Problem — Sci-fi physics: Boltzmann constant doubled

Problem

If the Boltzmann constant suddenly doubled, how would the following change: the pressure of a gas at fixed $T, V, N$, $v_\text{rms}$, and the latent heat of evaporation?

Solution

I analyse one quantity at a time using the relations of kinetic theory, keeping $T$, $V$, $N$ fixed and substituting $k_B \to 2k_B$. $pV = N k_B T \ \Rightarrow \ p = \frac{N k_B T}{V}$ The pressure is proportional to $k_B$, so it doubles: $p \to 2p$ The root-mean-square speed depends on $\sqrt{k_B}$ (since $\frac{1}{2}m v_\text{rms}^2 = \frac{3}{2}k_B T$): $v_\text{rms} = \sqrt{\frac{3 k_B T}{m}} \ \Rightarrow \ v_\text{rms} \to \sqrt{2}\,v_\text{rms}$ The latent heat $L$ is an energy tied to molecular bonds: it does not contain $k_B$, so its value per particle does not change directly. What matters for boiling, however, is the dimensionless ratio $L/(k_B T)$ that governs the Clausius–Clapeyron relation: $\frac{L}{k_B T} \to \frac{L}{2 k_B T} \quad (\text{halved})$ $\ev{p \to 2p; \quad v_\text{rms} \to \sqrt{2}\,v_\text{rms}; \quad L \text{ unchanged, but } L/(k_B T) \text{ halves} \Rightarrow \text{boiling at lower } T}$ Halving $L/(k_B T)$ makes it easier to overcome the bonds: the liquid would evaporate at lower temperatures.

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Topics: Kinetic theory of gases Concepts: Pressure · Latent heat and changes of state Skills: Analysis of limiting cases and sci-fi physics