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The derivation of entropy of perfect gas is given as follows in my textbook:

Suppose we have a vessel containing a volume $V$ of a perfect gas. The gas has $N$ molecules and is in a state of equilibrium at a temperature $T$. As the gas is in equilibrium it is in the most probable macrostate.

Let the $N$ molecules be divided into $k$ compartments (energy intervals) numbered $(1,2,...,k)$ having $g_1,g_2,..,g_k$ equal sized cells and $n_1,n_2,...n_k$ molecules respectively. As a perfect gas obeys Maxwell Boltzmann statistics, the thermodynamic probability of the macrostate ($n_1,n_2,...,n_k$) is given by

$$W(n_1,n_2,..,n_k)=\frac{N!(g_1)^{n_1}(g_2)^{n_2}...(g_k)^{n_k}}{n_1!n_2!...n_k!}$$

or $$W=N!\prod_{i=1}^{i=k}\frac{(g_i)^{n_i}}{n_i}$$

where $$n_i=g_ie^{-\alpha}e^{-\beta u_i}=g_ie^{-\alpha}e^{-u_i/kT}$$

Taking natural logarithms on both sides we have

$$\ln(W)=\ln(N!)+\sum_{i=1}^{i=k}n_i\ln(g_i)-\sum_{i=1}^{i=k}\ln(n_i!)$$

Now using $\ln(N!)=N\ln(N)-N$ for $N\to\infty$;

$$\ln(W)=N\ln(N)-\sum_{i=1}^{i=k}n_i\ln(\frac{n_i}{g_i})$$

Substituting $n_i=g_ie^{-\alpha}e^{-\beta > u_i}=g_ie^{-\alpha}e^{-u_i/kT}$

$$\ln(W)=N\ln(N)-N\ln(e^{-a})+\sum_{i=1}^{i=k}\frac{n_iu_i}{kT}\ln(e)$$

But $\sum_{i=1}^{i=k}n_iu_i=U$ where $U$ is total energy of system

$$\therefore > \ln(W)=N\ln(N)-N\ln(e^{-a})+U/kT=N\ln(N/e^{-a})+U/kT=N\ln(N)e^{a}+U/kT$$

For a system obeying Maxwell Boltzmann statistics $$e^{-\alpha}=\frac{Nh^3}{V}(\frac{\beta}{2\pi m})^{3/2} \implies Ne^{\alpha}=V/h^3(2\pi mkT)^{3/2}$$

Substituting the above value of $Ne^{\alpha}$ in equation (iii) we have $$\ln(W)=N\ln[\frac{V}{h^3}(2\pi mkT)^{3/2}]+\frac{U}{kT}$$

For mono-atomic gas $U=\frac{3}{2}NkT$.

Now, using the statistical definition of entropy:

$$S=k\ln(W)=\frac{3}{2}+Nk\ln[\frac{V}{h^3}(2\pi mkT)^{3/2}]$$


Now my question is:

Why is $e^{-\alpha}=\frac{Nh^3}{V}(\frac{\beta}{2\pi m})^{3/2}$ true for Maxwell Boltzmann statistics and how is the equation derived?

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Background: partition functions and entropy

So it's usually something of a mistake to be mixing $W$ and $e^{-U/(k_\text B T)}$ in one statistical mechanics expression, as the former generally refers to the microcanonical ensemble and the latter refers to the canonical ensemble. However it is worth exploring how they are related, and that is by this formula that $\sigma = S/k_\text B = -\sum_i p_i\log p_i,$ which is seen to be trivial in the case where $p_i = 1/W$ for a bunch of equivalent macrostates, but which can be extended to the canonical ensemble where $p_i = e^{-U_i/\tau}/Z$ where $Z = \sum_i e^{-U_i/\tau}$ is our normalizing constant and $\tau = k_\text B T$ is the temperature in units of energy. This allows us to find the entropy in the canonical ensemble as,$$\sigma = - \sum_{\text{all }i} \frac{e^{-U_i/\tau}}Z \left[-\frac{U_i}{\tau} - \log Z\right] = \frac1\tau\langle U\rangle + \log Z. $$ However another well-known relation is that ${\partial \over\partial\tau}\log Z = \frac1Z{\partial Z\over\partial\tau} = \sum_i p_i~\frac{U_i}{\tau^2} = \langle U\rangle/\tau^2$ and thus we can rewrite the above as $\tau \frac{\partial}{\partial \tau} \log Z + \log Z,$ which turns out to just be a derivative-of-a-product rule, $$\sigma = \frac\partial{\partial\tau}(\tau \log Z).$$

