Air has an average molar mass of 29.0 g/mol. The density of air at 1.00 atm and 30 °C is:?

Question

Air has an average molar mass of 29.0 g/mol. The density of air at 1.00 atm and 30 °C is:?

Answer 1

I think that the volume of 1 mole of a gas at 1 atm and 0C is 22.4 L. So at 30C it would be 303/273 X 22.4 = 24.9 L. Density would be 29.0/24.9 = 1.16 g/L

Answer 2

We can use the Ideal Gas Law (PV = nRT) to solve this problem, but we'll have to first determine the volume of air, V.

With the information given:

moles = mass / (molar mass)

=> mass = moles * molar mass, which I'll rewrite as:

m = n * M

density = mass / volume

=> volume = mass / density, which I'll rewrite as:

V = m / d

From above, we know that m = nM, so:

V = nM / d

Since PV = nRT:

P * (nM / d) = nRT

=> nPM / d = nRT

=> PM / d = RT

=> d = PM / RT

d = PM / RT

where,

d = Density in g/L

P = Pressure = 1.00 atm

M = Molar Mass = 29.0 g/mol

R = Ideal Gas Constant = 0.0821 L*atm/K*mol

T = Temperature = 30°C = 303.15 K

=> d = (1.00 * 29.0) / (0.0821 * 303.15)

=> d = 1.165 g/L

Therefore, the density of the air is: 1.165 g/L