Classical ideal gases

One gas, three ways to change it

A sealed sample contains one mole of a hypothetical monatomic gas at 300 K in 24.0 L. Its particles have assigned mass 6.64×10⁻²⁷ kg and negligible interactions. The model predicts an equilibrium pressure of about 103.931 kPa. If the volume doubles while temperature stays fixed, why does pressure halve even though the mean internal energy stays the same? If we instead heat the original volume to 600 K, what changes?

These questions connect the microscopic model to measurable thermodynamics. The model treats translational motion classically, counts identical particles without fictitious names, and assigns equilibrium thermal weights. It predicts pressure, energy, heat capacity, entropy and the tendency to exchange particles from one consistent state-counting expression.

On a first reading, follow the opening case through the partition integral, pressure and energy, entropy changes, chemical potential and mixing. Return to sections marked “Further” for exact finite-particle corrections, alternative energy-shell definitions, internal modes and the limits of the approximation. Every statistical term needed for the main argument is defined here.

Useful published preparation is “Thermodynamic systems, variables, and equilibrium”, “Equations of state and gases”, and “Heat, work, internal energy, and the first law”. The adjacent multiplicity, probability, ensemble and partition-function subjects provide deeper statistical preparation; their general machinery is not required without explanation here. Our assigned gas is a model example, not a measurement of a named real substance.

Name the state and the measure

Let N be the number of particles, ν=N/N_A the amount in moles, V the accessible volume and T the positive absolute temperature. Use n=N/V for number density and c=ν/V for molar density, so n and c have different units. The constant R=N_AkB converts particle formulas to molar formulas: NkB=νR.

A classical microscopic state specifies every particle position qᵢ and momentum pᵢ. A macroscopic state keeps controls such as N, V, T and measured averages. In a canonical preparation the system exchanges energy with a reservoir at T while N and V stay fixed. Its energy fluctuates; U means the ensemble mean internal energy.

Write β=1/(kBT). A microscopic alternative of energy ℋ receives relative weight exp(−βℋ); the partition function Z normalizes the weights using the specified state measure. It is dimensionless. Helmholtz energy F=−kBT ln Z then generates pressure and entropy through derivatives whose other controls are held fixed.

Entropy S describes the equilibrium state under this physical model and normalization. Chemical potential μ is an energy per added particle; its molar counterpart is μ_m=N_Aμ, in J/mol. These two quantities must not be interchanged inside formulas. All logarithms below are natural and have dimensionless arguments.

Ideal and classical are separate conditions

“Ideal” means that interaction contributions between particles can be neglected in the equilibrium properties being calculated. For structureless nonrelativistic particles inside a uniform box, ℋ=Σᵢ|pᵢ|²/(2m). The walls confine the sample and can exchange momentum; an ideal gas is not a gas that fails to exert a force on its container.

“Classical” here means that translational quantum level spacing and exchange corrections are negligible. For a single nondegenerate internal state, a useful test is nλ_T³≪1, where λ_T=h/√(2πmkBT) is the thermal wavelength. All relevant box dimensions must also be large compared with λ_T. Low density can help both classical and ideal limits, but the two tests concern different physics.

Equilibrium weighting is an additional preparation assumption. Exactly noninteracting isolated particles do not automatically randomize one another’s motion or establish a temperature distribution. A reservoir, initial preparation or weak processes that produce equilibrium can justify the statistical description while their contribution to equilibrium energy is negligible. No collision or relaxation time follows from the formulas below.

Integrate one particle’s translations

For one particle, the classical phase-space element d³q d³p has units of action cubed. Divide by h³ to obtain the dimensionless semiclassical measure. With zero potential inside the box, z₁=(1/h³)∫_box d³q∫_ℝ³ d³p exp[−β|p|²/(2m)]. The position integral is V; the momentum integrals factor into three independent Gaussian integrals.

Using ∫_−∞^∞exp[−βp_x²/(2m)]dp_x=√(2πm/β), obtain z₁=V(2πmkBT)^(3/2)/h³=V/λ_T³. The quantity z₁ is a weighted phase-space integral, not the total number of classical points in the box: a continuum contains infinitely many points, and high energies are suppressed by thermal weights.

