Problem Statement
Derive the thermal conductivity $\kappa$ of an ideal gas from kinetic theory and show it is independent of pressure.
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: Heat flux across a temperature gradient $dT/dz$:
Step 2 — Apply the relevant physical law or equation: $$q = -\kappa\frac{dT}{dz}$$
Step 3 — Solve algebraically for the unknown: Molecules carry thermal energy $C_v T$ per mole. The kinetic theory result:
Step 4 — Substitute numerical values with units: $$\kappa = \frac{1}{3}\rho\bar{v}\langle l\rangle C_v = \eta C_v$$
Step 5 — Compute and check the result: Since $\eta = \frac{1}{3}\rho\bar{v}\langle l\rangle$ and $\rho\langle l\rangle \propto m/d^2$ (independent of $n$, hence pressure), $\kappa$ is also independent of pressure.
Step 6: More precisely:
Worked Calculation
$$q = -\kappa\frac{dT}{dz}$$
$$\kappa = \frac{1}{3}\rho\bar{v}\langle l\rangle C_v = \eta C_v$$
$$\kappa = \frac{C_v\eta}{M} = \frac{C_v}{M}\frac{m\bar{v}}{3\sqrt{2}\pi d^2}$$
Heat flux across a temperature gradient $dT/dz$:
$$q = -\kappa\frac{dT}{dz}$$
Molecules carry thermal energy $C_v T$ per mole. The kinetic theory result:
$$\kappa = \frac{1}{3}\rho\bar{v}\langle l\rangle C_v = \eta C_v$$
Since $\eta = \frac{1}{3}\rho\bar{v}\langle l\rangle$ and $\rho\langle l\rangle \propto m/d^2$ (independent of $n$, hence pressure), $\kappa$ is also independent of pressure.
More precisely:
$$\kappa = \frac{C_v\eta}{M} = \frac{C_v}{M}\frac{m\bar{v}}{3\sqrt{2}\pi d^2}$$
Ratio $\kappa/\eta = C_v/M = c_v$ (specific heat). For monatomic gases, experiment gives $\kappa/\eta \approx 2.5c_v$ (Eucken correction accounts for internal degrees of freedom).
Answer
$$\boxed{\kappa = \frac{C_v\eta}{M} = \frac{C_v}{M}\frac{m\bar{v}}{3\sqrt{2}\pi d^2}}$$
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.
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