CL: Molecular Origins of Thermal Conductivity

Returning to our molecular picture, we can think of a heat flux as being related to the movement (swapping) of molecules across a plane. These molecules each have their own value of thermal energy (given as $\rho c T$) so that their motion represents a flux (or a flow per unit area). We might write this flux (for Z molecules that moved) as:

$\displaystyle{\frac{q_y}{A} = \sum_{n=1}^Zm_n c_p\left (T|_{y-}-T|_{y+}\right)}$

Again, the |y- refers to the value of the temperature at the location just below our plane. We can relate this temperature (and the + variety) to a local value of the slope of the temperature (i.e., a derivative of the temperature with respect to position) by:

$\displaystyle{T|_{y-} = T|_{y_0}-\left (\frac{dT}{dy} \right)|_{y_o}\delta}$

$\displaystyle{T|_{y+} = T|_{y_0}+\left (\frac{dT}{dy} \right)|_{y_o}\delta}$

we can then combine these equations to give

$\displaystyle{\frac{q_y}{A} = -2 \sum_{n=1}^Zm_n c_p\frac{dT}{dy}|_{y_o}\delta}$

Again, relating $\delta$ to the mean free path using $\delta = (2/3)\lambda$ gives (for molecules of the same mass):

$\displaystyle{\frac{q_y}{A} = -\frac{4}{3}\lambda Zm c_p\frac{dT}{dy}|_{y_o}}$

The proportionality constant between the heat flux and the temperature gradient, which we will call the conductivity, can then be written as

$\displaystyle{k = \frac{4}{3}m Z c_p \lambda}$

Z can be related to the molecular concentration N and the average random molecular velocity C as Z = NC/4. Using this and the statistical mechanics expressions for the hard-sphere mean free path and average velocity:

$\displaystyle{\lambda = \frac{1}{\sqrt{2}\pi N d^2}}$

$\displaystyle{C = \sqrt{\frac{8 \kappa T}{\pi m}}}$

where d is the molecular diameter. Finally, we can write an expression for the heat capacity of the material using the Boltzmann constant (which also showed up in the average velocity) to give:

$\displaystyle{c_p = \frac{4}{3}\frac{\kappa}{N}}$

Combining these expressions gives the conductivity as:

$\displaystyle{k = \frac{1}{\pi^{3/2}d^2}\sqrt{\frac{\kappa^3T}{m}}}$

OUTCOME:

Explain the molecular origins of thermal and mass diffusion/conduction