Question

Determine the density of air in grams per liter at a temperature of 20.5 deg. C and a pressure of 755 mmHg. Assume that air is approximately 20% O2 and 80% N2 (i.e. the mole fraction of oxygen is 0.20 and the mole fraction of nitrogen is 0.80)

Answer 1

Q: Determine the density of air in grams per liter at a temperature of 20.5 deg. C and a pressure of 755 mmHg. Assume that air is approximately 20% O2 and 80% N2 (i.e. the mole fraction of oxygen is 0.20 and the mole fraction of nitrogen is 0.80)

Solution:

pV = nRT, (eq. 1)

where P is the pressure of the gas, V is the volume of the gas(755 mmHg), n is the amount of substance of gas, T is the temperature of the gas(in Kelvins T=273+20.5=293.5K) and R is the ideal, or universal, gas constant (R = 62.364 L*mmHg*K⁻¹*mol⁻¹).

n = m/M, (eq. 2)

where m is the mass of a compound, M is the molar mass of the compound.

M(O₂) = 32 g/mol

M(N₂) = 28 g/mol

we can rewrite (eq. 1) together (eq. 2):

pV = mRT/M

since,

mᵢ = pVᵢMᵢ/RT. (eq. 3)

We can determine the density of air as mass of air divided by volume of air:

d(air) = m(air)/V(air) (eq. 4)

air mass is sum of oxygen mass and nitrogen mass

m(air) = m(O₂) + m(N₂) (eq. 5)

considering (eq. 3) and (eq. 5):

m(air) = pV₀₃₂ₗM₀₃₂/RT + pV₀₃₂ₗM₀₃₂/RT

considering (eq. 4):

d(air) = (p/RT)*(V₀₃₂ₗM₀₃₂/V) + (V₀₃₂ₗM₀₃₂/V)) (eq. 6)

the mole fraction of gas is

χᵢ = Vᵢ/V (eq. 7)

considering (eq. 6) and (eq. 7):

d(air) = (p/RT)*(χ₀₃₂ₗ*M₀₃₂ + χ₀₃₂ₗ*M₀₃₂) (eq. 6)

$$d(\text{air}) = \frac{p}{RT} \times \left( \chi_{\mathrm{O}_2} M_{\mathrm{O}_2} + \chi_{\mathrm{N}_2} M_{\mathrm{N}_2} \right)$$

d(air) = ((755 mmHg/((62.364 L*mmHg*K⁻¹*mol⁻¹)*(293.5K)))*(0.2*32 g/mol + 0.8*32 g/mol) = 1.188 g/L

Answer: the density of air is 1.188 grams per liter