$$
359 \times \frac{273.2 - 59.5}{273.2} \times \frac{1013.3}{120} = 2370 \text{ ft.}^3 \text{ (the mean molar volume at 15 km for Jan.1943 over Lakehurst, N.J.)}
$$
This volume data was computed for all levels given. Data was "borrowed" from other stations in the same latitude to piece out the 20 km soundings as needed.
II. Lifts were computed for various molar volumes for balloons between 7.5 and 75 feet diameter in the following manner:
Given
- purity of Hydrogen 99.7%
- impurity as oxygen 0.3%
- computed molecular wt. 2.11 #/1 mol
- Molecular weight of dry air as computed from data reported at 10 km. in Handbook of Chemistry and Physics. 28.764 #/1 mol
To find the lift of a 20' D balloon at an altitude where the molar volume is 1000 ft.³:
Volume 20 ft. D Balloon = 4190 ft.³
$$
\text{Lift/Balloon} = \frac{\text{Balloon Volume} \times (\text{Difference in molecular wts. of air\#hgn})}{\text{Molar Volume at a given altitude}}
$$
or
$$
\text{Total Lift of gas in \#/Balloon} = \frac{\text{ft.}^3/\text{Balloon} \times (\text{\#/mol})}{\text{ft.}^3/\text{mol}}
$$
for the 20 foot diameter balloon:
$$
\text{Lift} = \frac{4190 \times (28.76 - 2.11)}{1000} = 111.7\# \text{ lift from a 20 foot diameter sphere of hydrogen at an altitude where the molar volume is 1000 ft.}^3.
$$
The lifts were plotted against molar volume for each size balloon. The altitudes corresponding to various molar volumes for Lakehurst and Albuquerque in January and in August 1943 as computed above were plotted on the left margin of the chart.
The family of curves was plotted on log paper and is included in the appendix with the basic sounding data.
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