(2) From the mean sounding data at 49,200 ft. (15 km.)
over Lakehurst, N.J. (Jan. 1943).
Temperature =-59.5 $ ^{\circ} \mathrm{C} $
Pressure = 120 mb.
$$
\text{Molar volume}_{\text{standard}} \times \frac{\text{Temperature}(\text{observed})}{\text{Temperature}(\text{standard})} \times \frac{\text{Pressure}(\text{standard})}{\text{Pressure}(\text{observed})}
$$
= Molar volume at observed conditions.
$$
3 5 9 x \frac {2 7 3 . 2 - 5 9 . 5}{2 7 3 . 2} x \frac {1 0 1 3 . 3}{1 2 0} = 2 3 7 0 f t. ^ {3}
$$
This is the mean pound molar volume at 15 km for Jan. 1943 over Lakehurst, N. J. This volume data was computed for levels up to 100,000 ft. over several stations and may be found in Appendix 3, plotted on the left hand side of figures 19 and 20.
B. Load-Diameter Maximum Altitude Relationships
Molar volume is related to buoyancy in the following fashion. Using 98% hydrogen of molecular weight, 2.11 lb./mol. and dry air of molecular weight 28.76 lb./mol., a buoyancy equal to the difference, 26.65 lb/mol. (See Table IV) is available whenever one pound molecular weight of hydrogen displaces one pound molecular weight of dry air under the same conditions of temperature and pressure.
TABLE IV
Buoyancy per Pound-Mol.
Helium (98%)
24.6 #/#mol, or
11.1 kg/#mol
Hydrogen (98%)
26.6#/#mol, or
12.1 kg/#mol