This relationship is expressed as:
(5)
$$
\Delta p = \Delta z \left( \frac{dp}{dz} \right)_{\text{air}} (1 - B)
$$
where $ B = \frac{M_g}{M_a} $ , $ M_g $ & $ M_a $ are molecular weights of lifting gas and air, respectively.
Since, for a full balloon, $ \Delta z $ is constant at any altitude, and B (for our discussion) is a constant:
(6)
$$
Q \propto \sqrt{ \frac{ \left( \frac{dp}{dz} \right)_{\text{air}} }{ d_g } }
$$
The mass rate of flow is equal to the density of the lifting gas multiplied by the volumetric rate of flow:
(7)
$$
L = Q d_g \propto \sqrt{ \left( \frac{dp}{dz} \right)_{\text{air}} } \, d_g
$$
Since the number of openings will not change with altitude, equation (7) expresses the relationship for mass rate of flow from a full balloon for any altitude. The leakage at any altitude may be expressed as a function of leakage at sea level:
$$
\frac{L_z}{L_0} = \left( \frac{ \left( \frac{dp}{dz} \right)_{\text{air}-z} }{ \left( \frac{dp}{dz} \right)_{\text{air}-0} } \cdot \frac{ d_{g_z} }{ d_{g_0} } \right)^{\frac{1}{2}}
$$
As an example, let us compare the leakage rates of a lifting gas through a full balloon at sea level, at 40,000 feet and at 100,000 feet.
Altitude
$(dp/dz)_{\text{air}}$
$d_g$
0
$\frac{1}{27}$
$\frac{1013}{288R}$
40,000
$\frac{1}{112}$
$\frac{188}{218R}$
100,000
$\frac{1}{1880}$
$\frac{10.9}{218R}$
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