$$
\frac {\Delta L}{L _ {1}} = \frac {B}{1 - B} \left(\frac {T _ {g _ {2}} - T _ {1}}{T _ {g _ {2}}}\right)
$$
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 nonextensible balloon will be a slight increase of altitude.
Investigating the case where an initial amount of superheat is lost:
$$
\Delta L = V _ {2} \left(d _ {a _ {2}} - d _ {g _ {2}}\right) - V _ {1} \left(d _ {a _ {1}} - d _ {g _ {1}}\right)
$$
and since the balloon volume will decrease with cooling of the lifting gas:
$$
V _ {2} = V _ {1} \frac {T g _ {2}}{T g _ {1}}
$$
(assuming constant p)
Therefore, again making use of the assumptions that:
$$
P _ {a _ {1}} = P _ {a _ {2}} = P _ {g _ {1}} = P _ {g _ {2}} = P
$$
and:
$$
T g _ {2} = T a _ {1} = T a _ {2} = T _ {2}
$$
Combining equation (2) and equation (7), we have:
$$
\begin{array}{l} \Delta L = V _ {1} \left[ \frac {T _ {2}}{T _ {g _ {1}}} \left(\frac {p}{R _ {a} T _ {2}} - \frac {p}{R _ {g} T _ {2}}\right) - \left(\frac {p}{R _ {a} T _ {2}} - \frac {p}{R _ {g} T _ {g}}\right) \right] \\ = V _ {1} \left(\frac {p}{R _ {a} T _ {g _ {1}}} - \frac {p}{R _ {g} T _ {g _ {1}}} - \frac {p}{R _ {a} T _ {2}} + \frac {p}{R _ {g} T _ {g _ {1}}}\right) \\ \end{array}
$$