(2) Density:
$$
d = \frac {P}{R T}
$$
$$
p, R, T = \mathrm {p r e s s u r e}, \mathrm {s p e c i f i c g a s c o n s t a n t}, \mathrm {a n d} \mathrm {t e m p e r a t u r e} \mathrm {o f} \mathrm {t h e a i r o r l i f t i n g g a s}
$$
(3) Let:
$$
B = \frac {R _ {a}}{R _ {g}} \quad \left(= \frac {M _ {g}}{M _ {a}}\right)
$$
At any two positions:
$$
L _ {1} = V _ {1} \left(d _ {a _ {1}} - d _ {g _ {1}}\right)
$$
$$
L _ {2} = V \left(d _ {0 _ {2}} - d _ {g _ {2}}\right)
$$
Investigating the gain of superheat, since there is no change of volume $ V_{1}=V_{2} $ and:
$$
\Delta L = L _ {2} - L _ {1} = V _ {1} \left(d _ {a _ {2}} - d _ {a _ {1}} - d _ {g _ {2}} + d _ {g _ {1}}\right)
$$
Assume now that the balloon carries no internal pressure and that the difference in lift does not cause the balloon system to pass through any appreciable atmospheric pressure difference (in the case where the balloon is floating at 40,000 ft. MSL a change of 1000 ft. would be only 9 mb, or a 5% change).
Therefore:
$$
P _ {a _ {1}} = P _ {a _ {2}} = P _ {g _ {1}} = P _ {g _ {2}} = P
$$
Assume also that initially the air and lifting gas are at the same temperature and that the air passes through no appreciable temperature change. Then:
$$
T a _ {1} = T a _ {2} = T g _ {1} = T _ {1}
$$
Then, making use of our two assumptions and substituting equation (2) into equation (4), we have:
$$
\begin{array}{l} \Delta L = V _ {p} \left(\frac {1}{R _ {a} T _ {1}} - \frac {1}{R _ {a} T _ {1}} - \frac {1}{R _ {g} T _ {g _ {2}}} - \frac {1}{R _ {g} T _ {1}}\right) \\ = \frac {V _ {p}}{R _ {g}} \left(\frac {1}{T _ {1}} - \frac {1}{T _ {g _ {2}}}\right) \\ \end{array}
$$
and:
$$
\frac {\Delta L}{L _ {1}} = \frac {\frac {1}{R _ {g}} \left(\frac {1}{T _ {1}} - \frac {1}{T _ {q 2}}\right)}{\frac {1}{T _ {1}} \left(\frac {1}{R _ {a}} - \frac {1}{R _ {g}}\right)}
$$