The composition of dry air

Dry air is a mixture of two main component gases together with traces of a number of other gases. It is reasonable to consider all these as one homogeneous substance but to deal separately with the water vapour present because the latter is condensable at everyday pressures and temperatures whereas the associated dry gases are not.

One method of distinguishing between gases and vapours is to regard vapours as capable of liquefaction by the application of pressure alone but to consider gases as incapable of being liquefied unless their temperatures are reduced to below certain critical values. Each gas has its own unique critical temperature, and it so happens that the critical temperatures

Of nitrogen and oxygen, the major constituents of dry air, are very much below the temperatures dealt with in air conditioning. On the other hand, the critical temperature of steam (374.2°C) is very much higher than these values and, consequently, the water vapour mixed with the dry air in the atmosphere may change its phase from gas to liquid if its pressure is increased, without any reduction in temperature. While this is occurring, the phase of the dry air will, of course, remain gaseous.




Fig. 2.1 Pressure-volume diagrams for dry air and steam. /a is an air temperature of 21°C and ts is a steam temperature of 21°C. tc is the critical temperature in each case.

fig. 2.1 pressure-volume diagrams for dry air and steam. /a is an air temperature of 21°c and ts is a steam temperature of 21°c. tc is the critical temperature in each case.
Figures 2.1(a) and 2.1 (b) illustrate this. Pressure-volume diagrams are shown for dry air and for steam, separately. Point A in Figure 2.1(a) represents a state of dry air at 21°C. It can be seen that no amount of increase of pressure will cause the air to pass through the liquid phase, but if its temperature is reduced to -145°C, say, a value less than that of the critical isotherm, fc (-140.2°C), then the air may be compelled to pass through the liquid phase by increasing its pressure alone, even though its temperature is kept constant.

In the second diagram, Figure 2.1(b), a similar case for steam is shown. Here, point S represents water vapour at the same temperature, 21°C, as that considered for the dry air. It is evident that atmospheric dry air and steam, because they are intimately mixed, will have the same temperature. But it can be seen that the steam is superheated, that it is far below its critical temperature, and that an increase of pressure alone is sufficient for its liquefaction.

According to Threlkeld (1962), the dry air portion of the atmosphere may be thought of as being composed of true gases. These gases are mixed together as follows, to form the major part of the working fluid:

Gas Proportion (%) Molecular mass

TOC o "1-5" h z Nitrogen 78.048 28.02

Oxygen 20.9476 32.00

Carbon dioxide 0.0314 44.00

Hydrogen 0.00005 2.02

Argon 0.9347 39.91

A later estimate by the Scientific American (1989) of the carbon dioxide content of the atmosphere is 0.035% with a projection to more than 0.040% by the year 2030. ASHRAE (1997) quote the percentage of argon and other minor components as about 0.9368%. From the above, one may compute a value for the mean molecular mass of dry air:

M = 28.02 x 0.78084 + 32 x 0.209476 + 44 x 0.000314 + 2.02 x 0.0000005 + 39.91 x 0.009347 = 28.969 kg kmol-1

Harrison (1965) gives the weighted average molecular mass of all the components as 28.9645 kg kmol’1 but, for most practical purposes, it may be taken as 28.97 kg kmol-1.

As will be seen shortly, this is used in establishing the value of the particular gas constant for dry air, prior to making use of the General Gas Law. In a similar connection it is necessary to know the value of the particular gas constant for water vapour; it is therefore of use at this juncture to calculate the value of the mean molecular mass of steam.

Since steam is not a mixture of separate substances but a chemical compound in its own right, we do not use the proportioning technique adopted above. Instead, all that is needed is to add the masses of the constituent elements in a manner indicated by the chemical formula:

M = 2 x 1.01 + 1 x 16 = 18.02 kg kmol“1

More exactly, Threlkeld (1962) gives the molecular mass of water vapour as 18.01528 kg kmol-1.

Posted in Engineering Fifth Edition