Thermodynamic Temperature

Suppose that the systems $A$ and $A'$ are initially thermally isolated from one another, with respective internal energies $U_i$ and $U_i'$. If the two systems are subsequently placed in thermal contact, so that they are free to exchange heat energy, then, in general, the resulting state is an extremely improbable one [i.e., $P(U_i)$ is much less than the peak probability]. The configuration will, therefore, tend to evolve in time until the two systems attain final energies, $U_f$ and $U_f'$, which are such that $P(U_f)$ is maximized. In the special case where the initial energies, $U_i$ and $U_i'$, lie very close to the final energies, $U_f$ and $U_f'$, respectively, there is no change in the two systems when they are brought into thermal contact, because the initial state already corresponds to a state of maximum probability.

It follows from energy conservation that

$$U_f +U_f' = U_i + U_i'.$$ (5.306)

The energy change in each system is simply the net heat absorbed, so that

$$Q = U_f - U_i,$$ (5.307)

$$Q' =U_f' - U_i'.$$ (5.308)

The conservation of energy then reduces to

$$Q+ Q' =0.$$ (5.309)

In other words, the heat given off by one system is equal to the heat absorbed by the other. (In our notation, absorbed heat is positive, and emitted heat is negative.)

It is clear that if the systems $A$ and $A'$ are suddenly brought into thermal contact then they will only exchange heat, and evolve towards a new equilibrium state, if the final state is more probable than the initial one. In other words, the system will evolve if

$$P(U_f) > P(U_i),$$ (5.310)

or

$$\ln P(U_f)> \ln P(U_i),$$ (5.311)

because the logarithm is a monotonic function. The previous inequality can be written

$$\ln{\mit\Omega}(U_f) + \ln{\mit\Omega}'(U_f')>\ln{\mit\Omega}(U_i) + \ln{\mit\Omega}'(U_i'),$$ (5.312)

with the aid of Equation (5.304). Taylor expansion to first order yields

$$\frac{\partial\ln{\mit\Omega}(U_i)}{\partial U}\, (U_f- U_i) + \frac{\partial\ln{\mit\Omega}'(U_i')}{\partial U'}\,(U_f'-U_i') > 0,$$ (5.313)

which finally gives

$$[\beta(U_i) - \beta'(U_i)]\,Q >0,$$ (5.314)

where

$$\beta = \frac{\ln{\mit\Omega}}{\partial U},$$ (5.315)

$$\beta' = \frac{\ln{\mit\Omega'}}{\partial U},$$ (5.316)

and use has been made of Equations (5.307)–(5.309).

It is clear, from the previous analysis, that the parameter $\beta$ has the following properties:

1. If two systems separately in equilibrium have the same value of $\beta$ then the systems will remain in equilibrium when brought into thermal contact with one another.
2. If two systems separately in equilibrium have different values of $\beta$ then the systems will not remain in equilibrium when brought into thermal contact with one another. Instead, the system with the higher value of $\beta$ will absorb heat from the other system until the two $\beta$ values are the same. [See Equation (5.314).]

Let us define the dimensionless parameter, $T$, such that

$$\frac{1}{k_B\,T} \equiv \beta = \frac{\partial \ln{\mit\Omega}}{\partial E},$$ (5.317)

where $k_B$ is the Boltzmann constant. The parameter $T$ is termed the thermodynamic temperature, and controls heat flow in much the same manner as a conventional temperature. Thus, if two isolated systems in equilibrium possess the same thermodynamic temperature then they will remain in equilibrium when brought into thermal contact. However, if the two systems have different thermodynamic temperatures then heat will flow from the system with the higher temperature (i.e., the “hotter” system) to the system with the lower temperature, until the temperatures of the two systems are the same. In addition, suppose that we have three systems, $A$, $B$, and $C$. We know that if $A$ and $B$ remain in equilibrium when brought into thermal contact then their temperatures are the same, so that $T_A= T_B$. Similarly, if $B$ and $C$ remain in equilibrium when brought into thermal contact, then $T_B = T_C$. But, we can then conclude that $T_A=T_C$, so systems $A$ and $C$ will also remain in equilibrium when brought into thermal contact. Thus, we arrive at the following statement, which is sometimes called the zeroth law of thermodynamics:

If two systems are separately in thermal equilibrium with a third system then they must also be in thermal equilibrium with one another.

Let us test our scheme out on a monatomic ideal gas. We saw in Section 5.4.4 that the number of accessible states of an ideal monatomic gas consisting of $N$ particles is

$${\mit\Omega}(U,V)= B\,V^{\,N} U^{\,3N/2},$$ (5.318)

where $U$ is the internal energy, $V$ the volume, and $B$ is a constant that is independent of $U$ and $V$. According to the previous two equations, the thermodynamic temperature of such a gas is

$$\frac{1}{k_B\,T} = \frac{3\,N}{2\,U}.$$ (5.319)

However, $N=\nu\,N_A$, where $\nu$ is the number of moles of molecules in the gas, and $N_A$ is Avogadro's number. The previous equation can be rearranged to give

$$U= \frac{3}{2}\,\nu\,R\,T,$$ (5.320)

because $R=k_B\,N_A$. However, this is the correct expression for the internal energy of a monatomic ideal gas. (See Section 5.2.3.) Hence, it is clear that the thermodynamic temperature defined in Equation (5.317) corresponds to the more familiar absolute temperature associated with an ideal gas.