= Solution
A fully convective, monatomic <ideal gas> is approximately isentropic through its deep interior. Its <adiabatic index> is $5/3$, so $P=K\rho^{5/3}$ and $T\propto K\rho^{2/3}$. The $n=3/2$ <stellar polytrope> scaling, or direct substitution of the central homology scalings into $K=P_c/\rho_c^{5/3}$, gives $K\propto RM^{1/3}$.
Near the photosphere, hydrostatic balance expressed in optical depth gives $P_{\rm ph}\sim g/\kappa_{\rm ph}$, with $g\propto M/R^2$. Treat $T_{\rm ph}$ as proportional to $T_{\rm eff}$ and match to the convective adiabat, neglecting a varying surface entropy jump in this homology approximation. Since $\rho\propto P/T$ and $\kappa_{\rm ph}\propto\rho T^{10}\propto PT^9$,
$$
P_{\rm ph}^2T_{\rm eff}^9\propto g.
$$
The adiabat gives $P\propto K^{-3/2}T^{5/2}$, so the surface condition yields
$$
T_{\rm eff}^{14}\propto gK^3\propto M^2R.
$$
The <Stefan–Boltzmann law> consequently gives $L_{\rm surf}\propto M^{4/7}R^{16/7}$. The same specific nuclear law as before still gives $L_{\rm nuc}\propto M^7R^{-8}$. Equating these scalings produces $R^{72/7}\propto M^{45/7}$. Thus the <fully convective homology with tenth-power surface opacity> is
$$
\boxed{R\propto M^{5/8},\qquad L\propto M^2.}
$$
Then $T_{\rm eff}\propto M^{3/16}$, and \b[the theoretical HR slope is]
$$
\boxed{\frac{d\log L}{d\log T_{\rm eff}}=\frac{32}{3}.}
$$
Its steepness expresses the relatively weak change of effective temperature with mass. The graph using $-\log T_{\rm eff}$ has the negative of this slope.
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