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1 moleis equal to the amount of matter in a system containing the same structural elementshow many atoms are contained in carbon weighing 0.012 kg.
Moles of any gases at the same temperatures and pressures occupy the same volumes - avogadro's law... Under normal conditions ( r\u003d 1.013 · 10 5 Pa, T\u003d 273, 15 K) this volume is 22.41 · 10 -3 m 3 / mol.
The number of molecules (structural units) in 1 mole is equal to Avogadro's number: N A \u003d 6.02 · 10 23 mol -1.
mendeleev - Clapeyron equation:
Or (3.11)
where M is the molar mass of the gas, 
- the amount of substance, R
If N is the total number of gas molecules, dN is the number of molecules whose velocities are in the range from to + d, then maxwell's distribution lawwill be written as:
The rate at which the velocity distribution function of ideal gas molecules is maximum is called most likely speed:
. (3.14)
If we express the velocities of molecules not in ordinary units, but in relative ones, taking the most probable velocity of molecules as a unit of velocity, then the Maxwell distribution takes the form:
The dependence of atmospheric pressure on altitude at constant temperature is called barometric formula:
where n and n 0 is the concentration of molecules at the height h and h 0 \u003d 0.
Under the internal energy U in thermodynamicsunderstand the energy of the thermal motion of the particles forming the system, and the potential energy of their mutual position:
2) the heat imparted to the system in the process of changing its state is spent on changing its internal energy and on performing work against external forces:
where dU - small change in internal energy; δ Q is the elementary amount of heat; δ A - elementary work.
Expansion work performed with finite volume changes:
(3.23)
The heat capacity of the system of bodies (body) is called a physical quantity equal to the ratio of the amount of heat dQ, which must be spent to heat the system of bodies (body), to a change in temperature dT,characterizing this heating:
... [C] \u003d J / K. (3.24)
Specific heatsubstances from is called a scalar quantity equal to the ratio of the heat capacity of a homogeneous body FROM to its mass:
. [c] \u003d J / (kg.K) (3.25)
Molar heat capacity is a physical quantity that is numerically equal to the ratio of the heat capacity of the system FROM to the amount of substance n contained in it:
... \u003d J / (mol.K) (3.26)
Distinguish molar heat capacity at constant volume and constant pressure:
| , . | (3.27) |
The equation connecting the molar heat capacities at constant pressure and constant volume has the form ( mayer's equation):
C p - C V \u003d R. (3.28)
The first law of thermodynamics in the isochoric process (V\u003d const; dV \u003d 0, dA \u003d pdV \u003d0): - the heat imparted to the system during the isochoric process is used to change the internal energy.
, (3.29)
In this case, the work is not done.
The first law of thermodynamics in the isobaric process (p \u003d const):
.
(3.30)
Isobaric expansion work equals
. (3.31)
The first law of thermodynamics in the isothermal process(T\u003d const; dT \u003d0; 
): - the heat imparted to the system during the isothermal process goes to work against external forces:
(3.32)
Adiabaticis a process that occurs without heat exchange with the external environment: dQ \u003d 0.
From the first law of thermodynamics:
that is, work in an adiabatic process is performed due to the loss of internal energy.
Poisson's equations (equations of state for an adiabatic process):
The quantity g - adiabatic exponent - is determined by the number and nature of the degrees of freedom of the molecule (Table 4 of the Appendix):
. (3.34)
When comparing the adiabatic and isothermal processes (Fig. 3.4), it can be seen that the adiabat runs steeper than the isotherm.
Polytropicis called a thermodynamic process in which the heat capacity of a body is constant: FROM\u003d const.
Equations of a polytropic process in an ideal gas:
pV n\u003d cons t, TV n-1\u003d const, (3.35)
where is the polytrope index, which depends on the specific heat capacity of the gas.
There are several formulations second law of thermodynamics:
1. Clausius' wording: The process is not possible, the only end result which is the transfer of energy in the form of heat from a less heated body to a more heated one.
2. Thomson (Kelvin) formulation:A process is impossible, the only end result of which is the transformation of all the heat received from a certain body into work equivalent to it.
Circular process Is a set of thermodynamic processes, as a result of which the system returns to its original state. In state diagrams, circular processes are depicted by closed lines.
