= Solution
{c}
At the interface the <pressure> is continuous. Immediately below it the <temperature> is $T_s$ and mean <molar mass> is $M_d(1+ax)$; immediately above it the dry <hydrogen> has <temperature> $T_1$. The <ideal gas law> therefore gives
$$
\rho_- =\frac{p_1M_d(1+ax)}{RT_s},\qquad
\rho_+=\frac{p_1M_d}{RT_1}.
$$
An infinitesimal interchange is unstable to <convection> when the lower fluid is less dense than the upper fluid. Hence
$$
\boxed{\frac{T_s}{T_1}>1+a\frac{E(T_s)}{P(T_s)}.}
$$
Equality gives neutral interfacial <buoyancy>; reversing the inequality gives a stable <density> jump. This is an interfacial criterion, not a claim that the whole dry upper <atmosphere> is stable. Its vertical <temperature> profile has not been specified. Also, $\mathrm{RH}(p_1)=p_1/P<1$, so infinitesimal displacement at the interface initially involves no <condensation> or release of <latent heat>.
There are two competing effects. Increasing $T_s$ directly reduces the lower <density>, favouring <convection>. Increasing the <mole fraction> of heavy <water vapor> increases its mean <molar mass>, opposing <convection>. With $x\ll1$, the criterion is approximately $T_s>T_1$. With $x\to1$, it tends to $T_s>\epsilon T_1\simeq9T_1$: substantial surface heating need not overcome the compositional stabilization in <hydrogen>.
For a specified $P_d,\Delta p,T_1$, solve $T_s/T_1-1-ax(T_s)=0$ on the admissible branch to locate the transitions. Its <derivative> is $1/T_1-a x'(T_s)$, so there is no universal monotone trend: rapid growth of the <saturation pressure> can suppress an instability that was present at lower $T_s$. Without numerical parameter values, neither the number of crossings nor a unique interval of unstable <temperatures> can be asserted.
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