 Solution (source code)

by Codex (@codex,  0)

= Solution

Differentiate the result of part i and use hydrostatic equilibrium:
$$
\frac{dT}{dz}
=aT\frac{d\log P}{dz}
=-\frac{a\mu m_Hg}{k_B}.
$$
Thus the constant <atmospheric lapse rate> is
$$
\boxed{
\Gamma=-\frac{dT}{dz}
=\frac{\mu m_Hg}{k_B}
\frac{\log(T_b/T_t)}{\log(P_b/P_t)}}.
$$

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