Direct cycle is called a circular process in which the system does positive work. An example of a direct cycle is a cycle performed by a working fluid in a heat engine.
Reverse cycle is called a circular process in which the system performs negative work (for example, a cycle of a working fluid in a refrigeration unit).
Coefficient of performance (COP) of the heat engine is the ratio of the work performed per cycle ANDto the amount of heat received by the working fluid from the heater Q 1 :
, (3.36)
where Q 1 is the amount of heat received by the working substance, Q 2 is the amount of heat given by the working substance to the refrigerator.
 |
Carnot cycleis called a circular process in which the work performed by the system is maximum. The direct Carnot cycle consists of four successive reversible processes: isothermal expansion (1®2) at temperature T 1, adiabatic expansion and contraction (2®3, isothermal compression (3®4) at temperature T 2 and adiabatic compression (4®1) (Figure 3.5.).
A machine performing a Carnot cycle is called ideal heat engine.
Thermal efficiency of the direct Carnot cycle,committed by ideal gas:
. (3.37)
where T 1 and T 2 - the values \u200b\u200bof the temperature of the heater and refrigerator participating in the implementation of the cycle under consideration.
State function S whose differential
called entropy.Here dQ - an infinitely small amount of heat imparted to the system in an elementary reversible process.
Entropy change in any reversible process that transfers the system from state 1 to state 2 is equal to the reduced amount of heat transferred to the system in this process
, (3.39)
where S 1 and S 2 - values \u200b\u200bof entropy in states 1 and 2, DS- change in entropy during a reversible process.
Thermodynamic probabilitysystem W is the number of all possible particle distributions over coordinates and velocities corresponding to a given thermodynamic state.
Thermodynamic probability and entropy are related by the relation ( boltzmann formula):
When the balance is disturbed, the system tends to return to equilibrium state... This process is accompanied by an increase in entropy and, therefore, is irreversible. The imbalance is accompanied by the transfer of mass (diffusion), momentum (internal friction) or energy (thermal conductivity). These processes are called transport phenomena... Consequently, the transfer phenomena are irreversible processes.
The mean free path of molecules`l is the average distance that a molecule travels without collisions:
 | (3.41) |
The number of collisions experienced by a molecule per unit time can be different. Therefore, we should talk about the average value of this value:
, (3.42)
where n - concentration of molecules.
Average free path and average number of collisions per unit time are related by the equation:
where is the arithmetic mean speed.
Diffusion coefficient – is the mass transferred per unit of time through a unit area in the direction of the normal to this area in the direction of decreasing component density with a density gradient, equal to one
. (3.44)
internal friction coefficient (viscosity coefficient) is numerically equal to the momentum transferred per unit time through a unit area at a unit velocity gradient:
. (3.45)
Coefficient of thermal conductivitynumerically equal to the amount of heat transferred per unit of time through a unit area at a unit temperature gradient:
([TO] \u003d W / m.K), (3.46)
where ρ is the density of gases.
Examples of problem solving
Task 3.1. Define molar mass a mixture of oxygen with a mass of m 1 \u003d 25 g and nitrogen with a mass of m 2 \u003d 75 g.
The amount of substance in the mixture is equal to the sum of the number of components:
. (3)
Substituting expressions (2) and (3) into formula (1) and transforming, we get:
.
The molar masses of oxygen M 1 and nitrogen M 2 are determined from the periodic table:
M 1 \u003d 32 · 10 -3 kg / mol and M 2 \u003d 28 · 10 -3 kg / mol
Calculations:
Task 3.2. The two cylinders are connected by a closed valve tube, the volume of which is negligible. A cylinder with a volume of 0.02 m 3 contains gas under a pressure of 1.6 × 10 4 Pa, and a cylinder with a volume of 0.06 m 3 contains the same gas under a pressure of 1.2 × 10 4 Pa. What pressure will be established in the cylinders if the valve is opened? The gas temperature remains constant.
where r 1 "- gas pressure of the first vessel, r 2 "- gas pressure of the second vessel.
