Question

Because we are dealing in the Earth’s atmosphere (P = 1 atm, maximum), just about everything we discuss can be adequately described by the ideal gas law. This is important to remember as we proceed in this class.

a. In class, we will often use the fact that there are ~ 2.5 x 10¹⁹ molecules of air cm^-3 at sea level. Because you shouldn’t believe everything you hear, please show, mathematically, that this is the case. Start with the ideal gas law and state your assumptions

b. Carbon monoxide (CO) has a concentration of up to 10 ppm_v in an urban area. Express this in molecules cm^-3 if at sea level. Explain how you arrived at your answer. If you understand the concept, then no calculations are really necessary.

c. About how many molecules of air per cm³ are there at 15 km altitude? Use the scale height equation to estimate pressure and cite a source for the temperature at that altitude.

Answer 1

(a) PV = nRT

1atm× 1×10^-3L = n × 0.0821atm-L/K.mol × 298.15K

n = 4.0853×10^-5

Number molecules = number of moles × Avogadro's number = 4.0853×10^-5 × 6.022×10²³ = 2.46×10¹⁹ molecules/cm³

~ 2.5×10¹⁹molecules/cm³.

The assumptions considered are that temperature of atmosphere is 25℃ and gas follows ideal gas equation.

(b) concentration of CO = 10ppmv ( part per million by volume)

Means there are 10×10^-6= 10^-5 volume of CO in 1 volume of air.

Number of moles or molecules is directly proportional to the volume of the gas at same pressure and temperature.

Hence molecules of CO = 2.5×10¹⁹×10^-5molecule/cm³

= 2.5×10¹⁴ molecules/cm³. (Answer)

(c)

Temperature at the height of 15km is -56.5℃ or 216.65K.

P = P₀e^-(mgh/kT). Where m is average mass of air molecule, k = Boltzmann constant, P₀ is pressure at sea level.

Pressure at the height of 15km = 12044.6N/m²

(12044.6/101325) atm × 1×10^-3 L = n × 0.0821atm-L/K.mol × 216.65K

Number of moles in 1cm³ of air = 6.683×10^-6mol

Number of air molecules per cm³ at 15km above the sea level = 6.683×10^-6 × 6.022×10²³ = 4.02×10¹⁸

~ 4.0×10¹⁸molecules/cm³ . (Answer)