For an ideal Fermi gas of electrons at T = 0, which statements are correct?

Every single-particle state below E_F is occupied

The mean energy per electron is 3E_F/5

Every single-particle state below E_F is occupied

The mean energy per electron is 3E_F/5

The heat capacity is finite and non-zero at T = 0

Which of the following are features of a first-order phase transition?

A correlation length that diverges at the transition

Thermodynamics and Statistical Mechanics: Laws, Ensembles, Partition Functions, Quantum Gases, Blackbody Radiation and Phase Transitions

Section 4 of the GATE Physics paper. Engineering thermodynamics, which other papers read, is about cycles and devices; this section is about where thermodynamics comes from. It follows the syllabus: the laws of thermodynamics and the potentials; macrostates and microstates, phase space and the three ensembles; the partition function, the free energy and how every thermodynamic quantity follows from it; classical and quantum statistics; the degenerate Fermi gas; blackbody radiation and Planck’s law; Bose–Einstein condensation; and first- and second-order phase transitions, phase equilibria and critical phenomena. Constants used in the numericals are stated where they are used: h = 6.626 × 10⁻³⁴ J s, k_B = 1.381 × 10⁻²³ J/K, e = 1.602 × 10⁻¹⁹ C, m_e = 9.109 × 10⁻³¹ kg, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, Wien’s constant b = 2.898 × 10⁻³ m K and R = 8.314 J mol⁻¹ K⁻¹.

1. The laws of thermodynamics and the potentials

Zeroth law: thermal equilibrium is transitive, which makes temperature a well-defined property.
First law: dU = δQ − P dV. U is a state function; Q and W are not.
Second law: dS ≥ δQ/T, with equality for reversible changes; no engine beats the Carnot efficiency η = 1 − T_c/T_h, which is 0.4 between 500 K and 300 K.
Third law: S → a constant (zero for a non-degenerate ground state) as T → 0, so heat capacities vanish at absolute zero and T = 0 cannot be reached in a finite number of steps.

The potentials U(S, V), H = U + PV, F = U − TS and G = H − TS are minimised at equilibrium under the constraints named by their natural variables: F at fixed T and V, G at fixed T and P. Their second derivatives give the Maxwell relations, e.g. (∂S/∂V)_T = (∂P/∂T)_V from dF = −S dT − P dV. For an ideal gas ΔS = nC_V ln(T₂/T₁) + nR ln(V₂/V₁); in a free expansion of 1 mol to double its volume, T is unchanged and ΔS = R ln 2 = 5.76 J/K, although no heat flows — entropy is a state function, and the process is irreversible.

2. Microstates, phase space and ensembles

A macrostate (N, V, E) is compatible with Ω microstates, and the fundamental postulate — every accessible microstate equally likely in an isolated system — gives Boltzmann’s S = k_B ln Ω. Classically, a microstate is a point in the 6N-dimensional phase space of all qᵢ and pᵢ, and states are counted as phase-space volume divided by h^3N; for identical particles one further divides by N!, which removes the Gibbs paradox (a spurious entropy of mixing identical gases).

The three ensembles
Ensemble	Fixed	Distribution and potential
Microcanonical	N, V, E	all Ω states equally likely; S = k_B ln Ω
Canonical	N, V, T (heat bath)	Pᵢ = e^−βEᵢ/Z; F = −k_BT ln Z
Grand canonical	μ, V, T (particle reservoir)	Pᵢ ∝ e^{−β(Eᵢ − μNᵢ)}; Φ = −k_BT ln 𝒵 = −PV

For large systems the three give the same thermodynamics, because the relative fluctuations of E or N fall as 1/√N. The canonical energy fluctuation is ⟨ΔE²⟩ = k_BT²C_V, a response function measured as a fluctuation. Equipartition follows from the canonical ensemble: each quadratic term in the classical energy contributes ½k_BT, so a monatomic gas has C_V = 3R/2 and a diatomic gas at room temperature (vibration frozen) 5R/2.

3. The partition function and everything that follows from it

Z = Σᵢ gᵢ e^−βEᵢ, with β = 1/(k_BT) and gᵢ the degeneracy. Then F = −k_BT ln Z, U = −∂ln Z/∂β, S = (U − F)/T, P = −(∂F/∂V)_T and C_V = ∂U/∂T. Independent subsystems multiply their partition functions; for N identical, indistinguishable, weakly interacting particles in the classical limit Z_N = Z₁ᴺ/N!, with Z₁ = V/λ³ for translation and λ = h/√(2πmk_BT) the thermal wavelength.