The factor h fixes the state normalization inherited from the quantum theory. It cancels from the pressure and translational heat capacity, but remains in the entropy and chemical potential relative to the chosen energy convention. A change of coordinate units cannot change a dimensionless physical Z. The equality above is exact for the specified classical continuum integral; its use for a real quantum gas is an approximation.

Count identical particles once

Independent labeled coordinates would give z₁ᴺ. For identical gas particles in the dilute Maxwell–Boltzmann regime, the semiclassical partition function is Z_N=(1/N!h^(3N))∫d^(3N)q d^(3N)p exp(−βℋ)=z₁ᴺ/N!. Artificial names attached to otherwise identical atoms must not generate extra thermodynamic alternatives.

The factor N! does not make particles interact. It corrects the counting representation. It also is not a universal instruction to divide every finite discrete problem by N!: a localized site, a physical species label or a quantum occupation basis has its own state space. Exact Bose and Fermi state counts require their appropriate quantum statistics.

At fixed integer N, ln Z_N=N ln(V/λ_T³)−ln(N!) is a finite canonical expression. Retaining ln(N!) is different from immediately using Stirling’s leading form ln(N!)≈N ln N−N. The latter produces the leading extensive thermodynamic formulas and is excellent for macroscopic N, but finite-number questions can depend on the discarded terms.

Obtain pressure, energy and heat capacity

Since ∂ln Z_N/∂V at fixed T, N is N/V, p=−(∂F/∂V)_(T, N)=kBT(∂ln Z_N/∂V)_(T, N)=NkBT/V. This is the ideal gas equation pV=νRT. Changing the accessible volume changes the number of spatial alternatives and therefore the free-energy derivative that gives pressure.

Temperature enters the translational factor as T^(3N/2). Thus U=−(∂ln Z_N/∂β)_(V, N)=3NkBT/2. The three quadratic momentum components contribute kBT/2 per particle each. At fixed N, increasing the box volume changes pressure but not this translational mean energy.

The total heat capacity C_V=(∂U/∂T)_(V, N)=3NkB/2=3νR/2, in J/K. Enthalpy H=U+pV=5NkBT/2 gives C_p=5NkB/2 at fixed amount, and γ=C_p/C_V=5/3. These values require the structureless monatomic model; ideal molecular gases can have internal energy and additional heat capacity.

For the opening state, U≈3741.508 J and C_V≈12.472 J/K. States A and B have the same T and U, while state C has twice their T and U. State D is a different endpoint obtained by a reversible adiabatic expansion; it is colder and has less internal energy.

Entropy restores extensivity

The canonical entropy is S=kB(ln Z_N+βU). In the leading extensive approximation this becomes the Sackur–Tetrode expression S=NkB[ln(V/(Nλ_T³))+5/2]. It describes a three-dimensional monatomic classical ideal gas with the stated state normalization and one internal ground state.

If N and V are both multiplied by the same factor at fixed T, the density and λ_T remain fixed, and this leading entropy scales in proportion to N. Omitting N! instead leaves an artificial N ln N contribution. A correct pressure equation alone would not reveal this counting error: N! is independent of V and T at fixed N, so pressure and U would still look correct.

For state A the extensive formula gives S≈125.949 J/K. This absolute model value depends on particle mass, the state normalization and omitted internal degeneracies. Entropy differences at fixed species and amount are often easier to test. Extrapolating the classical expression toward T=0 eventually leaves its validity regime; it cannot establish a negative physical entropy or impose the third law on a classical continuum.

Compare expansion and heating

For fixed species and amount, subtracting the entropy formulas gives ΔS=NkB ln(V₂/V₁)+(3/2)NkB ln(T₂/T₁). Doubling V at unchanged T gives ΔS=νR ln 2≈5.763146 J/K for the one-mole sample. Doubling T at unchanged V gives ΔS=(3/2)νR ln 2≈8.644719 J/K.

