1, 2, 3 ... without omissions; (iv) partitions into the elements 1, 2, 3... in which each must appear twice at least, except the largest part, which need only appear once, but may appear any number of times; (v) the partitions described in the last paragraph. Taking Euler's example of x=10, omitting for brevity the signs connecting the parts, and writing, ex. gr., 1+1+2+2+4 as 12224, the partitions of the different kinds are for (i) 110, 173, 155, 137, 19, 133, 1432, 1235, 37, 52; for (ii) 10, 19, 28, 37, 46, 127, 136, 145, 235, 1234; for (iii) 110, 182, 162o, 1423, 1224, 1523, 18223, 1233, 1234; for (iv) 10, 128, 146, 164, 182, 324, 226, 12224, 1423, 25; for (v) 10, 19, 127, 1225, 1233, 37, 28, 136, 1234, 242. 1 + P(1, 2) 4 + P(1, . . . 4)3 + P(1,... 6) 2+1=1+3+3+2+1; while (5) gives P(1, 2)9+ P(1,2,3,4)4=5+5. There is a formula implicitly given by Euler (Opera minora collecta, vol. i. p. 93) which may be noticed here for the sake of completeness, viz. P(1, 3, 5...)x = P(1, 2, 3 ...)x-P(1, 2, 3 ...) (x-2) -P(1, 2, 3 ...) (x−4) + P(1, 2, 3 ...) (x −10) + &c., the terms being after the first alternately negative and positive in pairs, and the general term being P(1, 2, 3 ...) (x — 3n2±3n). This formula would, of course, be quite inappropriate for the calculation of P(1, 3, 5... x); for x=10 it gives 42-22-11+1=10. Cambridge, March 10, 1875. XXXIV. On the Work that may be gained during the Mixing of Gases. By LORD RAYLEIGH, M.A., F.R.S.* THE well-known fact that hydrogen tends to escape through fine apertures more rapidly than air enters to supply its place, even although the advantage of the greater pressure may be on the side of the air, proves that the operation of mixing the two gases has a certain mechanical value. In a common form of the experiment a tube containing hydrogen and closed at the upper end with a porous plug of plaster of Paris stands over water. In a short time the escape of hydrogen creates a partial vacuum in the tube, and the water rises accordingly. Whenever then two gases are allowed to mix without the performance of work, there is dissipation of energy, and an opportunity of doing work at the expense of low temperature heat has been for ever lost. The present paper is an attempt to calculate this amount of work. * Communicated by the Author. The result at which I have arrived is extremely simple. It appears that the work that may be done during the mixing of the volumes U1 and V2 of two different gases is the same as that which would be gained during the expansion of the first gas from volume v to volume v1+v, together with the work gained during the expansion of the second gas from v2 to v1 +v1⁄2, the expansions being supposed to be made into vacuum. Now these expansions may be considered actually to take place; and thus the rule is brought under Dalton's principle that each gas behaves to the other as a vacuum. It is understood that the gases follow the common law of independent pressures, so that the total pressure is always the sum of those which would be exerted by each gas in the absence of the other. We will take first the case when one gas is condensable, and estimate how much work must be done in order to separate the components of a mixture. Suppose, then, that a long cylinder, closed at the bottom, contains a uniform mixture of (for example) hydrogen and steam confined under a piston, and that the walls of the cylinder are maintained at a constant temperature. When the piston descends, heat will be generated; but the operation is supposed to proceed so slowly, that not only is the temperature rigorously constant throughout, but also the mixture is at any time in that condition which it would finally attain were the descent of the piston arrested. The pressure on the piston resisting the descent is by hypothesis the sum of those which it would experience from the hydrogen and steam separately*. When the space under the piston is reduced to that which the given quantity of steam is capable of saturating at the given temperature, condensation commences and continues as the steamspace is gradually diminished." By carrying this process sufficiently far, the condensation of the steam may be effected with any desired degree of completeness, and thus the water and hydrogen separated. A second movable piston may now be inserted immediately over the condensed water, and a very gradual expansion allowed until the original total volume is recovered. If the second piston be allowed free motion, the constituents of the original mixture are now separated, under equal pressures, and occupying the same total volume as before; and the question is, how much work has been expended in arriving at this state of things? In view of the fact that during the first part of the operation the hydrogen and steam press independently, it is clear that the total work done is the same as that which would be required to * For the sake of simplicity we may suppose a vacuum on the other side of the piston, though, of course, any constant pressure would give finally no result. to v2 if no compress the hydrogen from the original volume v1+v, to the volume if no steam were present, together with the work necessary to compress the steam from hydrogen were present. And since every step of this process is reversible, the same amount of work might be gained in making the mixture, and is dissipated if the mixture is allowed to take place by free diffusion. The same argument will apply when the condensation of one of the gases is effected by chemical means. Suppose, for example, that we have a mixture of carbonic anhydride and hydrogen at a