The equation of state

The partition function  can also be used to derive the equation of state of the perfect gas, which was given in Chapter 1 on the basis of experimental observations. To do that, we recall Eq.  for the pressure, $P=(-\partial F/\partial V)_T$, and using Eq.  for the Helmholtz free energy we obtain:

$$P = -\left ( \frac{\partial F(N,V,T)}{\partial V} \right )_T = N ... ...2\pi mk_{\rm B}T}{h^2}\right)^{3/2}Z_{int}(T) \right \}}{\partial V}\right )_T,$$ (4.38)

The logarithmic derivative of the r.h.s w.r.t $V$ is equal to $1/V$, ^4.5 and therefore we obtain at once the perfect gas equation of state:

$$P V = N k_{\rm B}T.$$ (4.39)

Eq.  was derived entirely from statistical physics principles and its detailed expression is a consequence of the choices that we made for the constant entering the definition of the entropy in Eq.  (the Boltzmann constant $k_{\rm B}$) and the detailed expression of the relation between $(\partial S/\partial E)_V$ and temperature in Eq.  [ $(\partial S/\partial E)_V = 1/T$]. With these (entirely arbitrary) choices we see that the perfect gas temperature scale discussed between Eqs.  and  is identical to the thermodynamic (or statistical physics) temperature scale.

Using Eq.  we can now obtain the energy of the system:

$$E = -\frac{\partial \ln Z}{\partial \beta} = -\frac{\partial}{\pa... ...ft ( -N \ln \beta^{3/2} \frac{Nh^3}{eV(2\pi m)^{3/2}} + N\ln Z_{int} \right ) =$$ (4.40)

$$= \frac{3N}{2\beta} - N\frac{\partial \ln Z_{int}}{\partial \beta}.$$

The first term is the translational contribution to the average internal energy

$$E_{tr} = \frac{3}{2} N k_{\rm B}T,$$ (4.41)

which is the same result obtained in Sec. . The second term is

$$E_{int} = -N\frac{\partial}{\partial \beta} \ln \sum_\alpha e^{-\beta \epsilon_\alpha^{int}} = N\bar{\epsilon}_{int},$$ (4.42)

where $\bar{\epsilon}_{int}$ is the average internal energy of each particle of gas We see that both $E_{tr}$ and $E_{int}$ are independent on the volume of the system.

We can also obtain the entropy:

$$S = -\left (\frac{\partial F}{\partial T}\right )_V = -N k_{\rm B... ...V}{N} \left( \frac{2\pi m k_{\rm B}T}{h^2} \right )^{3/2} Z_{int}(T)\right \} +$$ (4.43)

$$- Nk_{\rm B}T \left \{ \frac{3}{2 T} + \frac{\partial \ln Z_{int}(T)}{\partial T} \right \}.$$

We see that as $T \rightarrow 0$ the entropy diverges, and the heat capacity does not go to zero (the derivative of $E_{tr}$ w.r.t to $T$ is a constant). This difficulty appears because of the assumption in , where we have replaced the sum over the states with an integral. At $T=0$ only the ground state is populated, but in  this is excluded altogether, because it is given zero weight. Therefore, the expressions above for the Helmoltz free energy, the energy, the entropy, etc. are not valid in this limit.