Gas laws. Mendeleev-Clapeyron equation.
Experimental study of the properties of gases, carried out in the XVII-XVIII centuries. Boyle, Mariotte, Gay-Lussac, Charles, led to the formulation of gas laws.
1. Isothermal process – T= const .
Boyle-Mariotte law: pV=const.
Dependency graph p from V shown in Fig. 2.1. The higher the isotherm, the higher the temperature it corresponds to, T 2 >T 1 .
2. Isobaric process – p= const .
Gay-Lussac's Law: 
.
The graph of V versus T is shown in Fig. 2.2. The lower the isobar is inclined to the temperature axis, the greater the pressure it corresponds to, p 2 > p 1 .
3. Isochoric process – V=const .
Charles's Law: 
.
Dependency graph r from T shown in Figure 2.3. The lower the isochore is inclined to the temperature axis, the larger the volume it corresponds to, V 2 > V 1 .
Combining the expressions of gas laws, we obtain an equation relating p, V, T (combined gas law): 
.
The constant in this equation is determined experimentally. For the amount of gas substance 1 mole it turned out to be equal to R=8.31 J/(mol×K) and was called universal gas constant.
1 mole equal to the amount of substance in a system containing the same number of structural elements as there are atoms in carbon-12 weighing 0.012 kg. Number of molecules (structural units) in 1 mole equal to Avogadro's number: N A =6.02.10 23 mol -1. For R the relation is valid: R=k N A
So for one begging: 
.
For an arbitrary amount of gas n = m/m, Where m- molar mass of the gas. As a result, we obtain the equation of state of an ideal gas, or the Mendeleev-Clapeyron equation 
.
This equation is valid for all gases in any quantity and for all values of P, V and T at which gases can be considered ideal
where R is the universal gas constant;
R=8.314 J/mol k =0.0821 l amu/mol k
The composition of gas mixtures is expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the mixture
,
where is the volume fraction of component X, V(x) is the volume of component X; V is the volume of the system.
Volume fraction is a dimensionless quantity; it is expressed in fractions of a unit or as a percentage.
IV. Examples of problem solving.
Problem 1. What volume does 0.2 mole of any gas occupy at ground level?
Solution: The amount of substance is determined by the formula:
Problem 2. What is the volume at standard conditions? takes 11g. carbon dioxide?
Solution: The amount of substance is determined
Problem 3. Calculate the relative density of hydrogen chloride to nitrogen, to hydrogen, to air.
Solution: Relative density is determined by the formula:
;
;
Problem 4.Calculation of the molecular mass of a gas for a given volume.
The mass of 327 ml of gas at 13 0 C and a pressure of 1.04 * 10 5 Pa is equal to 828 g.
Calculate the molecular mass of the gas.
Solution: The molecular mass of a gas can be calculated using the Mendeleev-Clapeyron equation:
The value of the gas constant is determined by the accepted units of measurement. If pressure is measured in Pa and volume in m3, then .
Problem 5. Calculation of the absolute mass in a molecule of a substance.
1. Determine the mass of a gas molecule if the mass of 1 liter of gas at ground level. equal to 1.785g.
Solution: Based on the molecular volume of the gas, we determine the mass of a mole of gas
where m is the gas mass;
M – molar mass of gas;
Vm – molar volume, 22.4 l/mol;
V – volume of gas.
2. The number of molecules in a mole of any substance is equal to Avogadro’s constant ( 
). Therefore, the number of molecules m is equal to:
Problem 6. How many molecules are contained in 1 ml of hydrogen at standard conditions?
Solution: According to Avogadro's law, 1 mole of gas at no. occupies a volume of 22.4 l, 1 mole of gas contains 
(mol -1) molecules.
22.4 l contains 6.02 * 10 23 molecules
1 ml of hydrogen contains X molecules
Answer: 
Problem 7. Deriving formulas.
I. Organic matter contains carbon (mass fraction 84.21%) and hydrogen (15.79%). The vapor density of the substance in air is 3.93.
Determine the formula of the substance.
Solution: We represent the formula of the substance in the form CxHy.
1. Calculate the molar mass of a hydrocarbon using the air density.
2. Determine the amount of carbon and hydrogen substances
II. Determine the formula of the substance. With a content of 145 g of it, 330 g of CO 2 and 135 g of H 2 O are obtained. The relative vapor density of this substance with respect to hydrogen is 29.
