### Thermodynamics: Gas Volume

- Mole
- Gas Volume 🢀
- States of Matter
- Heat
- Enthalpy
- Thermodynamics
- Adiabatic Process
- Mass Energy Conservation
- Carnot Engine

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

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} m^{3}
(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:

where

*p* = pressure [Pa]

*V* = volume

*n* = number of moles

*R* = universal gas constant = 8.314462 [SI units]

*T* = temperature [Kelvin]

It follows that:

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 O_{2} 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

^{3}(at STP)

Should we conclude that 1 kg oxygen remains enclosed in 0.7 m^{3}? No, the oxygen will expand arbitrarily.
The interpretation should be: 1 kg oxygen in a container with 0.7 m^{3}
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].

As an example let us compute the volume of *m* kg air at room conditions.
The air is composed of about 80% nitrogen N_{2} and 20% oxygen O_{2}:

*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)

_{air}= (m / 0.029) ⋅ 8.314462 ⋅ 293.15 / 101325 = 0.829 m

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

_{spec}⋅ T)

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:

_{air}= p / (R

_{air}⋅ T) = 101325 / (287.058 ⋅ 273.15) = 1.29 (at STP)

_{air}= p / (R

_{air}⋅ T) = 101325 / (287.058 ⋅ 293.15) = 1.20 (at NTP)

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 m^{3}:

_{air}= 50 ⋅ 1.20 = 60 kg

We can keep in mind: 1 m^{3} of air weighs more than 1 kg.
Air is about 1000-times lighter than water.