Two-level system (0 and ε): Z = 1 + e^−βε, the upper-level population is e^−βε/(1 + e^−βε) = 1/(e^βε + 1), and U = ε/(e^βε + 1). With ε/k_BT = ln 3 the upper level holds 1/4. The heat capacity shows a peak near k_BT ≈ 0.42ε — the Schottky anomaly — and U never exceeds ε/2.
Degeneracy matters: levels 0 (g = 1) and ε (g = 2) with e^−βε = ½ give Z = 1 + 2 × ½ = 2 and U = ε × (2 × ½)/2 = ε/2.
Quantum oscillator: Z = e^−βħω/2/(1 − e^−βħω) and U = ħω/2 + ħω/(e^βħω − 1), which tends to k_BT at high T (equipartition) and to the zero-point ħω/2 at low T.

⚠️ Divide by N! only for indistinguishable particles

Localised particles in a crystal are distinguishable by their sites, and Z_N = Z₁ᴺ. Gas molecules are not, and Z_N = Z₁ᴺ/N!. The N! does not change U or C_V (it does not depend on T), but it changes F, S and μ, and without it the entropy is not extensive.

4. Classical and quantum statistics; the degenerate Fermi gas

Occupation of a single-particle state of energy ε
Statistics	Mean occupation	Particles
Maxwell–Boltzmann	e^{−β(ε − μ)}	classical limit, dilute and hot
Fermi–Dirac	1/(e^{β(ε − μ)} + 1)	half-integer spin; at most one per state
Bose–Einstein	1/(e^{β(ε − μ)} − 1)	integer spin; μ ≤ lowest level

Both quantum forms reduce to Maxwell–Boltzmann when e^{β(ε − μ)} ≫ 1, i.e. when the thermal wavelength is much smaller than the interparticle spacing, nλ³ ≪ 1. For electrons in a metal the opposite holds: the gas is degenerate. At T = 0 every state up to the Fermi energy is filled, E_F = (ħ²/2m)(3π²n)^2/3, and the mean energy is (3/5)E_F. Worked, for copper with n = 8.47 × 10²⁸ m⁻³: (3π²n)^2/3 = 1.85 × 10²⁰ m⁻², ħ²/2m = 6.10 × 10⁻³⁹ J m², so E_F = 1.13 × 10⁻¹⁸ J = 7.0 eV — a Fermi temperature near 8 × 10⁴ K.

Because only electrons within about k_BT of E_F can be excited, the electronic heat capacity is C_el = (π²/2)Nk_B(T/T_F), linear in T and far below the classical 3Nk_B/2. The Fermi gas also exerts a degeneracy pressure P = (2/5)nE_F at T = 0, which supports white dwarfs.

5. Blackbody radiation and Bose–Einstein condensation

Photons are bosons with μ = 0 and two polarisations. Counting modes (8πν²/c³ per unit volume per unit frequency) and multiplying by the mean energy per mode gives Planck’s law, u(ν) = (8πhν³/c³)/(e^hν/k_BT − 1). At low frequency it becomes the classical Rayleigh–Jeans law 8πν²k_BT/c³, whose integral diverges (the ultraviolet catastrophe); at high frequency it falls exponentially (Wien). Its consequences: Wien’s displacement law λ_max T = 2.898 × 10⁻³ m K (the Sun, near 5800 K, peaks near 500 nm); the Stefan–Boltzmann law, emitted flux σT⁴ (907 kW/m² at 2000 K); and a radiation pressure P = u/3.

For massive bosons μ rises towards the lowest level as T falls. Below the condensation temperature T_c = (2πħ²/mk_B)(n/2.612)^2/3 the excited states cannot hold all the particles, and a macroscopic fraction occupies the ground state: N₀/N = 1 − (T/T_c)^3/2 for a uniform ideal gas, so at T = T_c/2 about 65% are condensed. The condensation is a transition in momentum space, not a spatial separation, and was first seen in dilute alkali vapours at hundreds of nanokelvin.

6. Phase transitions, phase equilibria and critical phenomena

In Ehrenfest’s classification a first-order transition has discontinuous first derivatives of G — entropy (latent heat L = TΔS) and volume jump — while a second-order (continuous) transition has continuous S and V but discontinuous or divergent second derivatives (C_P, compressibility). Two phases coexist when their chemical potentials are equal, which gives the Clausius–Clapeyron equation dP/dT = L/(TΔv) along the coexistence line. Gibbs’ phase rule F = C − P + 2 gives the degrees of freedom: a one-component system has F = 0 at its triple point.

