Ideal Monatomic Gases

127. In the preceding chapter it was proven that the introduction of probability considerations into the mechanical and electrodynamical theory of heat is justifiable and necessary, and from the general connection between entropy $S$ and probability $W$, as expressed in equation (164), a method was derived for calculating the entropy of a physical system in a given state. Before we apply this method to the determination of the entropy of radiant heat we shall in this chapter make use of it for calculating the entropy of an ideal monatomic gas in an arbitrarily given state. The essential parts of this calculation are already contained in the investigations of L. Boltzmann42 on the mechanical theory of heat; it will, however, be advisable to discuss this simple case in full, firstly to enable us to compare more readily the method of calculation and physical significance of mechanical entropy with that of radiation entropy, and secondly, what is more important, to set forth clearly the differences as compared with Boltzmann’s treatment, that is, to discuss the meaning of the universal constant $k$ and of the finite region elements $G$. For this purpose the treatment of a special case is sufficient.

128. Let us then take $N$ similar monatomic gas molecules in an arbitrarily given thermodynamic state and try to find the corresponding entropy. The state space is six-dimensional, with the three coordinates $x$, $y$, $z$, and the three corresponding moments $m\xi$, $m\eta$, $m\zeta$, of a molecule, where we denote the mass by $m$ and velocity components by $\xi$, $\eta$, $\zeta$. Hence these quantities are to be substituted for the $\phi$ and $\psi$ in Sec. 126. We thus obtain for the size of a region element $G$ the sextuple integral

where, for brevity

If the region elements are known, then, since the macroscopic state of the system of molecules was assumed as known, the numbers $N_1$, $N_2$, $N_3,\dots$ of the molecules which lie in the separate region elements are also known, and hence the distribution densities $w_1$, $w_2$, $w_3,\dots$ (166) are given and the entropy of the state follows at once from (173).

129. The theoretical determination of $G$ is a problem as difficult as it is important. Hence we shall at this point restrict ourselves from the very outset to the special case in which the distribution density varies but slightly from one region element to the next—the characteristic feature of the state of an ideal gas. Then the summation over all region elements may be replaced by the integral over the whole state space. Thus we have from (176) and (167)

in which $w$ is no longer thought of as a discontinuous function of the ordinal number, $i$, of the region element, where $i=1$, $2$, $3,\dots n$, but as a continuous function of the variables, $x$, $y$, $z$, $\xi$, $\eta$, $\zeta$, of the state space. Since the whole state region contains very many region elements, it follows, according to (167) and from the fact that the distribution density $w$ changes slowly, that $w$ has everywhere a small value.

Similarly we find for the entropy of the gas from (173):

Of course the whole energy $E$ of the gas is also determined by the distribution densities $w$. If $w$ is sufficiently small in every region element, the molecules contained in any one region element are, on the average, so far apart that their energy depends only on the velocities. Hence:

where ${\xi }_1{\eta }_1{\zeta }_1$ denotes any velocity lying within the region element $1$ and $E_0$ denotes the internal energy of the stationary molecules, which is assumed constant. In place of the latter expression we may write, again according to (176),

130. Let us consider the state of thermodynamic equilibrium. According to the second principle of thermodynamics this state is distinguished from all others by the fact that, for a given volume $V$ and a given energy $E$ of the gas, the entropy $S$ is a maximum. Let us then regard the volume

and the energy $E$ of the gas as given. The condition for equilibrium is $\delta S=0$, or, according to (179),

and this holds for any variations of the distribution densities whatever, provided that, according to (167)  and (180), they satisfy the conditions

This gives us as the necessary and sufficient condition for thermodynamic equilibrium for every separate distribution density $w$:

or

where $\alpha$ and $\beta$ are constants. Hence in the state of equilibrium the distribution of the molecules in space is independent of $x$, $y$, $z$, that is, macroscopically uniform, and the distribution of velocities is the well-known one of Maxwell.

