Page:EB1911 - Volume 18.djvu/689

vessel, and let co-ordinate axes be taken such that the origin is in dS, and the axis of x is the normal at the origin into the gas. The number of molecules of the first kind of gas, whose components of velocity lie within the ranges between u and u+du, v and v+dv, w and w+dw, will, by formula (5), be

$v{\sqrt {(h^{3}m^{3}{\mbox{/}}\pi ^{3})}}e^{-hm(u^{2}+v^{2}+w^{2})}dudvdw$

(9)

per unit volume. Construct a small cylinder inside the gas, having dS as base and edges such that the projections of each on the coordinate axes are udt, vdt, wdt. Each of the molecules enumerated in expression (9) will move parallel to the edge of this cylinder, and each will describe a length equal to its edge in time dt. Thus each of these molecules which is initially inside the cylinder, will impinge on the area dS within an interval dt. The cylinder is of volume u dt dS, so that the product of this and expression (9) must give the number of impacts between the area dS and molecules of the kind under consideration within the interval dt. Each impinging molecule exerts an impulsive pressure equal to mu on the boundary before the component of velocity of its centre of gravity normal to the boundary is reduced to zero. Thus the contribution to the total impulsive pressure exerted on the area dS in time dt from this cause is

$mu\times udtd{\mbox{S}}\times v{\sqrt {(h^{3}m^{3}{\mbox{/}}\pi ^{3})}}e^{-hm(u^{2}+v^{2}+w^{2})}dudvdw$

(10)

The total pressure exerted in bringing the centres of gravity of all the colliding molecules to rest normally to the boundary is obtained by first integrating this expression with respect to u, v, w, the limits being all values for which collisions are possible (namely from −∞ to 0 for u, and from −∞ to +∞ for v and w), and then summing for all kinds of molecules in the gas. Further impulsive pressures are required to restart into motion all the molecules which have undergone collision. The aggregate amount of these pressures is clearly the sum of the momenta, normal to the boundary, of all molecules which have left dS within a time dt, and this will be given by expression (10), integrated with respect to u from 0 to ∞, and with respect to v and w from −∞ to +∞, and then summed for all kinds of molecules in the gas. On combining the two parts of the pressure which have been calculated, the aggregate impulsive pressure on dS in time dt is found to be

$\Sigma dt\ d{\mbox{S}}\iiint _{-\infty }^{+\infty }v{\sqrt {(h^{3}m^{3}{\mbox{/}}\pi ^{3})}}e^{-hm(u^{2}+v^{2}+w^{2})}mu^{2}dudvdw,$

where Σ denotes summation over all kinds of molecules. This is equivalent to a steady pressure p₁ per unit area where

$p_{1}-\Sigma dtd{\mbox{S}}\iiint _{-\infty }^{+\infty }v{\sqrt {(h^{3}m^{3}{\mbox{/}}\pi ^{3})}}e^{-hm(u^{2}+v^{2}+w^{2})}mu^{2}dudvdw$

Clearly the integral is the sum of the values of mu² for all the molecules of the first kind in unit volume, thus

$p=v{\overline {mu^{2}}}+v'{\overline {m'u^{2}}}+\cdots$

(11)

On substituting from equations (7) and (8), this expression assumes the forms

p＝(v+v ′+...)/2h

(12)

＝(v+v ′+...)RT

(13)

The number of molecules per unit volume in a gas at normal temperature and pressure is known to be about 2·75 X 10¹⁹. If in formula (13) we put p＝1·013×10⁶, (v+v ′+ ...)2·75×10¹⁹ T＝273, we obtain R＝1·35×10⁻¹⁶ and this enables us to determine the mean velocities produced by heatMolecular Velocities. motion in molecules of any given mass. For molecules of known gases the calculation is still easier. If ρ is the density corresponding to pressure p, we find that formula (11) assumes the form

p＝1/3ρC²,

where C is a velocity such that the gas would have its actual translational energy if each molecule moved with the same velocity C. By substituting experimentally determined pairs of values of p and ρ we can calculate C for different gases, and so obtain a knowledge of the magnitudes of the molecular velocities. For instance, it is found that

for hydrogen at	0° Cent. C＝183,900 cms. per sec.
for air⁠ ,,	15° Cent. C＝ 49,800 cms. per sec.
for mercury vapour at	0° Cent. C＝ 18,500 cms. per sec.

and other velocities can readily be calculated.