Critical point. The liquid–gas line ends at T_c, where the latent heat and Δv vanish. For a van der Waals gas T_c = 8a/(27Rb), P_c = a/(27b²), V_c = 3b, and P_cV_c/RT_c = 3/8.
Order parameter and exponents. Near a continuous transition an order parameter (magnetisation, density difference) vanishes as (T_c − T)^β, susceptibility diverges as |T − T_c|^−γ, and the correlation length diverges as |T − T_c|^−ν. Mean-field (Landau) theory gives β = ½, γ = 1.
Universality. Critical exponents depend only on dimension and the symmetry of the order parameter, which is why a liquid–gas critical point and a uniaxial magnet share them.

Key takeaways

F = −k_BT ln Z and U = −∂ln Z/∂β; everything else is a derivative. Divide Z₁ᴺ by N! only for indistinguishable particles.
S = k_B ln Ω; the entropy of a free expansion is R ln(V₂/V₁) per mole even though Q = 0.
E_F = (ħ²/2m)(3π²n)^2/3 depends only on n; ⟨E⟩ = 3E_F/5 and C_el ∝ T.
Planck gives Wien (λ_max T = 2.898 × 10⁻³ m K) and Stefan–Boltzmann (σT⁴); below T_c a Bose gas condenses with N₀/N = 1 − (T/T_c)^3/2.
First order: latent heat, jumps in S and V, Clausius–Clapeyron. Continuous: diverging response and correlation length, exponents set by universality.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A two-level system has energies 0 and ε, with ε/(k_BT) = ln 3. The fraction of systems in the upper level, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.25
e^−βε = 1/3, so P_upper = (1/3)/(1 + 1/3) = 1/4 = 0.25. Taking the Boltzmann factor itself, 1/3 = 0.33, forgets to normalise by Z = 4/3.
A system has a non-degenerate level at energy 0 and a doubly degenerate level at energy ε. At a temperature where e^−ε/k_BT = 0.5, the mean energy in units of ε is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
Z = 1 + 2 × 0.5 = 2, and U = (0 × 1 + ε × 2 × 0.5)/Z = ε/2 = 0.5ε. Ignoring the degeneracy gives Z = 1.5 and U = ε/3 = 0.33ε.
For N identical, indistinguishable, non-interacting gas molecules each with single-particle partition function Z₁, the classical N-particle partition function is:
1. Z₁ᴺ/N!
2. Z₁ᴺ
3. N Z₁
4. Z₁/N!
Show answer
Answer: A — Z₁ᴺ/N!
Permuting identical molecules does not produce a new state, so Z₁ᴺ overcounts by N!, and Z_N = Z₁ᴺ/N! — the correction that makes the entropy extensive and removes the Gibbs paradox. Z₁ᴺ is right for distinguishable, localised particles such as atoms on lattice sites.
Copper has n = 8.47 × 10²⁸ conduction electrons per m³. Using E_F = (ħ²/2m)(3π²n)^2/3 with h = 6.626 × 10⁻³⁴ J s, m_e = 9.109 × 10⁻³¹ kg and e = 1.602 × 10⁻¹⁹ C, the Fermi energy in eV, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 7.03
3π²n = 2.508 × 10³⁰ m⁻³, whose 2/3 power is 1.846 × 10²⁰ m⁻². ħ = h/2π = 1.0546 × 10⁻³⁴ J s, so ħ²/2m = 6.105 × 10⁻³⁹ J m². E_F = 1.127 × 10⁻¹⁸ J = 7.03 eV (7.04 if ħ is rounded to 1.055 × 10⁻³⁴). Using h in place of ħ multiplies the answer by 4π² ≈ 39.5.
For an ideal Fermi gas of electrons at T = 0, which statements are correct?
1. Every single-particle state below E_F is occupied
2. The mean energy per electron is 3E_F/5
3. The pressure is non-zero
4. The heat capacity is finite and non-zero at T = 0
Show answer
Answer: A — Every single-particle state below E_F is occupied; B — The mean energy per electron is 3E_F/5; C — The pressure is non-zero
At T = 0 the Fermi–Dirac function is a step at E_F. With g(E) ∝ √E, ⟨E⟩ = ∫E^3/2/∫E^1/2 = 3E_F/5, and the degeneracy pressure (2/5)nE_F is non-zero. The heat capacity is linear in T and vanishes at T = 0, as the third law requires.