131. The values of the constants $\alpha$ and $\beta$ may be found from those of $V$ and $E$. For, on substituting the value of $w$ just found in (178) and taking account of (177) and (182), we get

and on substituting $w$ in (181) we get

or

Solving for $\alpha$ and $\beta$ we have

From this finally we find, as an expression for the entropy $S$ of the gas in the state of equilibrium with given values of $N$, $V$, and $E$,

132. This determination of the entropy of an ideal monatomic gas is based solely on the general connection between entropy and probability as expressed in equation (164); in particular, we have at no stage of our calculation made use of any special law of the theory of gases. It is, therefore, of importance to see how the entire thermodynamic behavior of a monatomic gas, especially the equation of state and the values of the specific heats, may be deduced from the expression found for the entropy directly by means of the principles of thermodynamics. From the general thermodynamic equation defining the entropy, namely,

the partial differential coefficients of $S$ with respect to $E$ and $V$ are found to be

Hence, by using (186), we get for our gas

and

The second of these equations

contains the laws of Boyle, Gay Lussac, and Avogadro, the last named because the pressure depends only on the number $N$, not on the nature of the molecules. If we write it in the customary form:

where $n$ denotes the number of gram molecules or mols of the gas, referred to $O_2=32\text{ gr}.$, and $R$ represents the absolute gas constant

we obtain by comparison

If we now call the ratio of the number of mols to the number of molecules $\omega$, or, what is the same thing, the ratio of the mass of a molecule to that of a mol, $\omega =\frac{n}{N}$, we shall have

From this the universal constant $k$ may be calculated, when $\omega$ is given, and vice versa. According to (190) this constant $k$ is nothing but the absolute gas constant, if it is referred to molecules instead of mols.

From equation (188)

Now, since the energy of an ideal gas is also given by

where $c_v$ is the heat capacity of a mol at constant volume in calories and $A$ is the mechanical equivalent of heat:

it follows that

and further, by taking account of (193)

as an expression for the heat capacity per mol of any monatomic gas at constant volume in calories.43

For the heat capacity per mol at constant pressure, $c_p$, we have as a consequence of the first principle of thermodynamics:

and hence by (198)

as is known to be the case for monatomic gases. It follows from (195) that the kinetic energy $L$ of the gas molecules is equal to

133. The preceding relations, obtained simply by identifying the mechanical expression of the entropy (186) with its thermodynamic expression (187), show the usefulness of the theory developed. In them an additive constant in the expression for the entropy is immaterial and hence the size $G$ of the region element of probability does not matter. The hypothesis of quanta, however, goes further, since it fixes the absolute value of the entropy and thus leads to the same conclusion as the heat theorem of Nernst. According to this theorem the “characteristic function” of an ideal gas44 is in our notation

where $a$ denotes Nernst’s chemical constant, and $b$ the energy constant.

On the other hand, the preceding formulæ (186), (188), and (189) give for the same function $\Phi$ the following expression:

where for brevity $a\prime$ is put for:

From a comparison of the two expressions for $\Phi$ it is seen, by taking account of (199) and (193), that they agree completely, provided

This expresses the relation between the chemical constant $a$ of the gas and the region element $G$ of the probability.45

It is seen that $G$ is proportional to the total number, $N$, of the molecules. Hence, if we put $G=Ng$, we see that $g$, the molecular region element, depends only on the chemical nature of the gas.

Obviously the quantity $g$ must be closely connected with the law, so far unknown, according to which the molecules act microscopically on one another. Whether the value of $g$ varies with the nature of the molecules or whether it is the same for all kinds of molecules, may be left undecided for the present.

If $g$ were known, Nernst’s chemical constant, $a$, of the gas could be calculated from (201) and the theory could thus be tested. For the present the reverse only is feasible, namely, to calculate $g$ from $a$. For it is known that $a$ may be measured directly by the tension of the saturated vapor, which at sufficiently low temperatures satisfies the simple equation46

(where $r_0$ is the heat of vaporization of a mol at $0^{\circ }$ in calories). When $a$ has been found by measurement, the size $g$ of the molecular region element is found from (201) to be

Let us consider the dimensions of $g$.

According to (176) $g$ is of the dimensions $\left[{\text{ erg}}^3{\text{ sec}}^3\right]$. The same follows from the present equation, when we consider that the dimension of the chemical constant $a$ is not, as might at first be thought, that of $R$, but, according to (202), that of $Rlog\frac{p}{T^{\frac{5}{2}}}$.

134. To this we may at once add another quantitative relation. All the preceding calculations rest on the assumption that the distribution density $w$ and hence also the constant $\alpha$ in (183) are small (Sec. 129). Hence, if we take the value of $\alpha$ from (184) and take account of (188), (189) and (201), it follows that

When this relation is not satisfied, the gas cannot be in the ideal state. For the saturated vapor it follows then from (202) that $e^{-\frac{A{r}_0}{RT}}$ is small. In order, then, that a saturated vapor may be assumed to be in the state of an ideal gas, the temperature $T$ must certainly be less than $\frac{A}{R}{r}_0$ or $\frac{r_0}{2}$. Such a restriction is unknown to the classical thermodynamics.