From the value R＝1·35×10⁻¹⁶ it is readily calculated that a molecule, or aggregation of molecules, of mass 10⁻¹² grammes, ought to have a mean velocity of about 2 millimetres a second at 0° C. Such a velocity ought accordingly to be set up in a particle of 10⁻¹² grammes mass immersed in air or liquid“Brownian Movements.” at 0° C., by the continual jostling of the surrounding molecules or particles. A particle of this mass is easily visible microscopically, and a velocity of 2 mm. per second would of course be visible if continued for a sufficient length of time. Each bombardment will, however, change the motion of the particle, so that changes are too frequent for the separate motions to be individually visible. But it can be shown that from the aggregation of these separate short motions the particle ought to have a resultant motion, described . with an average velocity which, although much smaller than 2 mm. a second, ought still to be microscopically visible. It has been shown by R. von. S. Smoluchowski (Ann. d. Phys., 1906, 21, p. 756) that this theoretically predicted motion is simply that seen in the “Brownian movements” first observed by the botanist Robert Brown in 1827. Thus the “Brownian movements” provide visual demonstration of the reality of the heat-motion postulated by the kinetic theory.

Dalton’s Law.—The pressure as given by formula (12) can be written as the sum of a number of separate terms, one for each gas in the mixture. Hence we have Dalton’s law: The pressure of a mixture of gases is the sum of the pressures which would be exerted separately by the severalPressure, Volume and
Temperature Relations. constituents if each alone were present.

Avogadro’s Law.—From formula (13) it appears that v+v ′+ . . ., the total number of molecules per unit volume, is determined when p, T and the constant R are given. Hence we have Avogadro’s law: Different gases, at the same temperature and pressure, contain equal numbers of molecules per unit volume.

Boyle’s and Charles’ Laws.—If v is the volume of a homogeneous mass of gas, and N the total number of its molecules, N＝v(v+v ′+ . . .), so that

pv＝RNT.

(14)

In this equation we have the combined laws of Boyle and Charles: When the temperature of a gas is kept constant the pressure varies inversely as the volume, and when the volume is kept constant the pressure varies as the temperature.

Since the volume at constant pressure is exactly proportional to the absolute temperature, it follows that the coefficients of expansion of all gases ought, to within the limits of error introduced by the assumptions on which we are working, to have the same value 1/273.

Van der Waals’s Equation.—The laws which have just been stated are obeyed very approximately, but not with perfect accuracy, by all gases of which the density is not too great or the temperature too low. Van der Waals, in a famous monograph, On the Continuity of the Liquid and Gaseous States (Leiden, 1873), has shown that the imperfections of equation (14) may be traced to two causes:—

(i.) The calculation has not allowed for the finite size of the molecules, and their consequent interference with one another’s motion, and

(ii.) The calculation has not allowed for the field of inter-molecular force between the molecules, which, although small, is known to have a real existence. The presence of this field of force results in the molecules, when they reach the boundary, being acted on by forces in addition to those originating in their impact with the boundary.

To allow for the first of these two factors, Van der Waals finds that v in equation (14) must be replaced by v−b, where b is four times the aggregate space occupied by all the molecules, while to allow for the second factor, p must be replaced by p+a/v². Thus the pressure is given by the equation

(p+a/v²) (v−b)＝RNT,

which is known as Van der Waals’s equation. This equation is found experimentally to be capable of representing the relation between p, v, and T over large ranges of values. (See Condensation of Gases.)

Let us consider a single gas, consisting of N similar molecules in a volume v, and let the energy of each molecule, as in formula (2) be given byCalorimetry.

E＝1/2$\sum _{\mbox{N}}$ (mu2mv2+mw2+α1θ12+ . . . αnθn2)

(15)

＝N(n+3)/4h by equation (7)⁠

＝1/2(n+3)RNT⁠

(16)

Let a quantity dQ of energy, measured in work units, be absorbed by the gas from some external source, so that its pressure, volume and temperature change. The equation of energy is

dQ＝dE+pdv,

(17)

expressing that the total energy dQ is used partly in increasing the internal energy of the gas, and partly in expanding the gas against the pressure p. If we take p＝RNT/v from equation (14) and substitute for E from equation (16), this last equation becomes

dQ＝1/2 (n+3)RNdT+RNTdv/v,

(18)

which may be taken as the general equation of calorimetry, for a gas which accurately obeys equation (14).

Second Law of Thermodynamics.—If we divide throughout by T, we obtain

dQ/T ＝1/2(n+3)RNdT/T+RNdv/v.

showing that dQ/T is a perfect differential. This not only verifies that the second law of thermodynamics is obeyed, but enables us to identify T with the absolute thermodynamical temperature.

If the volume of the gas is kept constant, we put dv＝0 in equation (18) and dQ＝JC_vNmdT, where C_v, is the specific heat of the gas at constant volume and J is the mechanical equivalent of heat. We obtainSpecific Heats.

Cv＝1/2(n+3)R/Jm

(19)

On the other hand, if the pressure of the gas is kept constant