A star’s surface behaves as a blackbody at 5800 K. With Wien’s constant b = 2.898 × 10⁻³ m K, the wavelength of peak emission, in nm, to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 500
λ_max = b/T = 2.898 × 10⁻³/5800 = 4.997 × 10⁻⁷ m = 500 nm. The peak of u(ν) in frequency is at a different place (ν_max ≈ 2.82k_BT/h, corresponding to about 880 nm), so converting ν_max to a wavelength is not λ_max.
With σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, the power radiated per unit area by a blackbody at 2000 K, in kW/m², to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 907.2
σT⁴ = 5.67 × 10⁻⁸ × (2000)⁴ = 5.67 × 10⁻⁸ × 1.6 × 10¹³ = 9.072 × 10⁵ W/m² = 907.2 kW/m². Using T³ or forgetting that (2 × 10³)⁴ = 16 × 10¹² are the usual slips.
An ideal uniform Bose gas is held at half its condensation temperature. The fraction of particles in the ground state, to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.646
N₀/N = 1 − (T/T_c)^3/2 = 1 − (0.5)^1.5 = 1 − 0.354 = 0.646. The exponent 3/2 comes from the √E density of states; using 1 − T/T_c gives 0.5 and 1 − (T/T_c)³ gives 0.875.
One mole of an ideal gas expands freely into a vacuum, doubling its volume, in an insulated container. With R = 8.314 J mol⁻¹ K⁻¹, its entropy change in J/K, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 5.76
No work, no heat, so ΔU = 0 and T is unchanged. Entropy is a state function, so compute it along a reversible isotherm between the same states: ΔS = R ln 2 = 8.314 × 0.6931 = 5.76 J/K. Answering 0 because Q = 0 confuses δQ/T for an irreversible path with dS.
A reversible engine works between reservoirs at 500 K and 300 K. Its efficiency is:
1. 0.4
2. 0.6
3. 1.67
4. 0.25
Show answer
Answer: A — 0.4
η = 1 − T_c/T_h = 1 − 300/500 = 0.4. 0.6 is T_c/T_h itself, and 1.67 is T_h/T_c, which exceeds 1 and cannot be an efficiency.
Which of the following are features of a first-order phase transition?
1. A latent heat
2. A discontinuity in the entropy
3. A discontinuity in the volume
4. A correlation length that diverges at the transition
Show answer
Answer: A — A latent heat; B — A discontinuity in the entropy; C — A discontinuity in the volume
At a first-order transition S = −∂G/∂T and V = ∂G/∂P jump, and the jump in S is the latent heat L = TΔS. A diverging correlation length belongs to a continuous transition at a critical point; at a first-order transition the correlation length stays finite.
Along the coexistence curve of two phases, the slope dP/dT equals:
1. L/(T Δv)
2. L Δv/T
3. T Δv/L
4. L/Δv
Show answer
Answer: A — L/(T Δv)
Equal chemical potentials along the line, dμ₁ = dμ₂, give −s₁dT + v₁dP = −s₂dT + v₂dP, so dP/dT = Δs/Δv = L/(TΔv) — the Clausius–Clapeyron equation. For ice melting Δv < 0, which is why the slope is negative and pressure lowers the melting point.
By Gibbs’ phase rule, the number of degrees of freedom at the triple point of a pure substance is:
1. 0
2. 1
3. 2
4. 3
Show answer
Answer: A — 0
F = C − P + 2 = 1 − 3 + 2 = 0: the triple point is a single (P, T), which is why it serves as a fixed point of the temperature scale. A two-phase line has F = 1 and a single phase F = 2.
In the canonical ensemble the mean-square energy fluctuation ⟨ΔE²⟩ equals:
1. k_B T² C_V
2. k_B T C_V
3. C_V/k_B
4. zero, because T is fixed
Show answer
Answer: A — k_B T² C_V
⟨ΔE²⟩ = ∂²ln Z/∂β² = −∂U/∂β = k_BT² ∂U/∂T = k_BT²C_V. Fixing T fixes the mean energy, not the energy of each member, which exchanges heat with the bath. The relative fluctuation falls as 1/√N, which is why ensembles agree for large systems.