Use ΔU=Q−W_by, with work done by the gas positive. For a reversible isothermal expansion A→B, W_by=∫p dV=νRT ln 2≈1728.944 J and Q=W_by because ΔU=0. These equalities use mechanical balance and a reversible path, not merely the endpoint ideal gas equation.

An insulated free expansion into an evacuated second half can reach the same equilibrium endpoints A and B. It has Q=0 and W_by=0, hence unchanged U and T in this model, while the gas entropy increases by the same 5.763146 J/K. Entropy change is a state difference; Q/T represents its transfer only under the appropriate boundary and reversibility conditions.

For a reversible adiabatic expansion of this monatomic sample, ΔS=0 instead gives TV^(2/3)=constant. Doubling V leads to T_D≈188.988 K and p_D≈32.736 kPa, with work drawn from the gas’s internal energy. Adiabatic does not by itself mean isothermal or reversible; the process and first-law subjects develop those distinctions.

Chemical potential tracks density

Differentiate the leading F at fixed T, V to obtain μ=(∂F/∂N)_(T, V)=kBT ln(nλ_T³). The corresponding molar expression is μ_m=RT ln(cN_Aλ_T³). Adding particles at fixed T, V increases density and raises μ. Doubling V at fixed N, T lowers μ by kBT ln 2.

In the nondegenerate monatomic regime nλ_T³≪1, this μ is negative relative to our translational energy zero. It is not the positive kinetic energy 3kBT/2 of an incoming atom. Thermodynamically μ=(∂U/∂N)_(S, V) also holds: fixed entropy is a different constraint from fixed temperature, so the two verbal descriptions must keep their held variables.

At T=300 K, state A has μ/(kBT)≈−12.648148 and state B≈−13.341295. The pressure law is independent of m at fixed N, V, T, while λ_T and μ depend on m. A negative reference-relative chemical potential is therefore not evidence of an attractive interaction or a bound state.

If every particle energy is shifted by a constant ε₀ independent of T and V, then U and F shift by Nε₀ and μ by ε₀, while p and S do not change. For exchanging or reacting species, use consistent physical reference energies across the comparison. At finite N an exact addition difference is kBT ln[(N+1)λ_T³/V]; it approaches the continuous thermodynamic derivative as N grows.

Each species has its own count

For an ideal mixture of genuinely distinct species with fixed Nᵢ, the partition function is Z=∏ᵢ[(V/λᵢ³)^(Nᵢ)/Nᵢ!]. Permutations are removed within each species, not across different species. Each species explores the same final volume V, with its own mass and thermal wavelength.

The total pressure is p=ΣᵢNᵢkBT/V. Define xᵢ=Nᵢ/N_total and partial pressure pᵢ=xᵢp; in an ideal gas mixture pᵢ is also the pressure species i alone would exert at the same T and V. That latter physical interpretation generally fails for a real interacting mixture, even when pᵢ=xᵢp is retained as a composition convention.

In the thermodynamic limit, the particle chemical potential is μᵢ=kBT ln[(Nᵢ/V)λᵢ³]. In molar notation it can be written μ_m, i=μ°_m, i(T)+RT ln(pᵢ/p°), with a stated reference pressure p° and standard chemical potential at the same T. Equivalently, at fixed total p, μ_m, i=μ_m, i^pure(T, p)+RT ln xᵢ. The logarithm of a pressure ratio is dimensionless.

For our one-mole mixture at 300 K in 24 L, species A has mole fraction 0.25 and assigned mass 6.64×10⁻²⁷ kg; species B has mole fraction 0.75 and ten times that mass. Their partial pressures add to 103.931 kPa. Different masses affect their entropy and reference-relative chemical potentials, while the total ideal pressure depends on total amount and temperature.

Mixing depends on physical species

Initially separate the two pure ideal species at common T and common p, with volumes Vᵢ=xᵢV. Remove the partition in a rigid insulated container. Each species expands from Vᵢ into V, so ΔS_mix=ΣᵢNᵢkB ln(V/Vᵢ)=−kBΣᵢNᵢ ln xᵢ. With fixed species amounts, the ideal translational energy is unchanged and the final equilibrium temperature remains T.

