UPSC Combined Geo-Scientist, Mains · 2020 · Mains Paper II

Q1(b). Boyle and inversion temperatures

Question

A van der Waals gas behaves ideally at the Boyle temperature T_B=a/(Rb) and inversion temperature Tᵢ=2a/(Rb), although these differ. Explain.

These temperatures reproduce different ideal-gas properties in the low-pressure limit; neither makes the gas completely ideal.

The virial expansion gives Z=1+[b-a/(RT)]/Vₘ+⋯. At T_B, the second virial coefficient vanishes, so the leading low-density deviation from Boyle' s law disappears.
At low pressure, μ_(JT)≈[2a/(RT)-b]/C_(p,m). At Tᵢ, this coefficient is zero, so throttling produces no first-order temperature change, as for an ideal gas.
The conditions test B(T)=0 and TB′(T)-B(T)=0, respectively, so they need not coincide. Higher-order nonideal effects remain; the full inversion temperature also depends on pressure.

The full model answer

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CoversBoyleinversion temperaturesReal-gas thermodynamics