Thermal Conductivity

Consider a gas of uniform number density, $n$, that has a temperature gradient along the $z$-axis, such that

$$T(z)= T_0 + \frac{\partial T}{\partial z}\,dz.$$ (5.247)

Let each molecule in the gas have a mean thermal energy $\epsilon(T)$ (under most circumstances this energy is the sum of the molecule's translational and rotational kinetic energy). Slightly modifying the analysis of the previous section, the thermal energy per unit area, per second, carried by molecules whose speeds lie between $v$ and $v+dv$, and whose directions of motion subtend an angle lying between $\theta$ and $\theta+d\theta$ with the $z$-axis, that cross the $x$-$y$ plane, is

$$dq_z= [\epsilon']\left[n\,F(v)\,dv\right]\left[\frac{1}{2}\,\sin\theta\,d\theta\right][v\,\cos\theta],$$ (5.248)

where $F(v)$ is the distribution of molecular speeds, and $\epsilon'=\epsilon(T')$, where $T'$ is the temperature where the molecules last made a collision. On average, the molecules move a distance $l$ (i.e., the mean free path) between collisions. Hence, $dz =- l\,\cos\theta$, and

$$T' = T_0 - \frac{\partial T}{\partial z}\,l\,\cos\theta.$$ (5.249)

which implies that

$$\epsilon' \simeq \epsilon_0 - {\cal C}\,\frac{\partial T}{\partial z}\,l\,\cos\theta,$$ (5.250)

where

$${\cal C} = \frac{\partial \epsilon}{\partial T}$$ (5.251)

is the specific heat per molecule. Thus, the net heat flux across the $x$-$y$ plane is

$$q_z =\frac{n}{2}\int_0^\pi\left[\epsilon_0-{\cal C}\,\frac{\parti... ...z}\,l\,\cos\theta\right]\cos\theta\,\sin\theta\,d\theta \int_0^\infty F(v)\,dv,$$ (5.252)

which gives

$$q_z = -\kappa\,\frac{\partial T}{\partial z},$$ (5.253)

where

$$\kappa = \frac{1}{3}\,n\,l\,{\cal C}\,\langle v\rangle$$ (5.254)

is termed the thermal conductivity. Here, $\langle v\rangle$ is the mean molecular speed. (See Section 5.5.9.) Thus, we conclude that the heat flux in the $z$-direction is proportional to minus the local temperature gradient along the $z$-axis. This is another example of Fick's law.

Equations (5.219), (5.244), and (5.253) yield

$$\kappa = \frac{2}{3\,\pi^{3/2}}\,\frac{{\cal C}}{R^{\,2}}\sqrt{\frac{k_B\,T}{m}},$$ (5.255)

where $R$ is the molecular diameter, and $m$ the molecular mass. Moreover, for a diatomic gas, such as air, ${\cal C} = (5/2)\,k_B$. (See Section 5.3.6.) It can be seen that the thermal conductivity of an ideal gas scales as $T^{1/2}$, and is independent of the pressure at constant temperature. The previous formula yields the following estimate for the thermal conductivity of air (assuming that $T=15^\circ$ C, $R=2\times 10^{-10}\,{\rm m}$, and $m=29\,m_p$),

$$\kappa = 3\times 10^{-2}\,{\rm W\,{\rm m}^{-1}\,K^{-1}},$$ (5.256)

which turns out to be fairly accurate

Consider a slab of gas lying between $z$ and $z+dz$. The rate of change of the thermal energy contained in the slab is the difference between the flux of heat into the slab and the flux of heat out of the slab. In other words,

$$\frac{\partial (n\,A\,dz\,{\cal C}\,T)}{\partial t} = \left[q_z(z,t)- q_z(z+dz,t)\right]A,$$ (5.257)

where $A$ is the cross-sectional area of the slab. It follows that

$$\frac{\partial T}{\partial t} = -\frac{1}{n\,{\cal C}}\,\frac{\partial q_z}{\partial z}.$$ (5.258)

Making use of Equations (5.253) and (5.254), we obtain

$$\frac{\partial T}{\partial t} = D_\kappa\,\frac{\partial^2 T}{\partial z^2},$$ (5.259)

where

$$D_\kappa = \frac{\kappa}{n\,{\cal C}} = \frac{1}{3}\,l\,\langle v\rangle.$$ (5.260)

Of course, Equation (5.259) is the diffusion equation, and $D_\kappa$ is the associated diffusivity. Thus, we conclude, by comparison with the analysis in the previous section, that heat diffuses through an ideal gas at the same very slow rate at which molecules diffuse. Moreover, it is clear, from Sections 5.1.5 and 5.1.7, that heat diffusion is due to a random walk of molecules with excess energy under the action of molecular collisions.