According to the condition of the problem, the gas temperature remains unchanged, therefore, according to the Boyle-Mariotte law for two states of the gas, we can write:
, (2)
Solving the resulting system of equations
Check units physical quantities left and right of the equal sign
Calculations:
1.3 × 10 4 Pa.
Answer: r \u003d 1.3 × 10 4 Pa.
Task 3.3. The container contains m 1 \u003d 80 g oxygen and m 2 \u003d 320 g argon. Mix pressure r\u003d 1 MPa, temperature T\u003d 300 K. Taking these gases as ideal, determine the volume V balloon.
According to Dalton's law, the pressure of the mixture is equal to the sum of the partial pressures of the gases included in the mixture:
r = r 1 + r 2 (2)
Substituting equation (1) into equation (2), we get:
.
from the last expression we find the volume of the cylinder:
,
where M 1 \u003d 32 · 10 -3 kg / mol is the molar mass of oxygen, M 2 \u003d 40 · 10 -3 kg / mol is the molar mass of argon (from the periodic table).
Calculations:
Answer: V \u003d 0.0262 m 3
Task 3.4. Find the average kinetic energy of the rotational motion of one oxygen molecule at a temperature T \u003d 350 K, as well as the kinetic energy E to the rotational motion of all oxygen molecules with a mass m\u003d 4 g.
Since the rotational motion of a diatomic molecule (oxygen molecule is diatomic) corresponds to two degrees of freedom, the average energy of the rotational motion of an oxygen molecule is:
The kinetic energy of the rotational motion of all gas molecules:
We find the number of all gas molecules from the formula for the amount of substance:
, (3)
where N A \u003d 6.02 · 10 23 mol -1 is Avogadro's number, ν is the amount of substance, M \u003d 32 · 10 -3 kg / mol is the molar mass of oxygen.
Substituting equation (3) into formula (2), we get:
.
Calculations:
Answer: 
, 
Task 3.5. A certain amount of helium expands: first adiabatically and then isobaric. The final gas temperature is equal to the initial one. During adiabatic expansion, the gas performed work equal to 4.5 kJ. What is the work of the gas for the whole process?
In the diagram, process 1-2 is adiabatic, i.e. Q\u003d 0; process 2 - 3 - isobaric ( r \u003d const). Since the initial and final temperatures are equal (according to the problem statement), the process 3 - 1 will be isothermal (T \u003d const).
full work is equal to the sum of work on each of the sections:
A 123 \u003d A 12 + A 23 (1)
According to the first law of thermodynamics for the adiabatic process, given that the gas is monoatomic, the work A 12 of the gas in the section 1-2 is equal to the change in the internal energy, taken with a minus sign:
where M \u003d 4 · 10 -3 kg / mol is the molar mass of helium, T 1 and T 2 are the absolute temperatures of the gas in states 1 and 2, respectively, R\u003d 8.31 - universal gas constant.
The work of isobaric expansion, given that T 3 = T 1 is equal to
Solving the jointly obtained equations
A 123 \u003d A 12 + A 23
A 123 \u003d A 12.
Calculations:
A 123 \u003d 4.5 10 3 \u003d 7500 J
Answer: A 123 \u003d 7500 J.
Task 3.6.Diffusion coefficient Dand viscosity η hydrogen under some conditions are equal D\u003d 1.42 · 10 -4 m 2 / s η \u003d 8.5 μPa s. Hydrogen molecule diameter σ \u003d 0.3 nm . Find the number n hydrogen molecules per unit volume.
Average free path of molecules; ρ is the density of the gas.
Amount of substance:
,
where M \u003d 2 · 10 -3 kg / mol is the molar mass of hydrogen; m is the mass of the gas; N A \u003d 6.02 · 10 23 mol -1 is Avogadro's number, ν is the amount of substance.
Concentration of hydrogen molecules n determined by the number of molecules N per unit volume V:
Solving together the system of equations:
available:
Calculations:
m -3
Answer: n \u003d 1.8 10 25 m -3
task 3.7. The heat engine works according to the reversible Carnot cycle. The temperature of the heat sink T 1 \u003d 500 K. Determine the thermal efficiency of the cycle and the temperature T 2 of the heat sink of the heat machine, if at the expense of each kilojoule of heat received from the heat sink the machine performs work A \u003d 350 J.
