Question 2.169: (a) Assuming moderately dilute helium gas so that binary col...
(a) Assuming moderately dilute helium gas so that binary collisions of helium atoms determine the transport coefficients, derive an expression for the thermal conductivity of the gas.
(b) Estimate the ratio of the thermal conductivity of gaseous { }^{3} He to that of gaseous { }^{4} He at room temperature.
(c) Will this ratio become different at a temperature near 2 K? Why?
(a) Consider an area element dS of an imaginary plane at z=z_{0}.which divides the gas into two parts A and B (see Fig. 2 . 3 5 ) . We assume that the temperatures of A and B are T_{A} \text { and } T_{B} respectively. In the case of a small temperature difference, we can take approximately n_{A} \bar{v}_{A}=n_{B}\bar{v}_{B}=n \bar{v} then the number of molecules exchanged between A and B through d S in
time interval dt is n \bar{v} d S d t / 6. According to the principle of equipartition of energy, the average kinetic energy of the molecules of A is \frac{l}{2} k T_{A}, and that of B \text { is } \frac{l}{2} k T_{B} (l is the number of degrees of freedom of molecule). Therefore, the net energy transporting through d S in dt (or the heat transporting along the positive direction of the z-axis) is
d Q=l k\left(T_{A}-T_{B}\right) n \bar{v} d S d t / 12.
The temperature difference can be expanded in terms of the temperature gradient
T_{A}-T_{B}=-2 \bar{\lambda}\left(\frac{d T}{d z}\right)_{z_{0}}.
So
d Q=-\frac{1}{3} n \bar{v} \bar{\lambda} \frac{l}{2} k\left(\frac{d T}{d z}\right)_{z_{0}} d S d t.
giving the thermal conductivity
\kappa=\frac{1}{3} n \bar{v} \bar{\lambda} \frac{l}{2} k=\frac{1}{3} \rho \bar{v} \bar{\lambda} c_{v}. (*)
where c_{v} is the specific heat at constant volume.
\text { (b) Since } \rho \bar{\lambda} \propto m / \sigma^{2}, \bar{v} \propto m^{-1 / 2}, c_{v} \propto 1 / m, \text { with the formula }(*), we have \kappa \propto \frac{1}{\sqrt{m}} \sigma^{-2} \text {, where } \sigma is the atomic diameter. For { }^{3} He \text { and }{ }^{4} He , \sigma can be taken as the same, giving
\frac{\kappa_{3}}{\kappa_{4}}=\left(\frac{m_{3}}{m_{4}}\right)^{-1 / 2}=\left(\frac{3}{4}\right)^{-1 / 2} \approx 1.15.
(c) When the temperature is near 2K, { }^{3} He \text { is in liquid phase and }{ }^{4} He is in superfluid phase, so that the above model is no longer valid. The ratio of the thermal conductivities changes abruptly at this temperature.