For x_A=0.25 and x_B=0.75, ΔS_mix≈4.675515 J/K. A reversible isothermal separation that restores the initial pure gases would require at least TΔS_mix≈1402.654 J of work under the corresponding ideal constraints. Selective membranes illustrate the thermodynamic distinction between species; an actual separation device can require additional work and dissipation.

This is an entropy of mixing for distinct physical components, not a reward for assigning red and blue names to identical atoms. For the same species already at equal temperature and density, removing a wall produces no extensive thermodynamic mixing term. The N! bookkeeping and the choice of independent composition variables must reflect that identity.

The binary function −x ln x−(1−x)ln(1−x) reaches ln 2 at equal fractions and tends to zero at either pure endpoint. An absent species has zero amount and contributes the limiting value 0 ln 0=0 to this entropy sum. Its ideal-mixture chemical potential instead tends to minus infinity as its fraction tends to zero; do not assign it a finite endpoint value.

Further: release an exact number constraint

A finite system exposes what the extensive mixing statement leaves out. Put exactly N identical particles in each half of two equal boxes at the same T. Initially the left particle number is constrained to N. Removing the partition keeps the total 2N fixed but permits the left number to fluctuate. This change of an exact constraint is physically different from relabeling particles. The volumes are chosen large enough to preserve the stated classical dilute and continuum conditions.

Using the finite canonical factorials gives Z_final/Z_initial=2^(2N)(N!)²/(2N)!. The energy distribution is unchanged, so ΔS/kB=ln(Z_final/Z_initial). This positive correction grows approximately as ½ ln(πN); divided by the total particle number it tends to zero. It is subextensive, whereas mixing two different equal components gives an entropy proportional to 2N.

For N=10 per half, the released equilibrium spatial distribution is P(N_left=k)=C(20, k)/2²⁰ for k=0,…,20. Its central probability is 46189/262144≈0.176197. The exact constraint-release entropy is −kB ln P(N_left=10)≈1.736152kB. The 21 number outcomes and their exact probabilities provide a complete equivalent to the prospective discrete plot.

There is no contradiction with zero extensive mixing entropy for the same gas. Finite factorials retain information about exact local number preparation that ordinary macroscopic densities omit. The distribution states an equilibrium spatial measure, not how fast particles cross, how many crossings occur, or a universal trajectory after wall removal.

Further: check two different dilute limits

The quantum phase-space parameter η=nλ_T³ measures translational degeneracy for the one-state model. In state A, λ_T≈50.406 pm and η≈3.21351×10⁻⁶, which supports the nondegenerate approximation. Cooling increases λ_T and η; a noninteracting quantum gas can cease to obey the classical ideal equation even when interaction energy remains negligible.

Interaction effects need a separate test. In a molar virial expansion, p/(cRT)=1+B₂, m(T)c+B₃, m(T)c²+…, with B₂, m in m³/mol. Small |B₂, m c| is evidence that the first pressure correction is small, not a guarantee that all higher terms, molecular association or long-range effects are negligible. The number-density coefficient in an expansion using n has different units.

For an assigned illustrative B₂, m=4.00×10⁻⁵ m³/mol, the first relative pressure correction at c=40 mol/m³ is 0.0016. This is not measured data for our hypothetical gas, nor a controlled real-gas prediction at arbitrary density. A small translational degeneracy parameter cannot certify this interaction condition, and a small interaction coefficient cannot certify classical statistics.

The diagnostic comparison below uses separate dimensionless indicators, not a universal phase diagram. There is no exact boundary at which an ideal gas suddenly becomes a liquid or at which one approximate cutoff applies to every accuracy target. Near degeneracy, strong interactions, phase changes or confinement, select a more suitable physical model.

Further: internal modes change energy

A gas can be translationally classical and nearly noninteracting while its rotational, vibrational or electronic excitations remain quantum. If internal and translational energies separate and the internal one-particle partition function q_int(T) is independent of V, then z₁=Vq_int(T)/λ_T³ and Z_N=z₁ᴺ/N! in the corresponding dilute regime.

