Thermodynamics

Thermodynamics: Gas Volume

A specified amount of solid or fluid takes up a limited volume of space, that can be computed easily. In fact:

V = m / ρ

(1)

where V = volume, m = mass, ρ = density (i.e. specific weight). For example, the volume of 1 kg petrol is 1/720 = 1.4 · 10^-3 m³ (I will always use SI units except when otherwise stated).

But what is the volume of 1 kg oxygen? We have to use the ideal gas law:

p ⋅ V = n ⋅ R ⋅ T

(2)

where

p = pressure [Pa]
V = volume
n = number of moles
R = universal gas constant = 8.314462 [SI units]
T = temperature [Kelvin]

It follows that:

V = n ⋅ R ⋅ T / p

(3)

and we see that the volume depends not only on the amount of gas (n = number of moles) but also on the temperature and presssure. As an example, let us compute the volume of 1 kg oxygen at the so called Standard Temperature and Pressure (STP) conditions that are characterized by:

p = 1 atm = 101325 Pa.
T = 0° C = 273.15° K

One mole of oxygen O₂ has the mass 2 · 16 g = 0.032 kg. So, 1 kg oxygen contains n = 1 / 0.032 = 31.25 moles and its volume is

V = 31.25 ⋅ 8.314462 ⋅ 273.15 / 101325 = 0.7 m³ (at STP)

(4a)

Should we conclude that 1 kg oxygen remains enclosed in 0.7 m³? No, the oxygen will expand arbitrarily. The interpretation should be: 1 kg oxygen in a container with 0.7 m³ will execute 1 atm pressure on the container walls. It is in equilibrium with the enclosing air at 1 atm pressure.

The formula (3) can be easily generalized for the mass m of any gas with molar mass M [kg/mol].

V = (m / M) ⋅ R ⋅ T / p

(5)

As an example let us compute the volume of m kg air at room conditions. The air is composed of about 80% nitrogen N₂ and 20% oxygen O₂:

p = 1 atm = 101325 Pa
T = 21°C = 294.15°K (these conditions are called NTP - Normal Temperature and Pressure)
M_oxygen = 0.032 kg/mol
M_nitrogen = 0.028 kg/mol
M_air = 0.032 * 20% + 0.028 * 80% = 0.029 kg/mol (exact value is 0.0289647)

V_air = (m / 0.029) ⋅ 8.314462 ⋅ 293.15 / 101325 = 0.829 m

(6)

The equation (5) can be reformulated to get the density ρ of gas:

ρ = m / V = M ⋅ p / (R ⋅ T) = p / (R_spec ⋅ T)

(7)

Here I introduced the specific gas constant R_spec = R / M (M = molar mass). The specific gas constant for air is 287.058 [SI units] according to Wikipedia. The density of air is:

ρ_air = p / (R_air ⋅ T) = 101325 / (287.058 ⋅ 273.15) = 1.29 (at STP)

(8a)

ρ_air = p / (R_air ⋅ T) = 101325 / (287.058 ⋅ 293.15) = 1.20 (at NTP)

(8b)

Now we can calculate the density of gas and use the same equation (1) as for fluids and solids. As an example let us compute the weight of air in a living room with dimensions 5 x 4 x 2.5 = 50 m³:

mass m = V ⋅ ρ_air = 50 ⋅ 1.20 = 60 kg

We can keep in mind: 1 m³ of air weighs more than 1 kg. Air is about 1000-times lighter than water.