1. Determine the mass of the unknown substance:
2. Determine the mass of hydrogen:
2.1.
2.2. Determine the mass of carbon:
2.3. We determine whether there is a third element - oxygen.
That. m(O) = 40g
To express the resulting equation in integers (since this is the number of atoms in the molecule), we divide all its numbers by the smaller of them
Then the simplest formula of the unknown substance is C 3 H 6 O.
2.5. → the simplest formula is the unknown substance we are looking for.
Answer: C 3 H 5 O
Problem 8: (Decide on your own)
The compound contains 46.15% carbon, the rest nitrogen. The air density is 1.79.
Find the true formula of the compound.
Problem 9: (decide for yourself)
Are the number of molecules the same?
a) in 0.5 g of nitrogen and 0.5 g of methane
b) in 0.5 l of nitrogen and 0.5 l of methane
c) in mixtures of 1.1 g CO 2 and 2.4 g ozone and 1.32 g CO 2 and 2.16 g ozone
Problem 10: Relative density of hydrogen halide in air is 2.8. Determine the density of this gas in air and name it.
Solution: according to the gas law 
, i.e. the ratio of the molar mass of hydrogen halide (M (HX)) to the molar mass of air (M HX) is 2.8 →
Then the molar mass of the halogen is:
→ X is Br and the gas is hydrogen bromide.
Relative density of hydrogen bromide by hydrogen:
Answer: 40.5, hydrogen bromide.
As already indicated, the state of a certain mass of gas is determined by three thermodynamic parameters: pressure p, volume V and temperature T. There is a certain relationship between these parameters, called the equation of state, which is generally given by the expression
where each variable is a function of the other two.
The French physicist and engineer B. Clapeyron (1799-1864) derived the equation of state of an ideal gas by combining the Boyle-Mariotte and Gay-Lussac laws. Let a certain mass of gas occupy volume V 1 , has a pressure p 1 and is at a temperature T 1 . The same mass of gas in another arbitrary state is characterized by the parameters p 2, V 2, T 2 (Fig. 63). The transition from state 1 to state 2 occurs in the form of two processes: 1) isothermal (isotherm 1 - 1¢, 2) isochoric (isochore 1¢ - 2).
In accordance with the Boyle-Mariotte laws (41.1) and Gay-Lussac (41.5), we write:
(42.1) (42.2)
By excluding p¢ 1 from equations (42.1) and (42.2) , we get
Since states 1 and 2 were chosen arbitrarily, for a given gas mass the value pV/T remains constant, i.e.
Expression (42.3) is the Clapeyron equation, in which IN- gas constant, different for different gases.
Russian scientist D.I. Mendeleev (1834-1907) combined Clapeyron's equation with Avogadro's law, relating equation (42.3) to one mole, using the molar volume Vm. According to Avogadro's law, for equal r And T moles of all gases occupy the same molar volume Vm, therefore constant B will the same for all gases. This constant common to all gases is denoted R and is called the molar gas constant. Equation
(42.4)
satisfies only an ideal gas, and it is the equation of state of an ideal gas, also called the Clapeyron-Mendeleev equation.
We determine the numerical value of the molar gas constant from formula (42.4), assuming that a mole of gas is under normal conditions (p 0 = 1.013×10 5 Pa, T 0 = 273.15 K, V m = 22.41×10 -3 m e /mol): R = 8.31 J/(mol×K).
From equation (42.4) for a mole of gas one can go to the Clapeyron-Mendeleev equation for an arbitrary mass of gas. If at some given pressure and temperature one mole of gas occupies a molar volume Vm, then under the same conditions the mass m of gas will occupy the volume V= (t/M)× V m , Where M- molar mass (mass of one mole of a substance). The unit of molar mass is kilogram per mole (kg/mol). Clapeyron-Mendeleev equation for mass T gas
(42.5)
Where v=m/M- amount of substance.