red heat, and that it is proposed to absorb the carbonic gas with quicklime. It has been proved by Debray that at every temperature above a certain point carbonic gas in contact with quicklime and carbonate has a definite tension; any excess will be absorbed by the lime, and any deficiency supplied by a decomposition of the carbonate. If the tension of the carbonic gas in the given mixture be higher than that proper to the temperature, absorption will take place in an irreversible manner. In order to prevent dissipation, the mixture of gases must be first expanded until the tension of the carbonic gas is no higher than that corresponding to the temperature at which it is proposed to work. When the contact is made, the mixture may be very slowly condensed, so that after the point is passed at which chemical action commences, the tension of the carbonic anhydride remains constant. This process may be continued until nearly all the carbonic anhydride is absorbed. The hydrogen may then be separated. The space over the carbonate of lime must next be slowly increased until the original quantity of carbonic gas has been again evolved, when the connexion with the quicklime must be cut off. It now only remains to reduce the separated gases to the same pressure and to a total volume equal to that of the original mixture. From the preceding considerations we may, I think, infer that the law above stated is general whenever the gases really press independently; for it is difficult to see how its truth could depend on what would seem to be the accident of the existence or nonexistence of a chemical capable of absorbing one or other of the gases. It is worthy of notice that exactly the same rule applies for the mechanical value of the separation of two gases, even when the pressures are different; for we get the same result whether we first before mixing allow the pressures to become equal and add the work gained in this process to that due to the subsequent mixing, or whether we calculate at once the work due to the separate expansion of the two gases from their original volumes to the total volume of the mixture. Phil. Mag. S. 4. Vol. 49. No. 325. April 1875. Y In like manner the work that can be gained during the mixing of any number of pure and different gases, which press independently, is the sum of those due to the expansions of the several gases from their original to their final volumes, where the volume of a gas is understood to mean the space in which the is confined. gas The next problem which presents itself is that of finding the work that may be done during the mixture of two quantities of mixed gases-for example, oxygen and hydrogen. Suppose the two mixtures to be contained in a cylinder, and separated from one another by a piston which moves freely. The rule is that the work required to be estimated is that which would be gained during the equalization of the oxygen-pressures if the hydrogen were annihilated, together with that which would be gained during the equalization of the hydrogen-pressures if the oxygen were annihilated. If the proportions of the gases are the same in the two mixtures, and also the total pressures, there is, of course, no possibility of doing work. If, on the other hand, the gas on the one side of the piston be pure oxygen, and on the other side pure hydrogen, the more general rule reduces to that already given for pure gases. I now pass to another proof of the fundamental rule, depending on the possibility of separating two gases of different densities by means of gravity. In a vertical column maintained at a uniform temperature, two gases which press independently will arrange themselves each as if the other were absent. Consequently, if there be any difference in density, the percentage composition will vary at different heights, and a partial separation of the gases is thus effected. Imagine now a large reservoir containing gas at sensibly constant pressure, on which is mounted a tall narrow vertical tube; and first, in order to understand the operation more easily, let there be only one kind of gas present. If p be the pressure and p the density, p=kp, since the temperature is constant; and if z be the height measured from the reservoir in which the pressure is P, dp=-gpdz: =-updz, if μ=g+k; expresses the law of variation of pressure with height. Suppose now that a small quantity of gas of volume v is (1) removed from the top of the tube, (2) compressed to volume v until it is of the same pressure as the gas in the reservoir, (3) allowed to fall through the height z to the level of the reservoir, and (4) forced 0 into the reservoir. The effect of this series of operations is nil, and there can be neither gain nor loss of work. The work gained in the first operation is consumed in the four h, since pv=Pv, so that attention may be confined to the second and third operations. Now the work consumed in the compression from v to vo is Spdv=Poo Το dv and the work gained in the descent = P P And these are equal, since P so that on the whole no work is lost or gained. The case is different when there are two kinds of gases present. Although, as before, the work gained in the first operation is consumed in the fourth, there is no longer compensation in the second and third operations. If P, and P, are the partial pressures in the reservoir, the work required for the compression from v to v is On the other hand, the work gained in the descent is if Q, Q are the partial pressures of the abstracted gas after condensation to volume vo. Thus, on the whole, if W be the work done on the gases, since Q+Q2=P1+P2, |