Question

Write an expression for the partition function for an ideal gas made up of N
indistinguishable particles. Using this expression, obtain Sackur-Tetrode equation.

Answer 1

Question #61148, Physics / Molecular Physics | Thermodynamics

Write an expression for the partition function for an ideal gas made up of $N$ indistinguishable particles. Using this expression, obtain Sackur-Tetrode equation.

Solution

Let us consider first the ideal gas in microcanonical ensemble. In the microcanonical ensemble for $N$ non-interacting point particles of mass $M$ confined in the volume $V$ with total energy in $\delta E$ at $E$ we must calculate

$$\Delta \Gamma = \int d q _ {1} \dots d q _ {3 N} \int d p _ {1} \dots d p _ {3 N} = V ^ {N} \int d p _ {1} \dots d p _ {3 N}$$

where the momentum space integral is to be evaluated subject to the constraint that

$$E - \delta E \leq \frac {1}{2 M} \sum_ {i = 1} ^ {3 N} p _ {i} ^ {2} \leq E$$

by the construction of the ensemble. The accessible volume in momentum space is that of a shell of thickness $(\delta E)(M / 2E)^{1 / 2}$ on a hypersphere of radius $(2ME)^{1 / 2}$. If the result were sensitive to the value of $\delta E$ employed, we would have difficulty in deciding on a value $\delta E$.

Fortunately we can prove that for a system of large numbers of particles the value of $\ln \Delta \Gamma$ is not sensitive to the value of $\delta E$, we may even replace $\delta E$ by entire range from 0 to $E$.

The proof now follows. We write

$$V (R) = C R ^ {\nu}$$

for the volume of a $\nu$-dimensional sphere of radius $R$. The volume of a shell of thickness $s$ at the surface of this hypersphere is

$$V _ {s} = V (R) - V (R - s) = C \left[ R ^ {\nu} - \left(R - s\right) ^ {\nu} \right] = C R ^ {\nu} \left[ 1 - \left(1 - \frac {s}{R}\right) ^ {\nu} \right]$$

or, by the definition of the exponential function,

$$V _ {s} \cong C R ^ {\nu} \left[ 1 - e ^ {- s \nu / R} \right]$$

Therefore if $v$ is large enough so that $s^{\nu} >> R$, $V_s$ is practically the volume $V(R)$ of the whole sphere. If $v \sim 10^{23}$ as for a macroscopic system, the requirement $s >> R / 10^{23}$ may be satisfied without any practical imprecision in the specification of the energy of the microcanonical ensemble.

We can replace now the constraint (6.2) by the relaxed condition

$$\mathbf {O} \leq \frac {1}{2 M} \sum_ {i = 1} ^ {3 N} p _ {i} ^ {2} \leq E$$

because for any reasonable (not too thin) shell the volume of the shell is essentially equal to the volume of the entire hypersphere. In other words, we want to evaluate the volume of the $3\nu$-dimensional sphere of radius $R = (2ME)^{1/2}$. We can evaluate the volume $V_{\nu}$ of the hypersphere by the following argument: Consider the integral

$$\mathbf {G} = \int_ {- \infty} ^ {\infty} e ^ {- \left(x _ {1} ^ {2} + x _ {2} ^ {2} + \dots + x _ {r} ^ {2}\right)} d x _ {1} d x _ {2} \dots d x _ {r} = \left\{\int_ {- \infty} ^ {\infty} e ^ {- x ^ {2}} d x \right\} ^ {r} = \pi^ {r / 2}$$

We may also write

$$\mathbf {G} = \int_ {0} ^ {\infty} e ^ {- r ^ {2}} r ^ {r - 1} S _ {r} d r = \frac {1}{2} S _ {r} \int_ {0} ^ {\infty} e ^ {- t} t ^ {(r - 2) / 2} d t = \frac {1}{2} S _ {r} \left(\frac {r}{2} - 1\right)!$$

where $R^{r-1}S_r$ denotes the surface area of the $r$-dimensional sphere. On comparison of the two results (6.7) and (6.8) we find

$$S _ {r} = \frac {2 \pi^ {r / 2}}{\left(\frac {r}{2} - 1\right)!}$$ $$\mathbf {G} = \pi^ {\nu / 2}$$ $$\mathbf {G} = \frac {1}{2} S _ {\nu} \left(\frac {\nu}{2} - 1\right)!$$

so that the volume of the sphere is

$$V _ {v} = \int_ {0} ^ {R} S _ {v} R ^ {v - 1} d R = \frac {\pi^ {v / 2}}{(\nu / 2) !} R ^ {v}$$

In the 3-dimensional case, $V_{3} = 4\pi R^{3} / 3$ and surface are $a_3 = 4\pi R^2$. In the two-dimensional case, we obtain $V_{2} = \pi R^{2}$ and surface are $a_2 = 2\pi R$ and in the one-dimensional case $V_{1} = 2R$ and surface are $a_1 = 2$.

$$\Delta \Gamma = V ^ {N} V _ {v}$$

and using the Stirling approximation to evaluate factorial

$$\sigma = \ln \Delta \Gamma = N \ln \left[ V \pi^ {3 / 2} \left(2 M E\right) ^ {3 / 2} \right] - \frac {3 N}{2} \ln \frac {3 N}{2} + \frac {3 N}{2}$$

where in the expression for $V_{\nu}$ we have to put $\nu = 3N$ and $R = (2ME)^{1 / 2}$

It turns out that if the $N$ particles are identical we must not count as different conditions, which differ only by interchange of identical particles in phase space. We have to overestimate the volume of phase space by a factor which is $N!$ under classical conditions. Taking this factor into account $e$ as the base of natural logarithms

$$\sigma = \ln (\Delta \Gamma / N!) = N \ln [ (V / N) (4 \pi M / 3) ^ {3 / 2} (E / N) ^ {3 / 2} e ] + \frac {3 N}{2}$$

To complete the expression for the entropy of ideal gas we need to introduce the unit of volume in phase space $\hbar^{3N}$, so that

$$\sigma = \ln (\Delta \Gamma / N! \hbar^ {3 N}) = N \ln \left[ \frac {(2 M) ^ {3 / 2} \pi^ {3 / 2} e (V / N) (E / N) ^ {3 / 2}}{\left(\frac {3}{2}\right) ^ {3 / 2} \hbar^ {3}} \right] + \frac {3}{2} N$$

From (3.13) we have

$$1 / \theta = \left(\frac {\partial \sigma}{\partial E}\right) _ {V, N} = \frac {\partial}{\partial E} (3 N / 2) \ln E = 3 N / 2 E$$

so that

$$E = 3 N \theta / 2 = 3 N k T / 2$$

in agreement with the elementary result for the internal energy of a perfect monatomic gas. We can consider (6.16) as establishing the connection between $\theta$ and T. Further

$$p / \theta = \left(\partial \sigma / \partial V\right) _ {N, V} = \frac {\partial}{\partial V} N \ln V = N / V$$

whence

$$p V = N \theta = N k T$$

Using (6.16) and $S = k\sigma$, we have the famous Sackur-Tetrode formula for the entropy of an ideal gas:

$$S = N k \ln \left[ \left(V / N\right) e \left(2 \pi M k T / \hbar^ {2}\right) ^ {3 / 2} \right] + 3 N k / 2$$

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