Management of Mendeleev's clapeyron. Ideal gas

Gas 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 it 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 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.

1. The potential energy of interaction between molecules of an ideal gas is so small that it is neglected compared to kinetic energy.
2. 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.
3. 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 pressure of a gas 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

That's what it is 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. The pressure increases accordingly.

Graphically, an isothermal process is displayed on the plane of the curve, 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 : " Etc 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, it was clarified by Gay-Lussac. 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 .

How more 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.

1. An ideal gas is a gas in which there are no intermolecular interaction forces. With a sufficient degree of accuracy, gases can be considered ideal in cases where their states are considered that are far from the regions of phase transformations.
2. The following laws are valid for ideal gases:

a) Boyle’s Law - Mapuomma: at constant temperature and mass, the product of the numerical values ​​of pressure and volume of a gas is constant:
pV = const

Graphically, this law in PV coordinates is depicted by a line called an isotherm (Fig. 1).

b) Gay-Lussac's law: at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature:
V = V0(1 + at)

where V is the volume of gas at temperature t, °C; V0 is its volume at 0°C. The quantity a is called the temperature coefficient of volumetric expansion. For all gases a = (1/273°С-1). Hence,
V = V0(1 +(1/273)t)

Graphically, the dependence of volume on temperature is depicted by a straight line - an isobar (Fig. 2). At very low temperatures (close to -273°C), Gay-Lussac's law is not satisfied, therefore solid line replaced by a dotted line on the graph.

c) Charles’s law: at constant volume, the pressure of a given mass of gas is directly proportional to its absolute temperature:
p = p0(1+gt)

where p0 is the gas pressure at temperature t = 273.15 K.
The quantity g is called the temperature coefficient of pressure. Its value does not depend on the nature of the gas; for all gases = 1/273 °C-1. Thus,
p = p0(1 +(1/273)t)

The graphical dependence of pressure on temperature is depicted by a straight line - an isochore (Fig. 3).

d) Avogadro's law: at the same pressures and the same temperatures and equal volumes various ideal gases contained same number molecules; or, what is the same: at the same pressures and the same temperatures, the gram molecules of different ideal gases occupy the same volumes.
So, for example, under normal conditions (t = 0°C and p = 1 atm = 760 mm Hg), gram molecules of all ideal gases occupy a volume Vm = 22.414 liters. The number of molecules located in 1 cm3 of an ideal gas at under normal conditions, is called the Loschmidt number; it is equal to 2.687*1019> 1/cm3
3. The equation of state of an ideal gas has the form:
pVm = RT

where p, Vm and T are the pressure, molar volume and absolute temperature of the gas, and R is the universal gas constant, numerically equal to the work done by 1 mole of an ideal gas when heated isobarically by one degree:
R = 8.31*103 J/(kmol*deg)

For an arbitrary mass M of gas, the volume will be V = (M/m)*Vm and the equation of state has the form:
pV = (M/m) RT

This equation is called the Mendeleev-Clapeyron equation.
4. From the Mendeleev-Clapeyron equation it follows that the number n0 of molecules contained in a unit volume of an ideal gas is equal to
n0 = NA/Vm = p*NA /(R*T) = p/(kT)

where k = R/NA = 1/38*1023 J/deg - Boltzmann's constant, NA - Avogadro's number.

This equation is valid for all gases in any quantities 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 is the 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 liters, 1 mole of gas contains (mol -1) molecules.

22.4 l contains 6.02 * 10 23 molecules

1 ml of hydrogen contains X molecules

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:

2.2. 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 law of the gas state, 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.

As already indicated, the state of a certain mass 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.

French physicist B. Clapeyron derived the equation of state of an ideal gas by combining the Boyle-Mariotte and Gay-Lussac laws.

1) isothermal (isotherm 1-1¢),

2) isochoric (isochore 1¢-2).

In accordance with the Boyle-Mariotte laws (1.1) and Gay-Lussac (1.4), we write:

Eliminating p 1 " from equations (1.5) and (1.6), we obtain

Since states 1 and 2 were chosen arbitrarily, for a given mass of gas the value remains constant, i.e.

. (1.7)
Expression (1.7) is the Clapeyron equation, in which B is the gas constant, different for different gases.

The Russian scientist D.I. Mendeleev combined the Clapeyron equation with Avogadro's law, relating equation (1.7) to one mole, using the molar volume V m. According to Avogadro's law, at the same p and T, the moles of all gases occupy the same molar volume V m, therefore the constant B will be the same for all gases. This constant common to all gases is denoted by R and is called molar gas constant. Equation

satisfies only an ideal gas, and it is ideal gas equation of state, also called Mendeleev-Clapeyron equation.

Numeric value we determine the molar gas constant from formula (1.8), 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 (1.8) for a mole of gas one can go to the Clapeyron-Mendeleev equation for an arbitrary mass of gas. If at a certain given pressure and temperature one mole of gas occupies the volume V m, then under the same conditions the mass m of gas will occupy the volume, where M is molar mass(mass of one mole of substance). The unit of molar mass is kilogram per mole (kg/mol). Clapeyron-Mendeleev equation for gas mass m

where is the amount of substance.

