In the semi-classical treatment of the ideal gas, we write the partition function for the system as $$Z = \frac{Z(1)^N}{N!}$$ where $Z(1)$ is the single particle partition function and $N$ is the number of particles. It is semi-classical in the sense that we consider the indistinguishability of the particles, so we divide by $N!$.

The resulting expression for the entropy of the system is $$S = Nk \left(\ln \left[\left(\frac{V}{N}\right) \left(\frac{2\pi mkT}{h^2}\right)^{3/2} \right] + \frac{5}{2}\right)$$

Now consider a fully classical analysis. There $Z(1) = \sum_{E} \exp (-\frac{E}{kT})$, where $E = p^2/2m$ (assuming no interaction potentials). The problem can be mapped to an integral over the phase space of the Hamiltonian to give $$Z(1) \rightarrow \int \exp \left(-\frac{1}{2mkT} (p_x^2 + p_y^2 + p_z^2) \right)\text{d}^3 \underline{p} \,\text{d}^3 \underline{x}$$ This can then be rewritten like $$ \int \exp \left(-\frac{p_x^2}{2mkT}\right) \text{d}p_x \int \exp \left(-\frac{p_y^2}{2mkT}\right) \text{d}p_y \int \exp \left(-\frac{p_z^2}{2mkT} \right) \text{d}p_z \cdot V$$ where $V$ is the volume of the container. Those are Gaussian integrals and so evaluation is immediate. The result is that $Z(1) = (2\pi mkT)^{3/2} V$ The corresponding entropy can be calculated and the result is that $$S = Nk \left(\frac{3}{2} + \ln\left(\frac{(2\pi mkT)^{3/2}}{V}\right)\right)$$.

My question is: What is the significance of the factors $5/2$ in the semi-classical treatment and the factor $3/2$ in the classical treatment and why are they different? They look like the number of degrees of freedom a monatomic and diatomic molecule would have at room temperature, but I think this is a coincidence.

  • $\begingroup$ Just for your information, the first formula for the entropy is actually an approximation, even classically, because it employs Stirling's approximation of the gamma function. $\endgroup$
    – JamalS
    Commented May 14, 2014 at 13:48
  • $\begingroup$ Hi JamalS. Yes, I am aware of that - also dividing by $N!$ alone is an approximation. The correction only accounts for the case where the particles are all in different states. $\endgroup$
    – CAF
    Commented May 14, 2014 at 13:53
  • $\begingroup$ Why did you exclude the factor of $1/N!$ in the "classical" treatment? Identical particles can (and should) still be treated as identical particles in classical statistical mechanics in this way. $\endgroup$ Commented May 14, 2014 at 22:06
  • $\begingroup$ @joshphysics, the fact the particles are identical means they have the same physical properties, like mass or charge. Identity of particles by itself does not imply that the measure of states in phase space should be reduced by the corresponding number of their permutations. The reduction can be done on the grounds of connecting the statistical formula to conventional thermodynamic entropy, but it is not necessary. arxiv.org/abs/1012.4111 $\endgroup$ Commented Jun 15, 2015 at 0:47
  • $\begingroup$ @JánLalinský Interesting. I'll take a look at the reference. Thanks. $\endgroup$ Commented Jun 15, 2015 at 1:56

2 Answers 2


Why in the heck are you worrying about the difference between 3/2 and 5/2? Your formulae differ not just by 3/2 and 5/2 but also by $\ln(N)$. For the thermodynamic limit, $\ln(N)$ is infinity, and so a lot bigger than 1.

The number 3/2 has in this context no meaning, except that it is 5/2-1. The number 5/2 gives the correct entropy correct for the ideal, non-relativistic gas, composed of indistinguishable particles with a single spin state and no degrees of freedom other than the position and momentum. However, you should understand that the entropy is defined up to a constant only by the third law of thermodynamics. This third law, in turn, would not be valid if there were not an underlying quantum mechanics. So, this is an intrinsically QUANTUM notion, and can not really be understood without quantum mechanics.


Semi-classically, from Zemansky: $$ S = k \ln \Omega \quad , \quad \Omega = \prod_i \left( \frac{g_i^{N_i}}{N_i!} \right) \quad , \quad N_i = \frac{N}{Z} g_i e^{-\epsilon_i / kT} $$

one finds, applying Stirling's approximation: $\ln(x!) \approx x \ln x - x $:

$$ \ln \Omega = \sum_i N_i \ln \frac{g_i}{N_i} + N = \sum_i N_i \left( \ln \frac{Z}{N} + \frac{\epsilon_i}{kT} \right) + N = N \left(\ln \frac{Z}{N} + \frac{U}{NkT} + 1 \right) $$

$$ S = Nk \left( \ln \frac{Z}{N} + \frac{U}{NkT} + 1 \right) $$

The $U$ term for a monatomic gas contributes $3/2$.

[By the way, your classical $Z$ needs some kind of constant with the dimensions of momentum x distance in the denominator to make the result unit-less. Huang calls it "h"... ]


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