerate, so the population of the first excited bending level relative to the ground level is 2 × 0.0399 = 0.080, about 8 %. That is small but not negligible, and it is why CO₂ has a measurably larger heat capacity than a rigid linear molecule would.

Getting thermodynamic functions out of q

Once you have q, the thermodynamic functions follow by differentiation. The forms worth remembering are:

U − U(0) = NkT² (∂ ln q/∂T)_V
A − A(0) = −kT ln Q
p = kT (∂ ln Q/∂V)_T
S = [U − U(0)]/T + k ln Q (and S = k ln W at equilibrium)

For an ideal monatomic gas these collapse into the Sackur–Tetrode equation, which is the single most examinable result in the whole topic:

S_m = R [ ln( V_m / (N_AΛ³) ) + 5/2 ] with Λ = h/√(2πmkT)

Worked example 3 — standard molar entropy of argon from first principles

Calculate the standard molar entropy of argon gas at 298.15 K and 1 bar, treating it as an ideal monatomic gas.

Step 1 — mass of one atom. M(Ar) = 39.948 g mol⁻¹, and 1 u = 1.66054 × 10⁻²⁷ kg:
m = 39.948 × 1.66054 × 10⁻²⁷ = 6.6335 × 10⁻²⁶ kg

Step 2 — thermal wavelength.
2πmkT = 2π × 6.6335 × 10⁻²⁶ × 1.380649 × 10⁻²³ × 298.15
= 6.28319 × 2.7305 × 10⁻⁴⁶ = 1.7155 × 10⁻⁴⁵ kg² m² s⁻²
√(1.7155 × 10⁻⁴⁵) = 4.1419 × 10⁻²³
Λ = h ÷ 4.1419 × 10⁻²³ = 6.62607 × 10⁻³⁴ ÷ 4.1419 × 10⁻²³ = 1.5997 × 10⁻¹¹ m (about 16 pm)

Step 3 — cube it.
Λ³ = (1.5997 × 10⁻¹¹)³ = 4.0936 × 10⁻³³ m³

Step 4 — molar volume at 1 bar.
V_m = RT/p = (8.31446 × 298.15) ÷ 100 000 = 2478.96 ÷ 100 000 = 0.0247896 m³ mol⁻¹

Step 5 — the logarithm's argument.
N_AΛ³ = 6.02214 × 10²³ × 4.0936 × 10⁻³³ = 2.4652 × 10⁻⁹ m³ mol⁻¹
V_m ÷ N_AΛ³ = 0.0247896 ÷ 2.4652 × 10⁻⁹ = 1.00557 × 10⁷
ln(1.00557 × 10⁷) = 16.1237

Step 6 — the entropy.
S_m = 8.31446 × (16.1237 + 2.5) = 8.31446 × 18.6237 = 154.8 J K⁻¹ mol⁻¹

Cross-check against experiment. The tabulated standard molar entropy of argon at 298.15 K is 154.8 J K⁻¹ mol⁻¹. A calculation from nothing but the atomic mass and three fundamental constants reproduces a measured thermodynamic quantity to four significant figures — that agreement is the whole reason statistical thermodynamics is in the syllabus.

Note in passing how large the translational partition function itself is: q_trans = V_m/Λ³ = 0.0247896 ÷ 4.0936 × 10⁻³³ ≈ 6.06 × 10³⁰ accessible translational states in one mole's worth of volume.

Equipartition and where it fails

The classical equipartition theorem says every quadratic term in the energy contributes ½kT per molecule, that is ½R per mole to C_V,m. Translation gives three quadratic terms, each rotation about an axis gives one, and each vibrational mode gives two (kinetic and potential), hence R per mode.

C_V,m(classical) = (3/2)R + (rotational modes)(1/2)R + (vibrational modes)(1)R

Worked example 4 — why CO₂ does not obey equipartition at 298 K

CO₂ is linear with N = 3 atoms, so it has 3 translational, 2 rotational and 3N − 5 = 4 vibrational modes.

