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speed axis. As can be seen from Fig. 3.3, the relative number of molecules having speeds exceeding v₀, increases with increasing temperature.

In conclusion of this section, let us note that all the formulas for the distribution function and the characteristic speeds contain the ratio of the mass of a molecule to the Boltzmann constant

Multiplying the numerator and denominator by the Avogadro number N_A and taking into account that

— is the molar mass of the gas, and

— is the universal gas constant; we can use this relation everywhere in whichever form is most convenient for the particular problem

Distribution of molecules over dimensionless speed. If the Maxwell distribution function (3.29) is plotted with the molecular speed v, on the abscissa, the shape of the curve and the position of its maximum depend on the mass of the molecules and on the gas temperature. But if the ratio of the speed to the most probable speed is plotted on the horizontal axis, that is, the dimensionless speed

then the same curve is obtained for all temperatures and any molecular masses (any gases) (Fig. 3.4).

Fig. 3.4. Maxwell distribution over dimensionless speed

Making the change of variable

in (3.29) and taking into account that

we obtain the Maxwell distribution in the form

This formula and the corresponding graph (see Fig. 3.4) are convenient for solving many problems.

Example. Let us find what fraction of the total number of oxygen molecules at a temperature of 27 °C have speeds differing from the most probable speed by no more than 1%; and also speeds in the interval 562–572 m/s.

Let us carry out the necessary calculations. To answer the first question of the problem, note that u = 1 at v = v_mp. The size of the interval is du = 0.02. Therefore,

Let us calculate the most probable speed:

Let us find the ratio of v = 562 m/s to v_mp = 395 m/s

Let us determine from the curve (see Fig. 3.4) the value of the function f(u) at u = 1.42. We obtain f(u) = 0.62. The width of the interval is Dv = 10 m/s (Du = 10/395 = 0.0253). Therefore, the fraction of molecules in this interval is

It is interesting to note that an oxygen molecule travels on average a distance of 0.4 km per second. But one must not forget about molecular collisions. Because of them, a molecule moves in a straight line only for a very short time, and its path is a broken line. Therefore, a molecule moving at enormous speed along the individual segments of the broken trajectory travels from one layer of gas to another at a relatively low speed.

Mean (arithmetic average) speed. Knowing the molecular speed distribution function f(v) makes it possible to find the mean value of the speed, as well as of any quantity that is a function of speed, for example the square of the speed v² or the kinetic energy of a molecule mv²/2.

Let us divide the interval of all possible speed values from 0 to infinity into small intervals Dv_i. Each interval corresponds to a number of molecules

Since the intervals Dv_i are small, we can approximately regard the speeds of the molecules in a given interval as identical and equal to v_i. The sum of the speeds of the molecules in the interval is

The sum of the speeds of all molecules is

Dividing this sum by the number of molecules, we obtain the expression for the mean speed

Passing from the sum to an integral, we obtain

Evaluating the integral, we obtain the mean speed of the molecules

Root-mean-square speed. To find the mean value of an arbitrary function L(v) of the speed, one must multiply this function by the distribution function and integrate over all possible values of the speed:

In particular, for L(v) = v this gives .

The mean value of the square of the speed equals the ratio of the sum of the squares of the speeds of all molecules of the system to the total number of molecules. Thus,

It should be noted that the characteristic speeds differ from one another only by numerical factors, and

while their dependence on T and m₀ (or m) is the same.

The mean kinetic energy of the translational motion of molecules is expressed through the root-mean-square speed

This result agrees with formula (1.14) of the kinetic theory of ideal gases and with the law of equipartition of energy, which states that each degree of freedom of a molecule carries an energy of k_BT/2. The three translational degrees of freedom of a molecule correspond exactly to the result (3.44) obtained here. In essence, it was precisely to obtain this agreement that we chose the coefficient α in (3.26) appropriately.

An experiment to test the Maxwell distribution. It must be emphasized once more that the law of the distribution of molecules over speeds established by Maxwell, and all the consequences that follow from it, hold only for a gas in equilibrium.

The law holds for any number of molecules N, provided this number is sufficiently large. Maxwell's law is statistical, and the laws of statistics are satisfied the more accurately, the larger the number of identical objects to which they are applied. For a small number of objects, significant deviations from the predicted statistics may be observed — fluctuations.

The experimental determination of the molecular speed distribution was first carried out by O. Stern in 1920. He studied the speed distribution of monatomic molecules of metal vapors (Ag or Pt) from a filament made of that metal and placed on the axis of two cylinders. The filament was heated by an electric current, and the metal evaporated (see Fig. 3.5).

