Thermal Conductivity
Consider a gas in a containing vessel where it is subjected to a temperature gradient. In order to evaluate the coefficient of the thermal conductivity of the gas, it is necessary to solve the kinetic equation having a temperature gradient. Let us limit ourselves to a stationary state, though we have a non-equilibrium
state for which we write the Boltzmann equation. In our case = 0. From this it follows that
Here the solution is not all that simple. To consider dissipative processes (thermal conduction and viscosity) in a slightly inhomogeneous gas we do an approximation. This involves regarding the distribution function in each region of the gas as being not a local-equilibrium function f0 but one with a slight deviation of/« 1 from/0:
The quantity/ « lis a small correction due to the non-equilibrium state and is an odd function with respect to the velocity (/ = /(i')) ■ The function/must satisfy the transport equation as well as certain additional conditions. This is because the equilibrium distribution function /„ corresponds to given values (in given volume element) of the gas particle number
total energy
and momentum densities
So, from 21.92 it follows that
and from 21.93 then
Similarly, from 21.94 then
The expression in 21.91 corresponds to a small deviation from the equilibrium state. It is a linear approximation as/2 = Of/2). If we substitute 21.91 into 21.90 considering
Letting,
then
If there is no temperature gradient, then the equilibrium part is equal to zero and from this it follows that
It follows from 21.100 that
or
We examine the LHS of 21.90 i.e. (v,Vj./) and neglect V %asx « 1. It should be noted that the chemical potential // of the gas is a function of the temperature T:
If there is no external field, then the energy of the gas is independent of the coordinate but dependent on the temperature. Thus, as
then
It should be noted that
where S and H are the entropy and enthalpy of the gas, respectively. Thus
from where
We examine how with the knowledge of h we may evaluate the thermal flux and thermal conductivity. The energy flux in the direction of the i'h coordinate:
As
then
From here it follows that
where K(k is the tensor of the thermal conductivity:
in the general case as the flux moves in the reverse direction to VT. Considering the fact that in equilibrium, the gas is isotropic then there are no preferred directions in it and so
Hence, the energy flux:
and the scalar thermal conductivity:
This quantity is made positive by the transport equation so that the energy flux must be in the opposite direction to the temperature gradient. It is instructive to note that owing to the time-reversal properties of microscopic processes, the transport coefficient, and in particular scalar thermal conductivity, is observed to be not entirely independent.
