Page:Elementary Text-book of Physics (Anthony, 1897).djvu/250

236 by $$(C_{p} - C_{v})\Delta T.$$ We shall show (§ 332) that this work is also given by the product of the pressure by the increase in volume, or by $$p \Delta v.$$ Hence we have the relation $$p \Delta v = (C_{p} - C_{v})\Delta T.$$

The kinetic energy of molecular translation is $$\tfrac{1}{2}mnV^2,$$ and if $$C_{0}$$ represent its increase for arise in temperature of one degree, $$C_{0}\Delta T$$ represents its increase for the rise of temperature $$\Delta T.$$ But since $$pv = \tfrac{1}{3}mnV^2,$$ we have $$C_{0}\Delta T = \tfrac{3}{2}p\Delta v,$$ and hence $$C_{p} - C_{v} = \tfrac{2}{3}C_{0},$$ or Now on the supposition that the molecules are particles which have no energy except energy of translation, $$C_{0} = C_{v},$$ and hence $$\frac{C_{p}}{C_{v}} = \frac{5}{3}\cdot$$ We know by experiment that this is not always the case. For monatomic gases, such as mercury vapor, and possibly argon, $$\frac{C_{p}}{C_{v}} = 1.66;$$ but for the common diatomic gases it is more nearly $$\tfrac{7}{5} = 1.4,$$ and for gases with complex molecules, it is about $$\tfrac{4}{3} = 1.33.$$ Hence, in the case of gases with more than one atom in the molecule, the total energy is not merely the energy of translation, but includes other energy internal to the molecule. Boltzmann has shown that the ratio of the internal energy to the energy of translation is such as can be accounted for by supposing the monatomic molecules to be spheres or points, the diatomic molecules solids of revolution, and the more complex molecules irregular solids. It is likely that this is merely an artificial representation, since there is strong reason to believe that the atoms vibrate within the molecule and that the molecule is not rigid.

We have used $$C_{0}$$ to represent the increase in the energy of molecular translation in a unit of mass when the temperature rises one degree. If we represent the increase in the kinetic energy of a single molecule by $$\Delta \tfrac{1}{2}mV^2,$$ we have $$C_{0} = n\Delta \tfrac{1}{2}mV^2.$$ Now $$n$$ is the number of molecules in unit volume, which in this equation is the volume containing unit mass, so that $$\frac{1}{n}$$ is the mass of one molecule or $$m.$$ The gain in kinetic energy for a rise of temperature of