Knowing the efficiency of the cycle, it is possible from the formula for the efficiency of the Carnot cycle
find the temperature of the cooler T 2:
.
Calculations:
Answer: η \u003d 35%, T 2 \u003d 325 K
Task 3.8. The mass of 10 g of helium is at a temperature of 300 K. With isobaric heating, its volume has increased 3 times. Determine the change in internal energy, gas work and the amount of heat imparted to the gas.
To determine the temperature T 2, we use the Gay-Lussac law for the isobaric process
.
The work of the gas during its expansion is determined by the expression:
| A \u003d PDV \u003d P (V 2 - V 1). |
Using the Mendeleev-Cliperon equation, we find the difference between the volumes of the two states of the gas (V 2 - V 1):
.
.
To determine the amount of heat imparted to the gas, we use the first law of thermodynamics for the isobaric process:
.
Calculations:
3.3. Tasks for independent decision
201. In a cylinder long l\u003d 1.6 m filled with air at normal atmospheric pressure p 0, began to slowly push in the piston with the base area S\u003d 200 cm 2. Determine strength Facting on the piston if stopped at a distance l 1 \u003d 10 cm from the bottom of the cylinder.
202. The cylinder contains gas at a temperature T 1 \u003d 400 K. To what temperature T 2, it is necessary to heat the gas so that its pressure increases 1.5 times.
203. A container with a capacity V\u003d 20 l filled with nitrogen at a temperature T\u003d 400 K. When part of the gas was consumed, the pressure in the cylinder dropped by Δ p\u003d 200 kPa. Determine the mass m consumed gas. Consider the process isothermal.
204. In a cylinder with a capacity V\u003d 15 l argon is under pressure p 1 \u003d 600 kPa and at temperature T 1 \u003d 300 K. When a certain amount of gas was taken from the cylinder, the pressure in the cylinder dropped to p 2 \u003d 400 kPa, and the temperature is set T 2 \u003d 260 K. Determine the mass m argon taken from the cylinder.
205. Two vessels of the same volume contain oxygen. In one vessel pressure p 1 \u003d 2 MPa and temperature T 1 \u003d 800 K, in the other p 2 \u003d 2.5 MPa, T 2 \u003d 200 K. The vessels were connected by a tube and the oxygen in them was cooled to a temperature T\u003d 200 K. Determine the established pressure in the vessels p.
206. Calculate the density ρ nitrogen in the cylinder under pressure p\u003d 2 MPa and having a temperature T\u003d 400 K.
207. Determine the relative molecular weight M r gas if at a temperature T\u003d 154 K and pressure p\u003d 2.8 MPa it has a density ρ \u003d 6.1 kg / m 3.
208. Find the density ρ nitrogen at temperature T\u003d 400 K and pressure p\u003d 2 MPa.
209. In a vessel with a volume V\u003d 40 liters oxygen is at a temperature T\u003d 300 K. When some of the oxygen was consumed, the pressure in the cylinder dropped by Δ r\u003d 100 kPa. Determine the mass m consumed oxygen. Consider the process isothermal.
210. Determine the density ρ pressurized water vapor p\u003d 2.5 kPa and having a temperature T\u003d 250 K.
211. Determine the internal energy U hydrogen, as well as the average kinetic energy<ε \u003e molecules of this gas at a temperature T\u003d 300 K, if the amount of substance ν this gas is equal to 0.5 mol.
212. Determine the total kinetic energy E to the translational motion of all gas molecules in a vessel with a capacity V\u003d 3 liters under pressure p\u003d 540 kPa.
213. Amount of matter helium ν \u003d 1.5 mol, temperature T\u003d 120 K. Determine the total kinetic energy E to the translational motion of all molecules of this gas.
214. Molar Internal Energy U m of some diatomic gas is 6.02 kJ / mol. Determine the average kinetic energy<ε bp\u003e rotational motion of one molecule of this gas. The gas is considered ideal.
215. Determine the average kinetic energy<ε \u003e one molecule of water vapor at a temperature T\u003d 500 K.
216. Determine the root mean square speed<υ q\u003e gas molecules contained in a vessel with a capacity V\u003d 2 liters under pressure p\u003d 200 kPa. Gas mass m\u003d 0.3 g.