Pressure remains NkBT/V, while U=3NkBT/2+NkBT²d ln q_int/dT. The second term changes heat capacity. Thus the equation of state alone does not imply C_V=3NkB/2. At fixed composition a temperature-dependent ideal-gas U(T) still gives C_p−C_V=NkB, but γ need not be 5/3 or constant.

For a temperature-independent internal degeneracy g, q_int=g adds NkB ln g to entropy and replaces μ by kBT ln(nλ_T³/g), without adding an excitation energy. For thermally populated internal levels, keep their actual energies and weights. Classical exchange should be negligible in each relevant species/internal population; a total density criterion must account for the available internal states.

Molecular heat-capacity curves and quantum ideal-gas occupation laws belong to their adjacent subjects. Here the extension identifies precisely which part of the monatomic derivation changes and which volume derivative survives.

Further: shells and canonical averages

The same ideal Hamiltonian can be studied with fixed energy rather than a heat reservoir. Its dimensionless cumulative phase volume below E>0 is Φ(E, V, N)=Vᴺ(2πmE)^(3N/2)/[N!h^(3N)Γ(3N/2+1)], where Γ extends the factorial. This counts a volume below E; it is not a canonical Z or a finite-width energy-shell count.

A density of states ρ(E)=∂Φ/∂E has units J⁻¹. A thin positive interval δE gives a dimensionless shell count approximately ρ(E)δE when the density changes little across it. The shell convention and its width must be stated before taking a logarithm. A surface at exactly ℋ=E has zero ordinary 6N-dimensional volume.

For finite N, S_G=kB lnΦ gives 1/T_G=(3N/2)kB/E. A thin-shell entropy kB ln[ρ(E)δE] with fixed δE instead gives approximately 1/T_shell=(3N/2−1)kB/E. These finite-size definitions differ. At large N their leading extensive results agree with the canonical entropy at U≈3NkBT/2 under the stated regular limit.

In the canonical sample the microscopic energy ℋ fluctuates about its ensemble mean U, with variance Var(ℋ)=kBT²C_V=(3N/2)(kBT)². Its standard deviation divided by U is √[2/(3N)]. A narrow relative spread does not make every microscopic energy identical. The ensemble subject treats preparation and equivalence more generally; this explicit ideal-gas calculation shows where the finite corrections enter.

Further: change dimension or confinement

In a flat d-dimensional box with nonrelativistic quadratic momentum energy, z₁=V_d/λ_Tᵈ and U=dNkBT/2, where V_d has units of length to power d. The generalized pressure conjugate to V_d obeys p_dV_d=NkBT; in two dimensions it is a force per length, not an ordinary three-dimensional pressure.

An external potential changes the position integral to ∫d³q exp[−βV_ext(q)]. The spatial density is then nonuniform, and simply inserting the geometrical box volume into V/λ_T³ can fail. Wall-level quantization, strong confinement, relativistic energies and adsorption likewise require different integrals or spectra.

Ideal gas results are a baseline for dilute-gas thermometry, entropy and chemical-potential estimates, and checks on simulation free energies. Use the baseline to identify the missing effect, rather than fitting every deviation by changing a constant while retaining incompatible assumptions.

Carry the model to the next question

The opening pressure change comes from the volume dependence of spatial alternatives, while the translational energy comes from three quadratic momentum components. The state measure and identical-particle factor make entropy and particle-exchange predictions consistent with that same equilibrium model.

Before applying the formulas, specify species, amount, accessible volume, temperature, energy zero, internal states and ensemble preparation. Check translational degeneracy, interactions, confinement and the needed accuracy. Keep finite exact-number questions separate from leading extensive thermodynamics.

“Kinetic theory and molecular velocity distributions” develops molecular speeds, momentum transfer and dynamical interpretation. “Bose–Einstein and Fermi–Dirac statistics” develops degenerate ideal gases. “Fluctuations and response functions”, real-gas equations of state and chemical-potential/phase-equilibrium subjects extend the corresponding limits. These connections guide the next question without replacing the classical ideal gas model developed here.

