next up previous
Next: Gibb's Paradox Up: Applications of Statistical Thermodynamics Previous: Partition Function

Ideal Monatomic Gas

Let us now practice calculating thermodynamic relations using the partition function by considering an example with which we are already quite familiar: namely, an ideal monatomic gas. Consider a gas consisting of $ N$ identical monatomic molecules of mass $ m$ , enclosed in a container of volume $ V$ . Let us denote the position and momentum vectors of the $ i$ th molecule by $ {\bf r}_i$ and $ {\bf p}_i$ , respectively. Because the gas is ideal, there are no interatomic forces, and the total energy is simply the sum of the individual kinetic energies of the molecules:

$\displaystyle E = \sum_{i=1,N}\frac{ p_i^{ 2}}{2 m},$ (7.63)

where $ p_i^{ 2} = {\bf p}_i\!\cdot \!{\bf p}_i$ .

Let us treat the problem classically. In this approach, we divide up phase-space into cells of equal volume $ h_0^{ f}$ . Here, $ f$ is the number of degrees of freedom, and $ h_0$ is a small constant with dimensions of angular momentum that parameterizes the precision to which the positions and momenta of molecules are determined. (See Section 3.2.) Each cell in phase-space corresponds to a different state. The partition function is the sum of the Boltzmann factor $ \exp(-\beta  E_r)$ taken over all possible states, where $ E_r$ is the energy of state $ r$ . Classically, we can approximate the summation over cells in phase-space as an integration over all phase-space. Thus,

$\displaystyle Z = \int\cdots \int \exp(-\beta   E)  \frac{d^{ 3}{\bf r}_1\cdots d^{ 3}{\bf r}_N  d^{ 3}{\bf p}_1\cdots d^{ 3}{\bf p}_N}{h_0^{ 3N}},$ (7.64)

where $ 3N$ is the number of degrees of freedom of a monatomic gas containing $ N$ molecules. Making use of Equation (7.63), the previous expression reduces to

$\displaystyle Z = \frac{V^{ N}}{h_0^{ 3N}}\int\cdots \int \exp\left(-\frac{\b... p}_1 \cdots \exp\left(-\frac{\beta}{2 m} p_N^{ 2}\right)d^{ 3}{\bf p}_N.$ (7.65)

Note that the integral over the coordinates of a given molecule simply yields the volume of the container, $ V$ , because the energy, $ E$ , is independent of the locations of the molecules in an ideal gas. There are $ N$ such integrals, so we obtain the factor $ V^{ N}$ in the previous expression. Note, also, that each of the integrals over the molecular momenta in Equation (7.65) are identical: they differ only by irrelevant dummy variables of integration. It follows that the partition function $ Z$ of the gas is made up of the product of $ N$ identical factors: that is,

$\displaystyle Z = \zeta^{ N},$ (7.66)


$\displaystyle \zeta = \frac{V}{h_0^{ 3}}\int \exp\left(-\frac{\beta}{2 m} p^{ 2}\right)d^{ 3}{\bf p}$ (7.67)

is the partition function for a single molecule. Of course, this result is obvious, because we have already shown that the partition function for a system made up of a number of weakly interacting subsystems is just the product of the partition functions of the subsystems. (See Section 7.6.)

The integral in Equation (7.67) is easily evaluated:

$\displaystyle \int \exp\left(-\frac{\beta}{2 m} p^{ 2}\right)d^{ 3}{\bf p}$ $\displaystyle = \int_{-\infty}^{\infty}\exp\left(-\frac{\beta}{2 m} p_x^{ 2}...
...  \int_{-\infty}^{\infty}\exp\left(-\frac{\beta}{2 m} p_y^{ 2}\right)dp_y $    
  $\displaystyle \phantom{=}\times \int_{-\infty}^{\infty}\exp\left(-\frac{\beta}{2 m} p_z^{ 2}\right) dp_z$    
  $\displaystyle = \left(\sqrt{\frac{2\pi m}{\beta}}\right)^3,$ (7.68)

where use has been made of Equation (2.79). Thus,

$\displaystyle \zeta = V \left( \frac{2\pi  m}{h_0^{ 2} \beta}\right)^{3/2},$ (7.69)


$\displaystyle \ln Z = N\ln \zeta = N \left[\ln V - \frac{3}{2}\ln \beta +\frac{3}{2} \ln\left(\frac{2\pi  m}{h_0^{ 2}}\right)\right].$ (7.70)

The expression for the mean pressure, (7.50), yields

$\displaystyle \overline{p} = \frac{1}{\beta}\frac{\partial\ln Z}{\partial V} = \frac{1}{\beta} \frac{N}{V},$ (7.71)

which reduces to the ideal gas equation of state

$\displaystyle \overline{p}  V = N  k  T = \nu  R  T,$ (7.72)

where use has been made of $ N= \nu  N_A$ and $ R= N_A  k$ . According to Equation (7.35), the mean energy of the gas is given by

$\displaystyle \overline{E} = - \frac{\partial \ln Z}{\partial \beta} = \frac{3}{2} \frac{N}{\beta} = \frac{3}{2} \nu  R  T.$ (7.73)

Note that the internal energy is a function of temperature alone, with no dependence on volume. The molar heat capacity at constant volume of the gas is given by

$\displaystyle c_V = \frac{1}{\nu} \left(\frac{\partial \overline{E}}{\partial T} \right)_V = \frac{3}{2}  R,$ (7.74)

so the mean energy can be written

$\displaystyle \overline{E} = \nu  c_V  T.$ (7.75)

We have seen all of the previous results before. Let us now use the partition function to calculate a new result. The entropy of the gas can be calculated quite simply from the expression

$\displaystyle S = k\left(\ln Z + \beta  \overline{E}\right).$ (7.76)


$\displaystyle S = \nu  R \left[ \ln V -\frac{3}{2} \ln \beta + \frac{3}{2}\ln \left(\frac{2\pi  m }{h_0^{ 2}}\right) + \frac{3}{2} \right],$ (7.77)


$\displaystyle S = \nu  R \left( \ln V + \frac{3}{2}\ln T + \sigma\right),$ (7.78)


$\displaystyle \sigma = \frac{3}{2} \ln \left( \frac{2\pi  m k}{h_0^{ 2}}\right) + \frac{3}{2}.$ (7.79)

The previous expression for the entropy of an ideal gas is certainly a new result. Unfortunately, it is also quite obviously incorrect.

next up previous
Next: Gibb's Paradox Up: Applications of Statistical Thermodynamics Previous: Partition Function
Richard Fitzpatrick 2016-01-25