217. Hydrogen is at a temperature T\u003d 300 K. Find the average kinetic energy<ε bp\u003e rotational motion of one molecule, as well as the total kinetic energy E to all molecules of this gas; amount of hydrogen ν \u003d 0.5 mol.
218. At what temperature is the average kinetic energy<ε n\u003e the translational motion of a gas molecule is 4.14 · 10 -21 J?
219. The smallest grains of dust are suspended in nitrogen, which move as if they were very large molecules. The mass of each grain of dust is 6 · 10 -10 g. The gas is at a temperature T\u003d 400 K. Determine the mean square velocity<υ kv\u003e, as well as average kinetic energies<ε k\u003e translational motion of nitrogen molecules and dust particles.
220. Determine the average kinetic energy<ε k\u003e translational motion and<ε bp\u003e rotational motion of nitrogen molecule at temperature T\u003d 1 kK. Determine also the total kinetic energy E to a molecule under the same conditions.
221. Determine the molar mass M diatomic gas and its specific heat capacity, if it is known that the difference c p - c V the specific heat capacities of this gas is 260 J / (kg · K).
222. Find specific c p and c Vas well as molar C p and C V heat capacity of carbon dioxide.
223. Determine the adiabatic exponent γ ideal gas, which at a temperature T\u003d 350 K and pressure p\u003d 0.4 MPa occupies volume V\u003d 300 l and has a heat capacity C V\u003d 857 J / K.
224. In a vessel with a capacity V\u003d 6 l is a diatomic gas under normal conditions. Determine the heat capacity C V
225. Determine the relative molecular weight M r and molar mass of gas M, if the difference of its specific heat capacities c p - c V\u003d 2.08 kJ / (kg K).
226. Determine the molar heat capacity of the gas, if its specific heat c V\u003d 10.4 kJ / (kg K) and c p \u003d 14.6 kJ / (kg K).
227. Find specific c V and c pand molar C V and C p heat capacity of nitrogen and helium.
228. Calculate the specific heat of a gas, knowing that its molar mass M\u003d 4 · 10 -3 kg / mol and the ratio of heat capacities C p / C V= 1,67.
229. Triatomic gas under pressure p\u003d 240 kPa and temperature t\u003d 20 ° C occupies volume V\u003d 10 l. Determine the heat capacity C pthis gas at constant pressure.
230. Monatomic gas under normal conditions occupies a volume V\u003d 5 l. Calculate heat capacity C V of this gas at constant volume.
231. Find the average<z\u003e collisions over time t\u003d 1 s and free path<l\u003e mo
MOLECULAR PHYSICS
FOUNDATIONS OF MOLECULAR KINETIC THEORY
1. Basic provisions of molecular kinetic theory, structure of matter from the point of view of MKT
2. What is called an atom? A molecule?
3. What is called the amount of a substance? What is its unit (give a definition)?
4. What is called molar mass molar volume?
5. How can the mass of molecules be determined; molecular size - what is the approximate mass and size of the molecules?
6. Describe the experiments confirming the main provisions of the ICB.
7. What is called an ideal gas? What conditions should it satisfy? Under what conditions is a real gas close to it in its properties?
8. Write down the formulas for the arithmetic mean speed, mean square speed.
9. What do diffusion experiments prove? Brownian motion? Explain them based on MKT
10. What does Stern's experience prove? Explain based on MKT.
11. Derive and formulate the basic equation of the MKT. What assumptions are used to derive the basic equation of the MKT.
12. What characterizes body temperature?
13. Formulation and mathematical recording of the laws of Dalton, Boyle Marriott, Gay Lussac, Charles.
14. What is physical entity absolute zero temperature? Record the relationship between absolute temperature and Celsius temperature. Is absolute zero achievable, why?