Worked examples

Try it yourself

1. Explain the difference between n=N/V and c=ν/V. Which multiplies λ_T³?

   Show the answer and reasoning

   n has units m⁻³ andc has units mol/m³; nλ_T³ is dimensionless.

   N=N_Aν, so using molar density directly in nλ_T³ would omit a factor N_A. Use consistent particle or molar formulas throughout.

2. Why is the raw integral ∫d³q d³p not a valid dimensionless z₁?

   Show the answer and reasoning

   Its units are action cubed; divide by h³ before using it inside a logarithm.

   A numerical coordinate unit cannot define an absolute state count. The semiclassical normalization supplies the physical action scale and remains relevant to S and μ.

3. At the original 24 L, heat from 300 K to 600 K. Find p_C/p_A and U_C/U_A.

   Show the answer and reasoning

   Both ratios are 2.

   At fixed N, V, pressure is proportional to T; the monatomic mean translational energy also scales with T. The entropy change is 3νR ln 2/2, not zero.

4. A sample doubles both T and V at fixed N. Does unchanged pressure mean unchanged U?

   Show the answer and reasoning

   No. Pressure stays the same and U doubles.

   The equation of state depends on T/V, whereas translational U depends on T alone. Pressure is insufficient to specify a complete thermal state without other controls.

5. Why does omitting N! leave p and U correct at fixed N but corrupt extensive entropy?

   Show the answer and reasoning

   ln(N!) has no T or V derivative at fixed N, while S contains ln Z itself.

   The counting error enters the entropy and particle-number dependence even when the volume and temperature derivatives appear reasonable. Passing the ideal equation of state is not a complete statistical validation.

6. Find C_V and C_p for one mole of the stated gas.

   Show the answer and reasoning

   The molar values are C_V,m=3R/2≈12.471694 J/(mol K) and C_p,m=5R/2≈20.786157 J/(mol K). The one-mole sample therefore has C_V≈12.471694 J/K and C_p≈20.786157 J/K.

   Total and molar heat capacities have different units. The values assume three active translational quadratic components and no internal energy contribution.

7. The insulated free expansion has Q=0. Why can its gas entropy still increase?

   Show the answer and reasoning

   It is irreversible; entropy is produced within the gas rather than transferred as heat.

   The endpoint entropy depends on V and T. Setting ΔS=Q/T without checking reversibility and boundary temperatures confuses a state change with one transfer term.

8. What additional conditions make TV^(2/3)=constant valid here?

   Show the answer and reasoning

   Fixed amount, this monatomic ideal model, and a reversible adiabatic path between equilibrium states.

   Adiabatic alone sets no heat transfer. An irreversible adiabatic path can produce entropy and need not obey the reversible power law.

9. Does the negative μ of state A prove that the atoms attract or are bound?

   Show the answer and reasoning

   No; μ is negative relative to the stated translational energy zero in a dilute classical preparation.

   It is a constrained free-energy derivative, not a binding potential or the positive mean kinetic energy of an atom. A constant particle energy shift changes μ without changing p or S.

10. Compare the finite addition difference F(N+1)−F(N) with the thermodynamic μ.

    Show the answer and reasoning

    The difference is kBT ln(1+1/N), tending to 0 for large N.

    The exact factorial distinguishes N+1 from N. The continuous derivative uses the leading extensive form and should not be claimed as an exact finite integer addition law.

11. At mole fraction 0.25, what is p_A/p, and does p_A equal the pure-A pressure at the same T, V for any gas mixture?

    Show the answer and reasoning

    p_A/p=0.25; the pure-component interpretation requires an ideal gas mixture.

    The definition p_A=x_Ap is a composition convention. Interactions in real mixtures can change the pressure relative to what the isolated pure component would exert.

12. What is the molar chemical-potential change from pure A at the same T, p to mole fraction 0.25?

    Show the answer and reasoning

    RT ln 0.25=−2RT ln 2≈−3457.888 J/mol at 300 K.

    The standard or pure reference must remain at the same temperature. Multiplying by R gives a molar chemical potential; the per-particle form uses kB instead.

13. For a one-mole equal binary mixture initially separated at common T, p, find ΔS_mix.

    Show the answer and reasoning

    R ln 2≈5.763146 J/K.

