Now, it can be shown that the product of pressure and volume of a gas, pv, is equal to ⅔rds of the energy of translation of all molecules of the gas, or
pv = ⅔NR,
where N stands for the number of molecules in unit volume, and R for their energy of translation; inasmuch as a pressure diminishing a volume is of the nature of work, or energy. For one gram of air at O° C. and 76 cms. pressure (normal temperature and pressure), the pressure (p), measured in grams per square centimetre, is 1033, and the volume (v) is 773·3 cubic centimetres; and the raising of the temperature through 1°, as was shown before, requires 2927 gram-centimetres of work. Further, since the product of pressure into volume is equal to ⅔rds of the energy due to motion, or the translational energy of the gas,
NR = 3⁄2(pv) = 3⁄2 × 2927 = 4391 gram-centimetres.
Dividing this number by 42,380, the mechanical equivalent of heat, or the number of gram-centimetres corresponding to one calory, the quotient is 0·1040 calory. If the energy of the air were due to the translational motion of its molecules, we should expect this number, 0·1040, to stand for the specific heat of air at constant volume; but it has been found equal to 0·1683, as already shown.
We have seen that to convert specific heat at constant volume into specific heat at constant pressure 0·0692 must be added. Hence at constant pressure the specific heat of such an ideal gas should be 0·1732. And the relation between specific heat at constant volume and that at constant pressure should be 0·1040 to 0·1732, or 1 to 1⅔. The conclusion to be drawn from these numbers for air, 0·1683 and 0·2375, which bear to each other the ratio of 1: 1·41, is that air cannot be such an ideal gas; that in communicating heat to it some of that heat must be employed in performing some kind of work other than that of raising its temperature. What this work may possibly be we shall consider later.
But Kundt and Warburg found, from their experiments on the ratio between the specific heats of mercury gas, this ideal ratio, 1 to 1⅔; and Professor Ramsay obtained the same ideal ratio, or one very close to it indeed, 1 to 1·659, for argon. He subsequently found this ideal ratio also to hold for helium (1 to 1·652), and it must therefore be concluded that such gases possess only three degrees of freedom; or, in other words, their molecules, when heated, expend all the energy imparted to them in translational motion through space.
This is the consequence which we should infer from the supposition that such molecules are hard, smooth, elastic spheres. Were they each composed of two atoms, we should have to picture them as dumbbell-like structures; and here we enter on a theoretical conception put forward by Professor Boltzmann, but which has not been accepted universally by physicists.
| Fig. 5. | Fig. 6. |
Boltzmann imagines that to the three “degrees of freedom” of a single atom molecule there may be added, provided the molecule consists of two atoms, two other degrees of freedom, namely, freedom to rotate about two planes at right angles to each other. The knobs at the end of each imaginary dumb-bell may revolve round a central point in the handle joining them, and it is clear that they may revolve in one horizontal and in one vertical plane, as shown in Fig. 5. Such diatomic molecules are said to possess five “degrees of freedom.” They will not, it is supposed, rotate round the line joining the centres of the spheres, because, as before said, the atoms are pictured as perfectly smooth. But if the molecules are triatomic, as, for example, CO2 or N2O, they will have six degrees of freedom, for with the addition of an additional atom they have an additional plane of rotation (see Fig. 6). Boltzmann has attempted to show that the ratio of the specific heats of diatomic molecules should be as 1 to 1·4. In actual fact it approximates to that number. For the commoner gases it is—