15. How to explain the gas pressure from the point of view of MKT? What does it depend on?
16. What does Avogadro's constant show? What is its meaning?
17. What is the value of the universal gas constant?
18. What is the value of the Boltzmann constant?
19. Write the Mendeleev - Clapeyron equation. What quantities are included in the formula?
20. Write the Clapeyron equation. What quantities are included in the formula?
21. What is called partial gas pressure?
22. What is called an isoprocess, which isoprocesses you know.
23. Concept, definition, internal energy of an ideal gas.
24. Gas parameters. Conclusion of the unified gas law.
25. Derivation of the Mendeleev-Clapeyron equation.
26. What is called: molar mass of a substance, amount of a substance, relative atomic mass of a substance, density, concentration, absolute temperature of a body? What units are they measured in?
27. Gas pressure. Pressure units in SI. Formula. Pressure measuring instruments.
28. Describe and explain the two temperature scales: thermodynamic and practical.
30. Formulate the laws that describe all kinds of isoprocesses?
31. Plot the plot of ideal gas density versus thermodynamic temperature for the isochoric process.
32. Plot the plot of ideal gas density versus thermodynamic temperature for the isobaric process.
33. What is the difference between the Clapeyron-Mendeleev equation and the Clapeyron equation?
34. Write down the formula for the average kinetic energy of an ideal gas.
35. The mean square velocity of the thermal motion of molecules.
36. Average speed of chaotic movement of molecules.
2. The particles that make up substances are called molecules. The particles that make up molecules are called atoms.
3. The quantity that determines the number of molecules in a given sample of a substance is called the amount of a substance. one mole is the amount of a substance that contains as many molecules as there are carbon atoms in 12 grams of carbon.
4. Molar mass of a substance - the mass of one mole of a substance (g / mol) Molar volume - the volume of one mole of a substance, the value obtained by dividing the molar mass by density.
5. Knowing the molar mass, you can calculate the mass of one molecule: m0 \u003d m / N \u003d m / vNA \u003d M / NA The diameter of a molecule is considered to be the minimum distance at which repulsive forces allow them to approach. However, the concept of molecular size is arbitrary. The average molecular size is about 10-10 m.
7. An ideal gas is a model of a real gas that has the following properties:
Molecules are negligible compared to the average distance between them
Molecules behave like small solid balls: they collide elastically with each other and with the walls of the vessel, there are no other interactions between them.
Molecules are in continuous chaotic motion. All gases at not too high pressures and at not too low temperatures are close in their properties to an ideal gas. At high pressures, gas molecules approach so closely that their own dimensions cannot be neglected. With decreasing temperature, the kinetic energy of molecules decreases and becomes comparable to their potential energy; therefore, at low temperatures, the potential energy cannot be neglected.
At high pressures and low temperatures, the gas cannot be considered ideal. This gas is called real. (The behavior of a real gas is described by laws that differ from those of an ideal gas.)
The root mean square velocity of molecules is the root mean square value of the moduli of the velocities of all molecules of the considered amount of gas
And if we write down the universal gas constant, how, and for one molar mass, will we succeed?
In the Formula, we used:
Mean square velocity of molecules
Boltzmann constant
Temperature
Mass of one molecule
Universal gas constant
Molar mass
Amount of substance
Average kinetic energy of molecules
Avogadro's number
The arithmetic mean velocity of molecules is determined by the formula
where M - molar mass of a substance.
9. Brownian motion. Once in 1827, the English scientist R. Brown, studying plants with a microscope, discovered a very unusual phenomenon. Spores floating on the water (small seeds of some plants) moved in leaps and bounds for no apparent reason. Brown watched this movement (see figure) for several days, but he could not wait for it to stop. Brown realized that he was dealing with a phenomenon unknown to science, so he described it in great detail. Subsequently, physicists named this phenomenon by the name of the discoverer - brownian motion.
It is impossible to explain the Brownian motion if not assume that water molecules are in a disorderly, never-ending movement. They collide with each other and with other particles. Bumping into spores, the molecules cause their jump-like movements, which Brown observed through a microscope. And since the molecules are not visible in the microscope, the movement of the spores seemed to Brown unreasonable.
Diffusion
 |
How to explain the acceleration of these phenomena? There is only one explanation: an increase in body temperature leads to an increase in the speed of movement of its constituent particles.
So, what are the conclusions from the experiments? The independent movement of particles of substances is observed at any temperature. However, as the temperature rises, the motion of particles is accelerated, which leads to an increase in their kinetic energy... As a result, these more "energetic" particles accelerate diffusion, Brownian motion, and other phenomena such as dissolution or evaporation.