A slightly different form of the ideal gas equation of state is often used, introducing the Boltzmann constant:
Based on this, we write the equation of state (42.4) in the form
where N A /V m = n is the concentration of molecules (the number of molecules per unit volume). Thus, from Eq.
it follows that the pressure of an ideal gas at a given temperature is directly proportional to the concentration of its molecules (or gas density). At the same temperature and pressure, all gases contain the same number of molecules per unit volume. The number of molecules contained in 1 m 3 of gas at normal conditions, called the Loschmandt number*:
Basic equation
Molecular kinetic theory
Ideal gases
To derive the basic equation of molecular kinetic theory, consider a one-atomic ideal gas. Let us assume that gas molecules move chaotically, the number of mutual collisions between gas molecules is negligible compared to the number of impacts on the walls of the vessel, and the collisions of molecules with the walls of the vessel are absolutely elastic. Let us select some elementary area D on the wall of the vessel S(Fig. 64) and calculate the pressure exerted on this area. With each collision, a molecule moving perpendicular to the platform transfers momentum to it m 0 v -(- t 0) = 2t 0 v, where m 0 is the mass of the molecule, v is its speed. During time D t sites D S only those molecules that are contained in the volume of a cylinder with base D are reached S and height vDt (Fig. 64). The number of these molecules is equal to nDSvDt (n-concentration of molecules).
It is necessary, however, to take into account that in reality the molecules move towards the DS site at different angles and have different speeds, and the speed of the molecules changes with each collision. To simplify calculations, the chaotic movement of molecules is replaced by movement along three mutually perpendicular directions, so that at any moment of time 1/3 of the molecules move along each of them, and half of the molecules - 1/6 - move along a given direction in one direction, half - in the opposite direction . Then the number of impacts of molecules moving in a given direction on the platform D S will
l/6 nDSvDt . When colliding with the platform, these molecules will transfer momentum to it
Then the gas pressure exerted by it on the wall of the vessel is
If the gas volume V contains N molecules moving at speeds v 1 , v 2 , ..., v n , then it is advisable to consider the mean square speed
(43.2)
characterizing the entire set of molecules of the pelvis. Equation (43.1) taking into account (43.2) will take the form
(43.3)
Expression (43.3) is called the basic equation of the molecular-kinetic theory of ideal gases. An exact calculation taking into account the movement of molecules in all possible directions gives the same formula.
Considering that n=N/V, we get
Where E- the total kinetic energy of the translational motion of all gas molecules.
Since the mass of gas m=Nm 0 , then equation (43.4) can be rewritten as
For one mole of gas t = M (M- molar mass), therefore
where F m is the molar volume. On the other hand, according to the Clapeyron-Mendeleev equation, pV m = RT. Thus,
(43.6)
Since M = m 0 N A is the mass of one molecule, and N A is Avogadro’s constant, it follows from equation (43.6) that
(43.7)
where k=R/N A is Boltzmann's constant. From here we find that at room temperature, oxygen molecules have a mean square speed of 480 m/s, hydrogen molecules - 1900 m/s. At the temperature of liquid helium, the same speeds will be 40 and 160 m/s, respectively.
Average kinetic energy of translational motion of one ideal gas molecule
(we used formulas (43.5) and (43.7)) is proportional to the thermodynamic temperature and depends only on it. From this equation it follows that at T=0
As already indicated, the state of a certain mass of gas is determined by three thermodynamic parameters: pressure p, volume V and temperature T.
There is a certain relationship between these parameters, called equation of state, which is generally given by the expression
f(p,V,T)=0,
where each variable is a function of the other two.
The French physicist and engineer B. Clapeyron (1799-1864) derived the equation of state of an ideal gas by combining the Boyle-Mariotte and Gay-Lussac laws. Let a certain mass of gas occupy a volume V 1 , has pressure r 1 and is at a temperature T 1 . The same mass of gas in another arbitrary state is characterized by the parameters r 2 , V 2 , T 2 (Fig. 63). Transition from state 1 in a state 2 carried out in the form of two processes: 1) isothermal (isotherm 1 -1 "), 2) isochoric (isochor 1 "-2).
In accordance with the Boyle-Mariotte laws (41.1) and Gay-Lussac (41.5), we write:
p 1 V 1 =p" 1 V 2 , (42.1)
p" 1 /p" 2 =T 1 /T 2. (42.2)
Eliminating from equations (42.1) and (42.2) p" 1 , we get
p 1 V 1 /T 1 =p 2 V 2 / T 2 .
Since the states 1 And 2 were chosen arbitrarily, then for a given mass of gas
magnitude pV/T remains constant
pV/T =B=const.(42.3)
Expression (42.3) is Clapeyron's equation, in which IN- gas constant, different for different gases.
