transformation.” We thus have an instance of the property that every irreversible change leaves an indelible imprint somewhere or other on the progress of events in the universe. That the process of diffusion obeys the laws of irreversible thermodynamics (if these laws are properly stated) is proved by the fact that the compensating transformations required to separate mixed gases do not essentially involve anything but transformation of energy. The process of allowing gases to mix by diffusion, and then separating them by a compensating transformation, thus constitutes an irreversible cycle, the outside effects of which are that energy somewhere or other must be less capable of transformation than it was before the change. We express this fact by stating that an irreversible process essentially implies a loss of availability. To measure this loss we make use of the laws of thermodynamics, and in particular of Lord Kelvin’s statement that “It is impossible by means of inanimate material agency to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects.”
| Substances. | Temp. | K. | Author. |
| Carbon dioxide and air | 0°C. | 0·1423 cm2/sec. | J. Loschmidt. |
| ””hydrogen | 0°C. | 0·5558 ” | ” |
| ””oxygen | 0°C. | 0·1409 ” | ” |
| ””carbon monoxide | 0°C. | 0·1406 ” | ” |
| ””marsh gas (methane) | 0°C. | 0·1586 ” | ” |
| ””nitrous oxide | 0°C. | 0·0983 ” | ” |
| Hydrogen and oxygen | 0°C. | 0·7214 ” | ” |
| ”” carbon monoxide | 0°C. | 0·6422 ” | ” |
| ”” sulphur dioxide | 0°C. | 0·4800 ” | ” |
| Oxygen and carbon monoxide | 0°C. | 0·1802 ” | ” |
| Water and ammonia | 20°C. | 1·250 ” | G. Hüfner. |
| ”” | 5°C. | 0·822 ” | ” |
| ”common salt (density 1·0269) | 0·355 ” | J. Graham. | |
| ” ”” | 14·33°C. | 1·020, 0·996, 0·972, 0·932 cm2/day. | F. Heimbrodt. |
| ”zinc sulphate (0·312 gm/cm³) | 0·1162 cm2/day. | W. Seitz. | |
| ”zinc sulphate (normal) | 0·2355 ” | ” | |
| ”zinc acetate (double normal) | 0·1195 ” | ” | |
| ”zinc formate (half normal) | 0·4654 ” | ” | |
| ”cadmium sulphate (double normal) | · · | 0·2456 ” | ” |
| ”glycerin (18n, 12n, 78n, 1·5n) | 10·14°C. | 0·356, 0·350, 0·342, 0·315 cm2/day. | F. Heimbrodt. |
| ”urea”” | 14·83°C. | 0·973, 0·946, 0·926, 0·883 cm2/day. | ” |
| ”hydrochloric acid | 14·30°C. | 2·208, 2·331, 2·480 cm2/day. | ” |
| Gelatin 20% and ammonia | 17°C. | 127·1 cm2/day. | A. Hagenbach. |
| ” ” carbon dioxide | · · | 0·845 ” | ” |
| ” ” nitrous oxide | · · | 0·509 ” | ” |
| ” ” oxygen | · · | 0·230 ” | ” |
| ” ” hydrogen | · · | 0·0565 ” | ” |
Let us now assume that we have any system such as the gases above considered, and that it is in the presence of an indefinitely extended medium which we shall call the “auxiliary medium.” If heat be taken from any part of the system, only part of this heat can be converted into work by means of thermodynamic engines; and the rest will be given to the auxiliary medium, and will constitute unavailable energy or waste. To understand what this means, we may consider the case of a condensing steam engine. Only part of the energy liberated by the combustion of the coal is available for driving the engine, the rest takes the form of heat imparted to the condenser. The colder the condenser the more efficient is the engine, and the smaller is the quantity of waste.
The amount of unavailable energy associated with any given transformation is proportional to the absolute temperature of the auxiliary medium. When divided by that temperature the quotient is called the change of “entropy” associated with the given change (see Thermodynamics). Thus if a body at temperature T receives a quantity of heat Q, and if T0 is the temperature of the auxiliary medium, the quantity of work which could be obtained from Q by means of ideal thermodynamic engines would be Q(1 − T0/T), and the balance, which is QT0/T, would take the form of unavailable or waste energy given to the medium. The quotient of this, when divided by T0, is Q/T, and this represents the quantity of entropy associated with Q units of heat at temperature T.
Any irreversible change for which a compensating transformation of energy exists represents, therefore, an increase of unavailable energy, which is measurable in terms of entropy. The increase of entropy is independent of the temperature of the auxiliary medium. It thus affords a measure of the extent to which energy has run to waste during the change. Moreover, when a body is heated, the increase of entropy is the factor which determines how much of the energy imparted to the body is unavailable for conversion into work under given conditions. In all cases we have
increase of unavailable energy temperature of auxiliary medium ＝ increase of entropy.
When diffusion takes place between two gases inside a closed vessel at uniform pressure and temperature no energy in the form of heat or work is received from without, and hence the entropy gained by the gases from without is zero. But the irreversible processes inside the vessel may involve a gain of entropy, and this can only be estimated by examining by what means mixed gases can be separated, and, in particular, under what conditions the process of mixing and separating the gases could (theoretically) be made reversible.
9. Evidence derived from Liquefaction of one or both of the Gases.—The gases in a mixture can often be separated by liquefying, or even solidifying, one or both of the components. In connexion with this property we have the important law according to which “The pressure of a vapour in equilibrium with its liquid depends only on the temperature and is independent of the pressures of any other gases or vapours which may be mixed with it.” Thus if two closed vessels be taken containing some water and one be exhausted, the other containing air, and if the temperatures be equal, evaporation will go on until the pressure of the vapour in the exhausted vessel is equal to its partial pressure in the other vessel, notwithstanding the fact that the total pressure in the latter vessel is greater by the pressure of the air.
To separate mixed gases by liquefaction, they must be compressed and cooled till one separates in the form of a liquid. If no changes are to take place outside the system, the separate components must be allowed to expand until the work of expansion is equal to the work of compression, and the heat given out in compression is reabsorbed in expansion. The process may be made as nearly reversible as we like by performing the operations so slowly that the substances are practically in a state of equilibrium at every stage. This is a consequence of an important axiom in thermodynamics according to which “any small change in the neighbourhood of a state of equilibrium is to a first approximation reversible.”
Suppose now that at any stage of the compression the partial pressures of the two gases are p1 and p2, and that the volume is changed from V to V − dV. The work of compression is (p1 + p2)dV, and this work will be restored at the corresponding stage if each of the separated gases increases in volume from V − dV to V. The ultimate state of the separated gases will thus be one in which each gas occupies the volume V originally occupied by the mixture.
We may now obtain an estimate of the amount of energy rendered unavailable by diffusion. We suppose two gases occupying volumes V1 and V2 at equal pressure p to mix by diffusion, so that the final volume is V1 + V2. Then if before mixing each gas had been allowed to expand till its volume was V1 + V2, work would have been done in the expansion, and the gases could still have been mixed by a reversal of the process above described. In the actual diffusion this work of expansion is lost, and represents energy rendered unavailable at the temperature at which diffusion takes place. When divided by that temperature the quotient gives the increase of entropy. Thus the irreversible processes, and, in particular, the entropy changes associated with diffusion of two gases at uniform pressure, are the same as would take place if each of the gases in turn were to expand by rushing into a vacuum, till it occupied the whole volume of the mixture. A more rigorous proof involves considerations of the thermodynamic potentials, following the methods of J. Willard Gibbs (see Energetics).