10. Stern's experience - an experiment in which the speed of molecules was experimentally measured. It was proved that different molecules in a gas have different velocities, and at a given temperature we can talk about the distribution of molecules in terms of velocities and the average velocity of molecules.
Let us set ourselves the task: using simplified concepts of the motion and interaction of gas molecules, express the gas pressure in terms of the quantities characterizing the molecule.
Consider a gas enclosed in a spherical volume with a radius and volume. Abstracting from the collisions of gas molecules, we have the right to accept the following simple scheme of motion for each molecule.
The molecule moves in a straight line and evenly strikes the wall of the vessel with a certain speed and bounces off it at an angle, equal to the angle falling (fig. 83). Passing chords of the same length all the time, the molecule strikes the wall of the vessel in 1 s. At each impact, the momentum of the molecule changes to (see page 57). The pulse change in 1 s will be equal to
We see that the angle of incidence has decreased. If the molecule hits the wall at an acute angle, then the blows will be frequent, but weak; when falling at an angle close to 90 °, the molecule will hit the wall less often, but stronger.
The change in momentum at each collision of the molecule with the wall contributes to the total force of the gas pressure. It can be assumed in accordance with the basic law of mechanics that the force of pressure is nothing
otherwise as a change in the momentum of all molecules that occurs in one second: or, taking the constant term out of the parentheses,
Let the gas contain molecules, then we can introduce into consideration the average square of the molecular velocity, which is determined by the formula
The expression for the pressure force will now be written briefly:
We get the gas pressure by dividing the force expression by the area of \u200b\u200bthe sphere. We get
Replacing with we get the following interesting formula:
So, the gas pressure is proportional to the number of gas molecules and the average value of the kinetic energy of the translational motion of a gas molecule.
We come to the most important conclusion by comparing the obtained equation with the equation of the gas state. A comparison of the right-hand sides of the equalities shows that
that is, the average kinetic energy of the translational motion of molecules depends only on the absolute temperature and, moreover, is directly proportional to it.
This conclusion shows that gases obeying the law of the gas state are ideal in the sense that they approach an ideal model of a collection of particles, the interaction of which is not essential. Further, this conclusion shows that the empirically introduced concept of absolute temperature as a quantity proportional to the pressure of a rarefied gas has a simple molecular-kinetic meaning. The absolute temperature is proportional to the kinetic energy of the translational motion of the molecules. is Avogadro's number - the number of molecules in one gram-molecule, it is a universal constant: The reciprocal will be equal to the mass of a hydrogen atom:
The quantity
It is called the Boltzmann constant Then
If we represent the square of the speed through the sum of the squares of the components, obviously, any component will have an average energy
This quantity is called the energy per degree of freedom.
The universal gas constant is well known from experiments with gases. Determining the Avogadro number or Boltzmann constant (expressed in terms of each other) is a relatively difficult task requiring delicate measurements.
The conclusion made gives us useful formulas, which allow calculating the average velocities of molecules and the number of molecules per unit volume.
So, for the mean square of the speed, we get
Mean square velocity of molecules is the root-mean-square value of the velocity modules of all molecules of the considered amount of gas
Table of values \u200b\u200bof the mean square velocity of molecules of some gases
In order to understand where we get this formula, we will derive the mean square velocity of the molecules. The derivation of the formula begins with the basic molecular equation kinetic theory (MKT):
Where we have the amount of a substance, for an easier proof, we take 1 mole of a substance for consideration, then we get:
If you look, then PV is two-thirds of the average kinetic energy of all molecules (and we took 1 mole of molecules):
Then, if we equate the right-hand sides, we get that for 1 mole of gas, the average kinetic energy will be:
But the average kinetic energy is also found as:
But now, if we equate the right-hand sides and express the speed from them and take the square, Avogadro's number per molecule mass, we get the Molar mass, then we get the formula for the mean square velocity of a gas molecule:
And if we write down the universal gas constant, how, and for one molar mass, will we succeed?
In the Formula, we used:
Mean square velocity of molecules
Boltzmann constant