ATOC 3500/CHEM 3151 – Problem 2 Answers
(a) Repeat Problem 1, only this time express the answer in units of mass density, , in grams per
cubic centimeter (g cm-3). In other words, find an expression for  in terms of P in mbar and T
in K. Use 28.9 g mole-1 for the molecular weight of air.
Let’s change R into the units that we need, just like we did for Problem 1:
R = (0.08206 L atm K-1 mol-1) x (1000 cm3/1 L) x (mol/28.9 g) x (1013.25 mb/atm)
= 2877 cm3 mb K-1 g-1
[M] = P (in mb)/ (2877 cm3 mb K-1 g-1) T (in K)
= 3.46x10-4 P/T (cm-3 mb-1 K g)
Reality check - example - for 1013 mb, 298 K, density = 1.18 x 10-3 g cm-3
(b) By what factor must you multiply  in g cm-3 in order to convert it into grams per cubic
meter (g m-3)? Note that 1 m3 = (100 cm)3 = 1,000,000 cm3.
Answer - 1 x 106
Density (kg m-3) = 3.46x10-4 P(mb)/T(K) cm-3 mb-1 K g x (100 cm)3/m3
= 3.46x10-4 x cm-3 mb-1 K g (1 x 106 cm3/m3) x P(mb)/T(K)
= 34.6 x P/T (mb-1 m-3 K g)
Example - for 1013 mb, 298 K, density = 1.18 kg m-3
This is a useful (and easy) number to remember...essentially, a cubic meter of air at sea level
weighs about 1 kg!