Russian scientist D.I. Mendeleev (1834-1907) combined Clapeyron's equation with Avogadro's law, relating equation (42.3) to one mole, using the molar volume V T . According to Avogadro's law, for equal r And T moles of all gases occupy the same molar volume V m , therefore constant IN will the same for all gases. This constant common to all gases is denoted R and is called molar gas constant. Equation
pV m =RT(42.4)
satisfies only an ideal gas, and it is equation of state of an ideal gas, also called Clapeyron-Mendeleev equation.
We determine the numerical value of the molar gas constant from formula (42.4), assuming that a mole of gas is under normal conditions (p 0 = 1.013 10 5 Pa, T 0 = 273.15 K:, V m = 22.41 10 -3 m 3 /mol): R = 8.31 J/(mol K).
From equation (42.4) for a mole of gas one can go to the Clapeyron-Mendeleev equation for an arbitrary mass of gas. If at certain given pressures and temperatures one mole of gas occupies a molar volume l/m, then under the same conditions the mass t of gas will take up volume V = (m/M)V m , Where M- molar mass(mass of one mole of substance). The unit of molar mass is kilogram per mole (kg/mol). Clapeyron-Mendeleev equation for mass t of gas
Where v = m/M- amount of substance.
A slightly different form of the ideal gas equation of state is often used, introducing Boltzmann constant:
k=R/N A =1.38 10 -2 3 J/K.
Based on this, we write the equation of state (42.4) in the form
p = RT/V m = kN A T/V m = nkT,
Where N A/ V m = n-concentration of molecules (number of molecules per unit volume). Thus, from Eq.
p = nkT(42.6)
it follows that the pressure of an ideal gas at a given temperature is directly proportional to the concentration of its molecules (or gas density). At the same temperature and pressure, all gases contain the same number of molecules per unit volume. The number of molecules contained in 1 m 3 of gas at normal conditions, called numberLoschmidt :
N L = P0 /(kT 0 ) = 2.68 10 25 m -3 .
Details Category: Molecular kinetic theory Published 05.11.2014 07:28 Views: 13238Gas is one of four states of aggregation in which a substance can exist.
The particles that make up the gas are very mobile. They move almost freely and chaotically, periodically colliding with each other like billiard balls. Such a collision is called elastic collision . During a collision, they dramatically change the nature of their movement.
Since in gaseous substances the distance between molecules, atoms and ions is much greater than their sizes, these particles interact very weakly with each other, and their potential interaction energy is very small compared to the kinetic energy.
The connections between molecules in a real gas are complex. Therefore, it is also quite difficult to describe the dependence of its temperature, pressure, volume on the properties of the molecules themselves, their quantity, and the speed of their movement. But the task is greatly simplified if, instead of real gas, we consider its mathematical model - ideal gas .
It is assumed that in the ideal gas model there are no attractive or repulsive forces between molecules. They all move independently of each other. And the laws of classical Newtonian mechanics can be applied to each of them. And they interact with each other only during elastic collisions. The time of the collision itself is very short compared to the time between collisions.
Classical ideal gas
Let's try to imagine the molecules of an ideal gas as small balls located in a huge cube at a great distance from each other. Because of this distance, they cannot interact with each other. Therefore, their potential energy is zero. But these balls move at great speed. This means they have kinetic energy. When they collide with each other and with the walls of the cube, they behave like balls, that is, they bounce off elastically. At the same time, they change the direction of their movement, but do not change their speed. This is roughly what the motion of molecules in an ideal gas looks like.
- The potential energy of interaction between molecules of an ideal gas is so small that it is neglected compared to kinetic energy.
- Molecules in an ideal gas are also so small that they can be considered material points. And this means that they total volume is also negligible compared to the volume of the vessel in which the gas is located. And this volume is also neglected.
- The average time between collisions of molecules is much greater than the time of their interaction during a collision. Therefore, the interaction time is also neglected.
Gas always takes the shape of the container in which it is located. Moving particles collide with each other and with the walls of the container. During an impact, each molecule exerts some force on the wall for a very short period of time. This is how it arises pressure . The total gas pressure is the sum of the pressures of all molecules.
Ideal gas equation of state
The state of an ideal gas is characterized by three parameters: pressure, volume And temperature. The relationship between them is described by the equation:
Where r - pressure,
V M - molar volume,
R - universal gas constant,
T - absolute temperature (degrees Kelvin).
