Molecular-kinetic theory allows, by analyzing the microscopic behavior of the system and using the methods of statistical mechanics, to obtain important macroscopic characteristics of the thermodynamic system. One of the microscopic characteristics that is associated with the temperature of the system is the mean square velocity of the gas molecules. We give the formula for it and consider it in the article.
Perfect gas
Immediately, we note that the formula for the quadratic mean velocity of gas molecules will be given specifically for the ideal gas. Under it, in physics we mean a many-particle system in which particles (atoms, molecules) do not interact with each other (their kinetic energy is several orders of magnitude higher than the potential energy of interaction) and have no size, that is, they are points with a finite mass (distance between particles several orders of magnitude greater than their linear dimensions).
Any gas that consists of chemically neutral molecules or atoms, and that is under slight pressure and has a high temperature, can be considered ideal. For example, air is an ideal gas, and water vapor is no longer such (strong hydrogen bonds act between water molecules).
Molecular Kinetic Theory (MKT)
Studying the ideal gas in the framework of the ILC, you should pay attention to two important processes:
- Gas creates pressure due to the transfer to the walls of the vessel that contains it, the momentum in the collision of molecules and atoms with them. Such collisions are absolutely resilient.
- Molecules and gas atoms move randomly in all directions with different speeds, the distribution of which obeys Maxwell-Boltzmann statistics. The probability of collision between particles is extremely low, due to their negligible size and large distances between them.
Despite the fact that the individual velocities of gas particles are very different from each other, the average value of this quantity remains constant in time if there are no external influences on the system. The formula for the mean square velocity of gas molecules can be obtained by considering the relationship between kinetic energy and temperature. We will deal with this issue in the next paragraph of the article.
Derivation of the quadratic mean velocity formula for ideal gas molecules
Every student knows from the general course of physics that the kinetic energy of the translational motion of a body of mass m is calculated as follows:
E k = m * v 2/2
Where v is linear velocity. On the other hand, the kinetic energy of a particle can also be determined through the absolute temperature T using the conversion factor k B (Boltzmann constant). Since our space is three-dimensional, E k is calculated as follows:
E k = 3/2 * k B * T.
Equating both equalities and expressing v from them, we obtain the formula for the average velocity of a quadratic gas ideal:
m * v 2/2 = 3/2 * k B * T =>
v = √ (3 * k B * T / m).
In this formula, m - is the mass of the gas particle. Its value is inconvenient to use in practical calculations, since it is small (≈ 10 -27 kg). To avoid this inconvenience, let us recall the universal gas constant R and the molar mass M. The constant R with k B is related by the equality:
k B = R / N A.
The value of M is determined as follows:
M = m * N A.
Taking both equalities into account, we obtain the following expression for the mean square velocity of the molecules:
v = √ (3 * R * T / M).
Thus, the mean square velocity of the gas particles is directly proportional to the square root of the absolute temperature and inversely proportional to the square root of the molar mass.
Problem solving example
Everyone knows that the air we breathe is 99% nitrogen and oxygen. It is necessary to determine the differences in the average velocities of the molecules of N 2 and O 2 at a temperature of 15 o C.
This problem will be solved sequentially. First, translate the temperature into absolute units, we have:
T = 273.15 + 15 = 288.15 K.
Now we write out the molar masses for each molecule under consideration:
M N2 = 0.028 kg / mol;
M O2 = 0.032 kg / mol.
Since the values of molar masses differ slightly from each other, their average velocities at the same temperature should also be close. Using the formula for v, we obtain the following values for nitrogen and oxygen molecules:
v (N 2 ) = √ (3 * 8.314 * 288.15 / 0.028) = 506.6 m / s;
v (O 2 ) = √ (3 * 8.314 * 288.15 / 0.032) = 473.9 m / s.
Since nitrogen molecules are slightly lighter than oxygen molecules, they move faster. The difference in average speeds is:
v (N 2 ) - v (O 2 ) = 506.6 - 473.9 = 32.7 m / s.
The resulting value is only 6.5% of the average velocity of nitrogen molecules. We draw attention to the high values of the velocities of molecules in gases even at low temperatures.