Problem Statement
Solve the thermodynamics problem: Show that the viscosity of an ideal gas scales as $\eta \propto \sqrt{T}$ and is independent of pressure. Compare to liquids. $\eta = \frac{1}{3}\rho\bar{v}\langle l\rangle$. Substituting $\rho = nm$ and $\langle l\rangle = 1/(\sqrt{2}\pi d^2 n)$: $$\eta = \frac{m\bar{v}}{3\sqrt{2}\pi d^2}$$ Since $
Given Information
- $\langle l\rangle = 1/$
Physical Concepts & Formulas
Thermodynamics governs energy transformations involving heat and work. The First Law $\Delta U = Q – W$ expresses energy conservation. For an ideal gas, internal energy depends only on temperature ($U = nC_VT$), and the equation of state $PV = nRT$ links pressure, volume, and temperature.
- $\Delta U = Q – W$ — First Law of Thermodynamics
- $PV = nRT$ — ideal gas equation
- $C_P – C_V = R$, $\gamma = C_P/C_V$
- $W = \int P\,dV$ — work done by gas
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\eta = \frac{m\bar{v}}{3\sqrt{2}\pi d^2}$$
Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
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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