Because V M = V / n , Where V - volume, n - the amount of substance, and n= m/M , That
Where m - gas mass, M - molar mass. This equation is called Mendeleev-Clayperon equation .
At constant mass the equation becomes:
This equation is called united gas law .
Using the Mendeleev-Cliperon law, one of the gas parameters can be determined if the other two are known.
Isoprocesses
Using the equation of the unified gas law, it is possible to study processes in which the mass of a gas and one of the most important parameters - pressure, temperature or volume - remain constant. In physics such processes are called isoprocesses .
From The unified gas law leads to other important gas laws: Boyle-Mariotte law, Gay-Lussac's law, Charles's law, or Gay-Lussac's second law.
Isothermal process
A process in which pressure or volume changes but temperature remains constant is called isothermal process .
In an isothermal process T = const, m = const .
The behavior of a gas in an isothermal process is described by Boyle-Mariotte law . This law was discovered experimentally English physicist Robert Boyle in 1662 and French physicist Edme Mariotte in 1679. Moreover, they did this independently of each other. The Boyle-Mariotte law is formulated as follows: In an ideal gas at a constant temperature, the product of the gas pressure and its volume is also constant.
The Boyle-Marriott equation can be derived from the unified gas law. Substituting into the formula T = const , we get
p · V = const
This is it Boyle-Mariotte law . From the formula it is clear that the pressure of a gas at constant temperature is inversely proportional to its volume. The higher the pressure, the lower the volume, and vice versa.
How to explain this phenomenon? Why does the pressure of a gas decrease as the volume of a gas increases?
Since the temperature of the gas does not change, the frequency of collisions of molecules with the walls of the vessel does not change. If the volume increases, the concentration of molecules becomes less. Consequently, per unit area there will be fewer molecules that collide with the walls per unit time. The pressure drops. As the volume decreases, the number of collisions, on the contrary, increases. Accordingly, the pressure increases.
Graphically, an isothermal process is displayed on a curve plane, which is called isotherm . She has a shape hyperboles.
Each temperature value has its own isotherm. The higher the temperature, the higher the corresponding isotherm is located.
Isobaric process
The processes of changing the temperature and volume of a gas at constant pressure are called isobaric . For this process m = const, P = const.
The dependence of the volume of a gas on its temperature at constant pressure was also established experimentally French chemist and physicist Joseph Louis Gay-Lussac, who published it in 1802. That is why it is called Gay-Lussac's law : " Pr and constant pressure, the ratio of the volume of a constant mass of gas to its absolute temperature is a constant value."
At P = const the equation of the unified gas law turns into Gay-Lussac equation .
An example of an isobaric process is a gas located inside a cylinder in which a piston moves. As the temperature rises, the frequency of molecules hitting the walls increases. The pressure increases and the piston rises. As a result, the volume occupied by the gas in the cylinder increases.
Graphically, an isobaric process is represented by a straight line, which is called isobar .
The higher the pressure in the gas, the lower the corresponding isobar is located on the graph.
Isochoric process
Isochoric, or isochoric, is the process of changing the pressure and temperature of an ideal gas at constant volume.
For an isochoric process m = const, V = const.
It is very simple to imagine such a process. It occurs in a vessel of a fixed volume. For example, in a cylinder, the piston in which does not move, but is rigidly fixed.
The isochoric process is described Charles's law : « For a given mass of gas at constant volume, its pressure is proportional to temperature" The French inventor and scientist Jacques Alexandre César Charles established this relationship through experiments in 1787. In 1802, Gay-Lussac refined it. Therefore this law is sometimes called Gay-Lussac's second law.
At V = const from the equation of the unified gas law we get the equation Charles's law or Gay-Lussac's second law .
At constant volume, the pressure of a gas increases if its temperature increases. .
On graphs, an isochoric process is represented by a line called isochore .
The larger the volume occupied by the gas, the lower the isochore corresponding to this volume is located.
In reality, no gas parameter can be maintained unchanged. This can only be done in laboratory conditions.
Of course, an ideal gas does not exist in nature. But in real rarefied gases at very low temperatures and pressures not exceeding 200 atmospheres, the distance between the molecules is much greater than their sizes. Therefore, their properties approach those of an ideal gas.