    The two distinct components each expand by a factor 2. Summing their contributions gives the maximum binary mixing entropy at fixed total amount.

14. As x_A tends to 0, what happens to its mixing-entropy term and ideal chemical potential?

    Show the answer and reasoning

    −x_A ln x_A tends to 0, while μ_A tends to −∞ at fixed T, total p and reference.

    A vanishing amount can have a finite limiting contribution to an extensive sum even though its intensive ideal chemical potential diverges. The pure endpoint is not assigned a finite μ for an absent species.

15. Would recoloring identical atom labels create the distinct-species mixing term?

    Show the answer and reasoning

    No, unless the colors represent physically different accessible components or constraints.

    A bookkeeping name is not a new physical species. Separate thermodynamic composition variables and possible separation operations must correspond to physical distinctions.

16. Why can the 10+10 identical-particle wall example have a small positive ΔS without an extensive mixing term?

    Show the answer and reasoning

    Removing the wall releases an exact left-number constraint; its correction is subextensive.

    The final state permits number fluctuations that the initial preparation excluded. Keeping exact factorials retains this information, while the leading macroscopic density description discards it.

17. A model has nλ_T³=1 and |B₂, m c|=0.001. Which approximation is unsupported?

    Show the answer and reasoning

    Classical nondegenerate translational statistics.

    A small first interaction indicator does not make exchange negligible. An ideal quantum gas can differ from the classical equation even without a substantial interaction energy.

18. A molar density is 1000 mol/m³ and the assigned B₂, m is 4.00×10⁻⁵ m³/mol. Find B₂, m c.

    Show the answer and reasoning

    0.040, corresponding to a 4% first pressure correction.

    The units cancel. The result concerns the first interaction term only and does not bound neglected higher coefficients or establish applicability of the classical limit.

19. Does pV=NkBT require C_V=3NkB/2 for a molecular gas?

    Show the answer and reasoning

    No. Internal partition factors can add temperature-dependent energy without changing the ideal volume derivative.

    Rotational, vibrational or electronic levels may contribute to U(T). The relation C_p−C_V=NkB can remain valid at fixed composition even when γ differs from 5/3.

20. What distinguishes a cumulative Φ, density ρ and thin shell count?

    Show the answer and reasoning

    Φ is dimensionless below E, ρ=∂Φ/∂E has inverse-energy units, and ρδE is approximately a dimensionless thin-shell count.

    Their entropy definitions have finite-size differences. State the shell width and convention rather than swapping counts mid-derivation.

21. In a flat two-dimensional box, what is U and what are the units of its conjugate pressure?

    Show the answer and reasoning

    U=NkBT; the area-conjugate pressure has units J/m²=N/m.

    There are two quadratic momentum components per particle. The geometrical control is area, so its mechanical conjugate is force per length rather than 3D pressure N/m².

22. Can the equilibrium partition integral predict the gas’s collision time or diffusion rate?

    Show the answer and reasoning

    No. Dynamical laws and interaction or reservoir mechanisms are additional information.

    The equilibrium integral fixes normalized thermal weights and state functions. The kinetic-theory and transport subjects supply the motion and rate models needed for time predictions.

What to carry forward

• The classical ideal gas model combines negligible interaction effects, negligible translational exchange corrections and a stated equilibrium preparation; these are separate assumptions.
• The dimensionless measure and identical-particle factor give Z_N=(V/λ_T³)ᴺ/N!, with pressure and translational energy arising from different derivatives.
• For the structureless three-dimensional model, pV=NkBT, U=3NkBT/2 and C_V=3NkB/2; internal modes can change energy and heat capacity while preserving the ideal equation of state.
• Entropy and chemical potential retain the state normalization, energy convention and particle identity; leading extensive formulas should not erase finite-number constraints.
• Ideal-mixture partial pressures and entropy of mixing follow species-resolved counting. Fictitious labels on identical particles do not create an extensive mixing entropy.
• The equilibrium formulas supply state functions and fluctuation scales, not a collision clock or a guarantee that all real gases remain classical and ideal.