Fig. 3.5 Diagram of Stern's experiment: 1 — side view of the apparatus; 2 — top view of the apparatus

Molecules that passed through the slit in the inner cylinder flew in a straight line and settled on the wall of the cold outer cylinder. If the whole apparatus is set into rotation (with the slit always facing the point B₀), then molecules with a high speed v, will land at some point near B₀, while slower ones will spend more time on the way and will land at points farther from B₀. Note that the emitted molecules move in a straight line; they do not take part in the rotational motion. Since molecules land at different points of the outer cylinder depending on their speed, by examining the thickness of the metal layer deposited on its wall one can form an idea of the distribution of molecules over speeds.

Let us find the distribution of molecules over the distances S from the point B₀ to the place where they hit the cylinder wall. If R and r — are the radii of the large and small cylinders, respectively (see figure), then the flight time from the slit to the cylinder wall is

During this time the cylinder will turn through an angle

where ω is the angular velocity of rotation of the apparatus. Accordingly, the point of impact will be displaced relative to B₀ by a distance

Substituting the flight time here, we obtain the relation between the speed of the molecule and the distance S:

Substituting, in turn, the resulting expression into the Maxwell distribution and taking into account that

we find the distribution of molecules over the distances S:

(we omit the expression for the normalization constant C).

Stern's experiments confirmed the validity of the law established by Maxwell.

3.4. Distribution of molecules over coordinates

In this section we turn to the analysis of the distribution of gas molecules over coordinates. Obviously, if no external forces act on the gas molecules, then in a state of thermodynamic equilibrium the gas is uniformly distributed over the given volume. In this case the pressure and density of the gas are the same at all points. If, however, the gas is in a force field (like, for example, atmospheric air, which is attracted by the Earth), then the pressure and density of the gas will no longer be the same everywhere, but will vary from point to point.

Video 3.4. Distribution of gas molecules over the height of a vessel in a uniform gravitational field.

The barometric formula. Let us find the law according to which the atmospheric pressure (or the air density) changes with distance from the Earth's surface. Let us select a vertical column of air with horizontal cross-sectional area S.

Let us assume that

• this column is in thermal equilibrium, that is, the temperature is the same everywhere (in the real atmosphere this is not so, but for simplicity of analysis we will assume T = const);
• the gas is ideal, that is, the Clapeyron–Mendeleev equation holds for it

• the variation of the acceleration due to gravity g with height can be neglected (which is valid for not very large heights).

Atmospheric pressure at height h is due to the weight of the overlying layers of gas. Let the pressure at height h be p, then at height h + dh the pressure is p + dp (Fig. 3.6). Moreover, if dh > 0, then the pressure decreases, dp < 0, since the weight of the overlying layers of the atmosphere decreases.

Fig. 3.6. A vertical air cylinder (for the derivation of the barometric formula)

The selected layer of gas, of height dh and mass m, is in equilibrium. Therefore, the sum of the forces acting on it is zero:

In projection onto the vertical axis we obtain

where r is the gas density at height h. Expanding the brackets and collecting like terms, we arrive at the equation

Let us use the Clapeyron–Mendeleev equation for the selected mass of gas m and express the density in terms of the pressure:

Substituting (3.47) into (3.46), we finally obtain

This equation can be integrated in the case of an isothermal atmosphere (T = const):

Video 3.5. The barometric formula: a candle flame as a very sensitive indicator of the decrease of pressure with height.

The constant of integration p₀ equals the pressure at the surface (h = 0). The resulting dependence is called the barometric formula. It describes the distribution of gas pressure with height in a uniform gravitational field at constant temperature. Note that the distribution depends on the kind of gas. The smaller m, the smaller the absolute value of the exponent, and the more slowly the pressure of such a gas decreases with increasing height. Figure 3.7 shows the dependence of pressure on height at temperature T = 300 K (27 °C) for three gases of different molar mass — hydrogen H₂ (m₁ = 2.016 g/mol), nitrogen N₂ (m₂ = 28.013 g/mol) and oxygen O₂ (m₃ = 31.999 g/mol).

Fig. 3.7. Dependence of the pressure of three different gases H₂, N₂ and O₂ on height

Example. Let us determine at what height the pressure of oxygen decreases by a factor of two (at T = 300 K).

We apply the barometric formula.

Then

whence

Using the ideal gas equation in the form

from the barometric formula one can easily obtain the law of variation with height of the number n of molecules per unit volume:

It follows from (3.51) that the composition of air changes with increasing height: the concentration of gases with small molar mass, for example hydrogen and helium, increases.

