Worked exercise — Helium in a braking truck

Problem

A truck carries a rigid, thermally insulated cylinder containing $n = 10$ mol of helium gas (monatomic, $f=3$) at initial temperature $T_i = 300$ K. The truck is travelling at $v = 100$ m/s when it suddenly brakes hard and stops. By how much has the gas temperature changed?

Solution

When the truck stops, the cylinder stops with it, but the gas molecules continue to move with their thermal motion plus the drift motion they had before ($v = 100$ m/s). Since the cylinder is insulated (no heat) and rigid (no work), this macroscopic kinetic energy is converted into internal energy of the gas: the temperature rises.

Total mass of the gas (molar mass of helium $M_\text{He}=4$ g/mol): $M = n\, M_\text{He} = 10\cdot 4\cdot 10^{-3} = 0{,}04\ \text{kg}$ Initial macroscopic kinetic energy of the entrained gas: $K_\text{macro} = \tfrac{1}{2}Mv^2 = \tfrac{1}{2}\cdot 0{,}04\cdot 100^2 = 200\ \text{J}$ All of this energy becomes an increase in internal energy (monatomic gas, $f=3$): $\Delta E_\text{int} = \tfrac{3}{2}nR\,\Delta T = K_\text{macro} = 200\ \text{J}$ Solving for $\Delta T$: $\Delta T = \frac{2\,K_\text{macro}}{3nR} = \frac{2\cdot 200}{3\cdot 10\cdot 8{,}31} \approx 1{,}6\ \text{K}$ The temperature rises by less than two degrees: the braking is an almost imperceptible warming. It is the same principle by which a car’s brakes heat up: macroscopic kinetic energy → microscopic kinetic energy of the molecules. $\ev{\Delta T \approx 1{,}6\ \text{K}}$

Collegamenti

Argomenti: Teoria cinetica dei gas Concetti: Energia interna · Temperatura Competenze: Interpretazione micro-macro