A slightly different form of the ideal gas equation of state is often used, introducing Boltzmann constant:

Based on this, we write the equation of state (1.8) in the form

where is the concentration of molecules (the number of molecules per unit volume). Thus, from Eq.

р=nkT (1.10)
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 under normal conditions is called Loschmidt number:

Basic equation of molecular kinetics

Ideal gas theories

To derive the basic equation of molecular kinetic theory, consider a monatomic ideal gas. Let us assume that gas molecules move chaotically, the number of mutual collisions between them 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 DS on the wall of the vessel (Fig. 50) and calculate the pressure exerted on this area.

During the time Dt of the DS site, only those molecules that are enclosed in the volume of a cylinder with a base DS and a height Dt have reached (Fig. 50).

The number of these molecules is equal to nDSDt (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, with half (1/6) moving 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 DS will be 1/6nDS Dt. When colliding with the platform, these molecules will transfer momentum to it

The Mendeleev-Clapeyron equation is an equation of state for an ideal gas, referred to 1 mole of gas. In 1874, D.I. Mendeleev, based on the Clapeyron equation, combining it with Avogadro’s law, using the molar volume V m and relating it to 1 mole, derived the equation of state for 1 mole of an ideal gas:

pV = RT, Where R- universal gas constant,

R = 8.31 J/(mol. K)

The Clapeyron-Mendeleev equation shows that for a given mass of gas it is possible to simultaneously change three parameters characterizing the state of an ideal gas. For an arbitrary mass of gas M, the molar mass of which is m: pV = (M/m) . RT. or pV = N A kT,

where N A is Avogadro's number, k is Boltzmann's constant.

Derivation of the equation:

Using the equation of state of an ideal gas, one can study processes in which the mass of the gas and one of the parameters - pressure, volume or temperature - remain constant, and only the other two change, and theoretically obtain gas laws for these conditions of change in the state of the gas.

In order for the gas temperature to remain unchanged during the process, it is necessary that the gas can exchange heat with an external large system - a thermostat. The role of a thermostat can be played by the external environment (atmospheric air). According to the Boyle-Mariotte law, gas pressure is inversely proportional to its volume: P 1 V 1 =P 2 V 2 =const. The graphical dependence of gas pressure on volume is depicted in the form of a curve (hyperbola), which is called an isotherm. Different isotherms correspond to different temperatures.Isobaric process - the process of changing the state of a system at constant pressure. For a gas of a given mass, the ratio of gas volume to its temperature remains constant if the gas pressure does not change. This Gay-Lussac's law. According to Gay-Lussac's law, the volume of a gas is directly proportional to its temperature: V/T=const. Graphically, this dependence is V-T coordinates

Isochoric processis depicted as a straight line coming from the point T=0. This straight line is called an isobar. Different pressures correspond to different isobars. Gay-Lussac's law is not observed in the region of low temperatures close to the temperature of liquefaction (condensation) of gases.

So, from the law pV = (M/m). RT derives the following laws:

T = const=> PV = const- Boyle's law - Mariotta.

p = const => V/T = const- Gay-Lussac's law.

If an ideal gas is a mixture of several gases, then according to Dalton’s law, the pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the gases entering it. Partial pressure is the pressure that a gas would produce if it alone occupied the entire volume equal to the volume of the mixture.

Some may be interested in the question of how it was possible to determine Avogadro’s constant N A = 6.02·10 23? The value of Avogadro's number was experimentally established only at the end of the 19th and beginning of the 20th centuries. Let us describe one of these experiments.

A sample of the radium element weighing 0.5 g was placed in a vessel with a volume V = 30 ml, evacuated to a deep vacuum and kept there for one year. It was known that 1 g of radium emits 3.7 x 10 10 alpha particles per second. These particles are helium nuclei, which immediately accept electrons from the walls of the vessel and turn into helium atoms. Over the course of a year, the pressure in the vessel increased to 7.95·10 -4 atm (at a temperature of 27 o C). The change in the mass of radium over a year can be neglected. So, what is N A equal to?

First, let's find how many alpha particles (that is, helium atoms) were formed in one year. Let's denote this number as N atoms:

N = 3.7 10 10 0.5 g 60 sec 60 min 24 hours 365 days = 5.83 10 17 atoms.

Let us write the Clapeyron-Mendeleev equation PV = n RT and note that the number of moles of helium n= N/N A . From here:

N A = NRT = 5,83 . 10 17 . 0,0821 . 300 = 6,02 . 10 23

PV 7.95. 10 -4. 3. 10 -2

At the beginning of the 20th century, this method of determining Avogadro's constant was the most accurate. But why did the experiment last so long (a year)? The fact is that radium is very difficult to obtain. With its small amount (0.5 g), the radioactive decay of this element produces very little helium. And the less gas in a closed vessel, the less pressure it will create and the greater the measurement error will be. It is clear that a noticeable amount of helium can be formed from radium only over a sufficiently long time.