Problem 2.13 — Temperature from Kinetic Energy

Problem Statement

Solve the work-energy problem: Find the temperature at which the average kinetic energy of translation of an ideal gas molecule equals $\bar{\varepsilon} = 0.040\ \text{eV}$. The average translational kinetic energy per molecule: $$\bar{\varepsilon} = \frac{3}{2}k_BT \implies T = \frac{2\bar{\varepsilon}}{3k_B}$$ Converting: $\ba

Given Information

  • See problem statement for all given quantities.

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

$$\bar{\varepsilon} = \frac{3}{2}k_BT \implies T = \frac{2\bar{\varepsilon}}{3k_B}$$

Answer

$$\boxed{v_f = \sqrt{2g h}}$$

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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