Density of Air

Problem

Figure 1. The Earth's atmosphere

The composition of air is roughly 80% nitrogen gas and 20% oxygen gas. Calculate the density of air under normal temperature and pressure conditions (NTP).

Normal temperature and pressure (NTP) is defined as the condition when the temperature is 20 degrees Celsius and the pressure is 1 atmosphere.

Hints

(1) Nitrogen gas and oxygen gas are diatomic.

(2) Gases in the Earth's atmosphere is of low concentration; this means the atmosphere can be modelled as an ideal gas.

Solution

Mass

Nitrogen and oxygen gas are diatomic, so the molar mass is

$$0.80 \left(2 \times 14.01 \; \frac{g}{mol} \right) + 0.20 \left(2 \times 16.00 \; \frac{g}{mol} \right) = 28.80 \; \frac{g}{mol}.$$

.

Calculate the mass of 1 mole of air:

$$1 \; mol \times 28.80 \; \frac{g}{mol} \times \frac{1 \; kg}{1000 \; g} = 0.02880 \; kg.$$

.

Therefore, 1 mole of air weighs 0.02880 kg.

Volume

To calculate the volume of 1 mole of air at NTP, the pressure and temperature are:

$$P = 1 \; atm = 101325 \; Pa$$

$$T = {20} \;^{\circ}C + 273.15 = 293.15 \; K$$

Substitute the known values into the ideal gas law formula and solve for the volume:

$$PV = nRT$$

$$(101325 \; Pa)V = (1 \; mol)\left(8.31 \; \frac{J}{K\cdot mol}\right)(293.15 \; K)$$

$$V = 0.024 \; m^3 .$$

Density

Simply divide the mass by the volume to get the density of air :

$${\rho}_{air} = \frac{0.02880 \, kg}{0.024 \, m^3}$$

$${\rho}_{air} = 1.2 \, \frac{kg}{m^3}$$