Problem Statement
Show that for an ideal gas, $\kappa = \eta C_v/M$ (per unit mass), and that $D\rho = \eta$ to first approximation.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: All three transport coefficients come from kinetic theory with $l = \langle l\rangle$ (mean free path) and $\bar{v}$ (mean speed):
Step 2 — Apply the relevant physical law or equation: $$\eta = \frac{1}{3}\rho\bar{v}\langle l\rangle, \quad D = \frac{1}{3}\bar{v}\langle l\rangle, \quad \kappa = \frac{1}{3}\rho\bar{v}\langle l\rangle c_v$$
Step 3 — Solve algebraically for the unknown: Therefore:
Step 4 — Substitute numerical values with units: $$\frac{\kappa}{\eta} = \frac{\frac{1}{3}\rho\bar{v}\langle l\rangle c_v}{\frac{1}{3}\rho\bar{v}\langle l\rangle} = c_v \implies \kappa = \eta c_v$$
$$\frac{\eta}{\rho} = \frac{1}{3}\bar{v}\langle l\rangle = D \implies D\rho = \eta$$
Step 5 — Compute and check the result: These give $Sc = \nu_{kin}/D = 1$ (Schmidt number) and $Pr = \nu_{kin}/\chi = 1$ (Prandtl number) in the simplest kinetic theory. Realistic corrections (Eucken, Chapman-Enskog) improve the Prandtl number prediction.
Worked Calculation
$$\eta = \frac{1}{3}\rho\bar{v}\langle l\rangle, \quad D = \frac{1}{3}\bar{v}\langle l\rangle, \quad \kappa = \frac{1}{3}\rho\bar{v}\langle l\rangle c_v$$
$$\frac{\kappa}{\eta} = \frac{\frac{1}{3}\rho\bar{v}\langle l\rangle c_v}{\frac{1}{3}\rho\bar{v}\langle l\rangle} = c_v \implies \kappa = \eta c_v$$
$$\frac{\eta}{\rho} = \frac{1}{3}\bar{v}\langle l\rangle = D \implies D\rho = \eta$$
All three transport coefficients come from kinetic theory with $l = \langle l\rangle$ (mean free path) and $\bar{v}$ (mean speed):
$$\eta = \frac{1}{3}\rho\bar{v}\langle l\rangle, \quad D = \frac{1}{3}\bar{v}\langle l\rangle, \quad \kappa = \frac{1}{3}\rho\bar{v}\langle l\rangle c_v$$
Therefore:
$$\frac{\kappa}{\eta} = \frac{\frac{1}{3}\rho\bar{v}\langle l\rangle c_v}{\frac{1}{3}\rho\bar{v}\langle l\rangle} = c_v \implies \kappa = \eta c_v$$
$$\frac{\eta}{\rho} = \frac{1}{3}\bar{v}\langle l\rangle = D \implies D\rho = \eta$$
These give $Sc = \nu_{kin}/D = 1$ (Schmidt number) and $Pr = \nu_{kin}/\chi = 1$ (Prandtl number) in the simplest kinetic theory. Realistic corrections (Eucken, Chapman-Enskog) improve the Prandtl number prediction.
Answer
$$\boxed{\frac{\eta}{\rho} = \frac{1}{3}\bar{v}\langle l\rangle = D \implies D\rho = \eta}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
Leave a Reply