355 APPENDIX I PRESSURE AND ENERGY In the first part of this book it has several times been pointed out that we can correctly calculate the pressure exerted by a gas if, instead of ascribing to the molecules, as is actually the case, unequal speeds that are constantly varying, we assume that they have all the same mean speed. This mean value is of such size that the kinetic energy on the assumption of equal speeds in the gas has the same value as it really has with the actual inequalities that exist. The justification of this simplifying assumption rests on the fact that the pressure exerted by a gas is dependent on the speed of the molecules only in so far as it increases in proportion to their kinetic energy. Against the validity of the reasons marshalled in Chapter II. no objection can be raised. But it will not be superfluous to calculate the pressure without this simplifying assumption. There will therefore be made in the following investigation no assumption of any kind with respect to the distribution of unequal speeds. But the result of the calculation will be the same, viz. that the pressure depends only on the mean kinetic energy. 1*. Number of Molecules and their Paths Let the number of molecules of a gaseous medium in unit volume be N. These N molecules do not all move with the same speed, nor even in the same direction; the components u, v, w of the velocity of a molecule, reckoned along three fixed rectangular axes, assume for different molecules values which vary from 81 to +∞. The number of molecules for which the values of the components differ infinitely little from u, v, w, so that they lie between the limits u and u + du, v and v + dv, w and w + dw, is an infinitesimal of the order du dv dw; it may be expressed by NF(u, v, w)du dv dw, where F denotes a kind of probability-function, with the determination of which we shall proceed in Appendix II. §§ 10*–15*. Just as the values of the velocity for different particles are in general different, so also will the times be different during which the molecules move in the same direction with unchanged speed. Let the length of this time-that is, of the interval between two successive collisions-be t, and let the probability of a molecule's moving for the time t in a straight line with unchanged velocity, whose components are u, v, w, and then colliding in the following infinitesimal interval dt be f(t, u, v, w)dt. This function f, like the other F, I will not here more closely examine, as an obvious property of it is sufficient for our present purpose. Since every path must certainly come to an end, the sum of all the probabilities must be certainty, or [of (t, u, v, w)dt = 1. With this notation, therefore, the number of molecules in unit volume, which move with the velocities u, v, w, but only for the time t without collision, is NF(u, v, w) f (t, u, v, w)dt du dv dw. At the end of time t this group begins a new path in a new direction with a new velocity. But if we assume that the state of the gas does not change with the time, then for every molecule that at any moment loses the velocity-components u, v, w another molecule must at the same moment gain these velocities as the result of collision. After the interval t, therefore, the number of molecules above given move again out of the unit volume under the same circumstances and with the same probability of collision. Thus the number of paths which in the unit of time are newly begun by the N molecules in unit volume with velocities u, v, w, and which are ended after time t, is We use this formula to determine the number of particles which in a unit of time cross any surface-element within the gas with a given speed in a given direction. Take the axis of a perpendicular to the plane, and investigate the number of those particles which in a unit of time cross an infinitely small rectangle dy dz at the point (x, y, z) with a velocity the components of which are u, v, w. Since this specification of the components determines the direction of the motion of the particles in question, it follows that all these particles must come from a limited region, an oblique parallelopiped in shape, which has the rectangle dy dz as base and its length along the direction of motion. Denote the coordinates of any point in this region relatively to the given point (x, y, z) by I, 1, 3; these must satisfy the condition x: y: z=u: v: w, Divide up also this region, from which all the molecules in question come, into infinitely small oblique parallelopipeds dr dy dz = dr dy dz by planes drawn parallel to that of yz. In one of these volume-elements it will, by § 1*, happen NF(u, v, w) ƒ (t, u, v, w)dt du dv dw dx dy dz t times per time-unit that a particle begins a new path, with a velocity whose components are u, v, w, in the direction towards the surface-element dy dz, this path ending after the lapse of time t. The particles will actually reach their mark, the element dy dz, and pass through it if the time t is sufficient for the path to be traversed, and therefore, in the case of particles moving in the positive direction of x, if rut. We consequently obtain the total number of the particles which start from the elements of that oblique parallelopiped standing on the base dy dz, and pass through the element dy dz in unit time with a velocity whose components are u, v, w, by summing up the above expression for all the volume-elements with the condition x ut, that is, by integrating it with respect to dr from the initial value. r = 0 to the limiting value = ut. In the second place, to obtain |