Problem — Maxwell-Boltzmann at two temperatures

Problem

Maxwell-Boltzmann distribution at two temperatures. Sketch $f(v)$ for a gas at $T_1 < T_2$. What stays unchanged between the two curves, and what changes?

Solution

The Maxwell-Boltzmann distribution $f(v)$ describes how the molecules’ speeds are distributed. Raising the temperature from $T_1$ to $T_2 > T_1$:

• What stays unchanged: the area under the curve, which represents the total number $N$ of molecules (or the total probability $=1$). Molecules do not disappear: they merely redistribute. $\int_0^{\infty} f(v)\,\dd v = N = \text{constant}$
• What changes: the peak (most probable speed, $\propto \sqrt{T}$) shifts to the right, towards higher speeds. Since the area must stay fixed, the curve lowers and broadens: the maximum becomes lower and the exponential tail at high $v$ grows. It is this thickening of the tail that explains, for example, the atmospheric escape of fast molecules. $\ev{T_2 > T_1:\ \text{peak further right and lower, tail at high } v \text{ thickened; area} = N \text{ fixed}}$

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Topics: Kinetic theory of gases Concepts: Maxwell-Boltzmann distribution · Temperature