The quantum results of a particle in a box

The Schrödinger equation in free space (zero potential $U=0$) basically just says that you have plane waves $e^{ikx}$ with all $k$ allowed, with energy $E=\hbar^2 k^2/(2m).$ This corresponds to our classical equation $E=p^2/(2m)$ with the de Broglie momentum $p=\hbar k,$ directly. But to get the bound states for the particle in a box $0<x<L$, we have a superposition of forward and backward plane waves $e^{ikx}-e^{-ikx} = 2i~\sin(kx)$ to get the wavefunction to be zero at $x=0$, and in addition we find that $kL=n\pi$ for integer $n$ to get the wavefunction to be zero at $x=L.$ Conclusion: momentum in the 1D box actually comes in lumps of $\hbar\pi/L;$ but energy is still $p^2/2m$ where $p=n\hbar\pi/L$. In the 3D box we use a nifty trick for partial differential equations called "separation of variables" and we find the same basic expression.

Now let's compute this partition function $Z$ for a single particle in a 3D quantum box.$$Z = \sum_{n_x=1}^\infty\sum_{n_y=1}^\infty\sum_{n_z=1}^\infty \exp\left({-\frac1{2m\tau} (\hbar\pi/L)^2 (n_x^2 + n_y^2 + n_z^2)}\right).$$ The first thing that stands out is that this exponential turns into 3 independent sums which are all the same: $Z = Z_1^3.$. The second thing that stands out is that this sum is quadratic in $n$ and would therefore look like half of the Gaussian integral if we fudge the summation sign into an integral:$$Z_1 \approx \int_0^\infty dn~\exp\left(-\frac{\hbar^2\pi^2}{2m\tau L^2}~n^2\right) = \frac{\sqrt\pi}{2} \cdot \sqrt{\frac{2m\tau L^2}{\hbar^2\pi^2}}=L~\sqrt{m\tau\over2\pi\hbar^2}.$$ Recognizing that $Z_1$ is dimensionless the rest of that square root is actually a number that we usually abbreviate as the "thermal wavelength,"$$\lambda = \frac h{\sqrt{2\pi~m~\tau}};~~~Z_1 = \frac{L}{\lambda}.$$ This is the best answer that I can give to your question about where your term comes from: it is $N \lambda^3/V.$ The particle-in-a-box Hamiltonian when it goes through this procedure to create a $Z$ turns out to be mathematically equivalent to parceling up our length scales into chunks of size $\lambda.$

The Sackur-Tetrode equation

Let's quickly complete the above argument. As we've discussed for one particle the partition function is just $Z = Z_1^3 = V/\lambda^3.$ What we haven't discussed is $N$ identical particles, which turns out to be $\bar Z = Z^N/(N!).$ We therefore find the entropy as $$\sigma = \frac{\partial}{\partial \tau}(\tau \log \bar Z) = \frac{\partial}{\partial \tau} \left(\tau \cdot \left(N \log\left(\frac{V}{\lambda^3}\right) - N \log N + N\right)\right).$$ Let me note that everything is proportional to $N$ which commutes through the derivative; the per-particle entropy is by the product rule merely $$\frac{\sigma}{N} = 1\cdot\left(\log\left(\frac{V}{\lambda^3}\right) - \log N + 1\right) + \tau\cdot\left(\left(\frac{V}{\lambda^3}\right)^{-1}~\frac{-3V}{\lambda^4}~\frac{\partial\lambda}{\partial\tau}\right).$$ So we see a term $(-3/\lambda)(\partial\lambda/\partial\tau)$ which works out to $3/(2\tau)$ and the per-particle entropy is just $$\frac{\sigma}{N} = \log\left(\frac{V}{\lambda^3}\right) - \log N + \frac{5}{2}.$$This is known as the Sackur-Tetrode equation.

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