At the surface, air is a mixture of gases: N₂ — 78.08%, O₂ — 20.95%, CO₂ — 0.03%, inert gases — 0.94%. Let us see how the ratio of the concentrations of oxygen and nitrogen changes in an isothermal atmosphere (T = 300 K) at a height of 10 km.

The ratio of the concentrations of oxygen and nitrogen decreases from 0.27 to 0.23. Our calculation is valid only for an isothermal atmosphere and relatively small heights, for which the acceleration due to gravity changes insignificantly: g = const, T = const.

The Boltzmann distribution. The number of molecules per unit volume depends on the height h and the temperature T, and both variables enter the exponent. Equation (3.51) can be written in the form

where k_B is the mass of one gas molecule. Here the expression m₀gh, in the numerator, is nothing other than the potential energy of one molecule in the Earth's gravitational field. Therefore we can say that we have a distribution of molecules over values of potential energy. The greater the potential energy, the fewer such molecules there are. In the denominator of the exponent stands k_BT — a quantity proportional to the mean energy of the thermal motion of a molecule. The higher the temperature, that is, the greater the energy of the thermal motion of the molecules, the more slowly the exponential factor, which is proportional to the concentration of molecules, decreases with height. Figure 3.8 shows curves of the relative concentration of oxygen molecules O₂ at different heights for two different temperatures T₁ = 300 K and T₂ = 1 300 K (the latter case is, of course, unrealistic and is used merely as an illustration).

Fig. 3.8. Dependence of the relative concentration of oxygen molecules on height at different temperatures T₁ = 300 K and T₂ = 1 300 K

We see that at a higher temperature the number of particles per unit volume decreases more slowly with height. As the temperature decreases, most of the particles are located at lower heights. And at T = 0 all the particles would be located on the Earth's surface. This fact has a simple physical explanation. Each particular distribution of molecules over height is established as the result of two competing tendencies:

• the attraction of the molecules to the Earth, characterized by the potential energy m₀gh, tends to place them on the Earth's surface;
• thermal motion, characterized by the energy k_BT, tends to scatter the molecules uniformly over all heights.

Denoting E_p = m₀gh, we obtain

that is, the concentration of molecules is greater where their potential energy is smaller. Particles are more likely to be found at those points of space where the potential energy is lower.

Boltzmann showed that such a distribution holds in the field of any forces, not only in a gravitational field. Therefore the distribution (3.53), where n — is the concentration of particles with potential energy E_p is called the Boltzmann distribution.

3.5. The Maxwell–Boltzmann distribution

We have established the function describing the distribution of molecules over speeds (the Maxwell distribution) and the relation characterizing the distribution of molecules over values of potential energy (the Boltzmann distribution). Both relations can be combined into a single generalized distribution.

Consider an infinitesimal volume dV of gas located at the point with position vector in a large system representing an ideal gas at constant temperature in external force fields. The number of molecules in the selected volume is n( ) d³r. Since the volume is small, the particle density within it can be considered constant. This means that the condition for the validity of the Maxwell distribution is satisfied. Then, by combining relations (3.11) and (3.27), we obtain the following formula for the number of molecules dN with speeds from v to v + dv located in the volume d³r :

But the concentration of molecules n(r) depends on the location of this volume in the external force fields:

where n₀ — is the concentration of molecules at the point where E_p = 0. Then

Since the expression

represents the total energy of a particle in an external potential force field, we arrive at the generalized Maxwell–Boltzmann distribution over the energies of molecules:

where N — is the total number of particles in the system, and dN — is the number of particles with coordinates between r and r + dr and (simultaneously) with speeds between v and v + dv.

Mean energy of a quantum oscillator. The Maxwell–Boltzmann distribution was obtained in classical physics, but it turned out to be valid in quantum mechanics as well, where many seemingly unshakable propositions were subjected to revision. As an example, consider the problem of a load of mass m, attached to the end of a spring of stiffness k. The equation of motion is well known, and its solution is harmonic oscillation of the body with angular frequency

The classical energy of the system, which models the vibrations of atoms in a molecule, is given by formula (3.62) and can take any value depending on the amplitude of the oscillations. As we know from quantum mechanics, the vibrational energy is quantized, that is, it takes a discrete series of values determined by the formula:

In accordance with the general principles of statistical physics, the probability P_n of finding the oscillator in a state characterized by some value n of the vibrational quantum number is determined by the formula

where A — is the normalization constant. To determine it, one must use the probability normalization condition

To do so, into the well-known formula for a geometric progression