Worked exercise — Fictitious gas of two balls

Problem

Two balls of mass $m_1 = m_2 = 2\;\text{kg}$ have velocities $\vec{v}_1 = (3, 0)$ and $\vec{v}_2 = (-5, 4)$. Assuming (fancifully) that they form a monatomic ideal gas, calculate the associated “temperature”.

Solution

CM velocity: $\vec{v}_\text{CM} = \left(\frac{3-5}{2},\; \frac{0+4}{2}\right) = (-1,\; 2)\;\text{m/s}$

Velocities relative to the CM ($\vec{v}_i' = \vec{v}_i - \vec{v}_\text{CM}$): $\vec{v}_1' = (3,0) - (-1,2) = (4, -2), \qquad \vec{v}_2' = (-5,4) - (-1,2) = (-4, 2)$ As expected, $m_1\vec{v}_1' + m_2\vec{v}_2' = \vec{0}$: the momentum in the CM frame is zero.

Internal kinetic energy (with $|\vec{v}_i'|^2 = 4^2+2^2 = 20$ for both): $E_\text{cin,int} = \tfrac{1}{2}\cdot 2\cdot(16+4) + \tfrac{1}{2}\cdot 2\cdot(16+4) = 20 + 20 = 40\;\text{J}$

Equivalent temperature (monatomic gas, $E_\text{cin,int} = \tfrac{3}{2}N k_B T$, with $N=2$): $T = \frac{2\,E_\text{cin,int}}{3 N k_B} = \frac{2\cdot 40}{3\cdot 2\cdot 1{,}38\cdot 10^{-23}} = \frac{80}{8{,}28\cdot 10^{-23}}$

$\ev{T \approx 9{,}7\cdot 10^{23}\;\text{K}}$

Astronomical! But this is an artefact of the toy model: with only $N=2$ particles, sharing out $40\;\text{J}$ means an enormous energy per particle. With a real gas of $\sim 10^{23}$ particles at speeds of a few m/s, the same formula would give a reasonable temperature.

Links

Topics: Centre of mass Concepts: Internal energy · Temperature · Kinetic energy Skills: Micro–macro interpretation