(5)
$$
\frac{\Delta L}{L_1} = \frac{B}{1 - B} \left( \frac{T_{g2} - T_1}{T_{g2}} \right)
$$
or, for small temperature differences, we have:
(6)
$$
\frac{\Delta L}{L} = \frac{B}{1 - B} \left( \frac{\Delta T}{T} \right)
$$
With increasing temperatures, there will be an unbalance in the direction of greater altitude. While climbing to a greater altitude the balloon will valve gas and come to equilibrium at a new level. Thus the effect of gain of superheat with a full non-extensible balloon will be a slight increase of altitude.
Investigating the case where an initial amount of superheat is lost:
(7)
$$
\Delta L = V_2 (d_{a2} - d_{g2}) - V_1 (d_{a1} - d_{g1})
$$
and since the balloon volume will decrease with cooling of the lifting gas:
$$
V_2 = V_1 \frac{T_{g2}}{T_{g1}} \quad \text{(assuming constant } p\text{)}
$$
Therefore, again making use of the assumptions that:
$$
p_{a1} = p_{a2} = p_{g1} = p_{g2} = p
$$
and:
$$
T_{g2} = T_{a1} = T_{a2} = T_2
$$
Combining equation (2) and equation (7), we have:
$$
\Delta L = V_1 \left[ \frac{T_2}{T_{g1}} \left( \frac{p}{R_a T_2} - \frac{p}{R_g T_2} \right) - \left( \frac{p}{R_a T_2} - \frac{p}{R_g T_g} \right) \right]
$$
$$
= V_1 \left( \frac{p}{R_a T_{g1}} - \frac{p}{R_g T_{g1}} - \frac{p}{R_a T_2} + \frac{p}{R_g T_{g1}} \right)
$$
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