Classical prediction:
C_V,m = 1.5R + 1.0R + 4R = 6.5R = 6.5 × 8.314 = 54.0 J K⁻¹ mol⁻¹

Translation and rotation alone:
2.5R = 2.5 × 8.314 = 20.8 J K⁻¹ mol⁻¹

Measured value. The molar heat capacity at constant pressure for CO₂ near 298 K is about 37.1 J K⁻¹ mol⁻¹, so C_V,m = C_p,m − R ≈ 37.1 − 8.3 = 28.8 J K⁻¹ mol⁻¹.

Reading the result. The true value sits between the two predictions, much closer to the translation-plus-rotation figure. Vibration contributes about 28.8 − 20.8 = 8.0 instead of the classical 33.3, because three of the four modes are essentially frozen and only the two degenerate bending modes at 667 cm⁻¹ are partly excited — exactly the 8 % population computed earlier. Equipartition is the high-temperature limit, not a general law.

The convention traps — state which one you are using

Two places in this topic have two legitimate conventions, and textbooks genuinely differ. In an exam, state which you have used.

Zero of vibrational energy. Measuring from the v = 0 level (the usual spectroscopic choice) gives q_vib = 1/(1 − e^−x). Measuring from the bottom of the potential well, which includes the zero-point energy, gives q_vib = e^−x/2/(1 − e^−x). The two differ by the factor e^−x/2. Both appear in standard textbooks; energies computed from them differ by the zero-point energy, and heat capacities computed from them are identical.
Distinguishable versus indistinguishable particles. Q = q^N holds only for localised, distinguishable particles such as atoms fixed on lattice sites. Q = q^N/N! is required for a gas. The N! affects the entropy and the free energy but not the internal energy or the heat capacity, because ln(N!) has no temperature dependence.

Common mistakes that cost marks

Forgetting the symmetry number σ. Every homonuclear diatomic needs σ = 2, CH₄ needs σ = 12. Omitting it multiplies your answer by σ and changes the entropy by R ln σ.
Using the high-temperature rotational formula when it does not apply. It needs T ≫ θ_R. For H₂, θ_R ≈ 87.6 K, so at 298 K the ratio is only about 3.4 and the simple formula is a poor approximation. For CO or N₂ at 298 K the ratio is around 100 and the formula is excellent.
Mixing wavenumber and energy units. B̃ and ν̃ are usually tabulated in cm⁻¹. Multiply by hc before comparing with kT, or use the shortcut hc/k = 1.4388 cm K.
Dropping the N! in Q for a gas. This gives an entropy that is not extensive and is the classic Gibbs paradox error.
Counting vibrational modes wrongly. Linear molecules have 3N − 5 modes, non-linear ones 3N − 6. CO₂ is linear and has 4, not 3.
Giving each vibration ½R in equipartition. A vibration has both kinetic and potential quadratic terms, so it contributes R per mole, not ½R.
Assuming q_elec = 1 always. It is g₀, the ground-state degeneracy. NO has a low-lying excited electronic state and O₂ has a triplet ground term, so both need more care.

Where this appears in GATE Chemistry

Statistical thermodynamics sits in the physical chemistry portion of the GATE Chemistry (CY) syllabus and connects to spectroscopy (which supplies B̃ and ν̃), to quantum chemistry (which supplies the energy levels) and to classical thermodynamics (which the results must reproduce). For the exact syllabus wording, question count and marking scheme in your year, read the current official GATE information brochure.

Question type	What you must do
Evaluate q_rot (NAT)	Convert B̃ to θ_R, apply σ, divide T by σθ_R
Population ratio of two levels	Boltzmann factor with degeneracies; use hc/k = 1.4388 cm K
Sackur–Tetrode entropy (NAT)	Compute Λ, then V_m/(N_AΛ³), take ln, add 5/2, multiply by R
Heat capacity from modes	Count 3N − 5 or 3N − 6; apply equipartition; comment on frozen modes
Symmetry number MCQ	Count rotations that map the molecule onto itself
Conceptual MCQ on q	q as the number of accessible states; behaviour as T → 0 and T → ∞

Every one of these calculations needs constants to full precision. The Scientific Constants tool gives h, k_B, N_A, R, c and the atomic mass unit with their accepted values and units, so a Sackur–Tetrode or partition-function numerical does not fail on a mis-typed exponent.

Open the Scientific Constants Tool →

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