Question

Forty miles above the earth's surface the temperature is 290 K and the pressure is only 0.19 mm Hg. What is the density of air (in grams per liter) at this altitude? (Assume the molar mass of air is 29 g/mol.)

Answer 1

We can use the ideal gas law to solve this problem:

PV = nRT

Where: P = pressure = 0.19 mm Hg = 0.000252 kPa (convert to kPa) V = volume (we'll assume 1 liter to make the density calculation easier) n = number of moles of air R = gas constant = 8.31 J/(mol*K) T = temperature = 290 K

First, let's convert the pressure:

0.19 mm Hg = 0.19/760 kPa 0.19/760 kPa = 0.000252 kPa

Now we can rearrange the ideal gas law to solve for density:

n/V = P/RT

n/V = (0.000252 kPa)/(8.31 J/(mol*K) * 290 K)

n/V = 1.204 * 10^(-5) mol/L

To get density, we need to multiply by the molar mass of air:

density = (1.204 * 10^(-5) mol/L) * 29 g/mol

density = 0.000349 g/L

Therefore, the density of air at this altitude is approximately 0.000349 grams per liter (g/L).