Gas laws. Avogadro's law. Molar volume of gas

Where m is mass, M is molar mass, V is volume.

4. Avogadro's law. Established by the Italian physicist Avogadro in 1811. Identical volumes of any gases, taken at the same temperature and the same pressure, contain the same number of molecules.

Thus, we can formulate the concept of the amount of a substance: 1 mole of a substance contains a number of particles equal to 6.02 * 10 23 (called Avogadro’s constant)

The consequence of this law is that Under normal conditions (P 0 =101.3 kPa and T 0 =298 K), 1 mole of any gas occupies a volume equal to 22.4 liters.

5. Boyle-Mariotte Law

At constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is located:

6. Gay-Lussac's Law

At constant pressure, the change in gas volume is directly proportional to temperature:

V/T = const.

7. The relationship between gas volume, pressure and temperature can be expressed combined Boyle-Mariotte and Gay-Lussac law, which is used to convert gas volumes from one condition to another:

P 0 , V 0 , T 0 - pressure of volume and temperature under normal conditions: P 0 =760 mm Hg. Art. or 101.3 kPa; T 0 =273 K (0 0 C)

8. Independent assessment of the molecular value masses M can be done using the so-called ideal gas equations of state or Clapeyron-Mendeleev equations :

pV=(m/M)*RT=vRT.(1.1)

Where R - gas pressure in a closed system, V- volume of the system, T - gas mass, T - absolute temperature, R- universal gas constant.

Note that the value of the constant R can be obtained by substituting values ​​characterizing one mole of gas at normal conditions into equation (1.1):

r = (p V)/(T)=(101.325 kPa 22.4 l)/(1 mol 273K)=8.31J/mol.K)

Examples of problem solving

Example 1. Bringing the volume of gas to normal conditions.

What volume (n.s.) will be occupied by 0.4×10 -3 m 3 of gas located at 50 0 C and a pressure of 0.954×10 5 Pa?

Solution. To bring the volume of gas to normal conditions, use a general formula combining the Boyle-Mariotte and Gay-Lussac laws:

pV/T = p 0 V 0 /T 0 .

The volume of gas (n.s.) is equal to , where T 0 = 273 K; p 0 = 1.013 × 10 5 Pa; T = 273 + 50 = 323 K;

m 3 = 0.32 × 10 -3 m 3.

At (norm) the gas occupies a volume equal to 0.32×10 -3 m 3 .

Example 2. Calculation of the relative density of a gas from its molecular weight.

Calculate the density of ethane C 2 H 6 based on hydrogen and air.

Solution. From Avogadro's law it follows that the relative density of one gas to another is equal to the ratio of molecular masses ( M h) of these gases, i.e. D=M 1 /M 2. If M 1 C2H6 = 30, M 2 H2 = 2, the average molecular weight of air is 29, then the relative density of ethane with respect to hydrogen is D H2 = 30/2 =15.

Relative density of ethane in air: D air= 30/29 = 1.03, i.e. ethane is 15 times heavier than hydrogen and 1.03 times heavier than air.

Example 3. Determination of the average molecular weight of a mixture of gases by relative density.

Calculate the average molecular weight of a mixture of gases consisting of 80% methane and 20% oxygen (by volume), using the relative densities of these gases with respect to hydrogen.

Solution. Often calculations are made according to the mixing rule, which states that the ratio of the volumes of gases in a two-component gas mixture is inversely proportional to the differences between the density of the mixture and the densities of the gases that make up this mixture. Let us denote the relative density of the gas mixture with respect to hydrogen by D H2. it will be greater than the density of methane, but less than the density of oxygen:

80D H2 – 640 = 320 – 20 D H2; D H2 = 9.6.

The hydrogen density of this mixture of gases is 9.6. average molecular weight of the gas mixture M H2 = 2 D H2 = 9.6×2 = 19.2.

Example 4. Calculation of the molar mass of a gas.

The mass of 0.327×10 -3 m 3 gas at 13 0 C and a pressure of 1.040×10 5 Pa is equal to 0.828×10 -3 kg. Calculate the molar mass of the gas.

Solution. The molar mass of a gas can be calculated using the Mendeleev-Clapeyron equation:

Where m– mass of gas; M– molar mass of gas; R– molar (universal) gas constant, the value of which is determined by the accepted units of measurement.

If pressure is measured in Pa and volume in m3, then R=8.3144×10 3 J/(kmol×K).

Molecular physics studies the properties of bodies based on the behavior of individual molecules. All visible processes occur at the level of interaction of the smallest particles; what we see with the naked eye is only a consequence of these subtle deep connections.

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Basic Concepts

Molecular physics is sometimes seen as a theoretical complement to thermodynamics. Having emerged much earlier, thermodynamics dealt with the study of the transition of heat into work, pursuing purely practical goals. She did not provide a theoretical justification, describing only the results of experiments. The basic concepts of molecular physics emerged later, in the 19th century.

She studies the interaction of bodies at the molecular level, guided by a statistical method that determines patterns in the chaotic movements of minimal particles - molecules. Molecular physics and thermodynamics complement each other, looking at processes from different points of view. At the same time, thermodynamics does not concern atomic processes, dealing only with macroscopic bodies, and molecular physics, on the contrary, considers any process precisely from the point of view of the interaction of individual structural units.

All concepts and processes have their own designations and are described by special formulas that most clearly represent the interactions and dependencies of certain parameters on each other. Processes and phenomena intersect in their manifestations; different formulas can contain the same quantities and be expressed in different ways.

Quantity of substance

The amount of a substance determines the relationship between (mass) and the number of molecules that mass contains. The fact is that different substances with the same mass have different numbers of minimal particles. Processes taking place at the molecular level can only be understood by considering precisely the number of atomic units participating in the interactions. Unit of measurement of the amount of substance, adopted in the SI system, - mole.

Attention! One mole always contains the same number of minimal particles. This number is called Avogadro's number (or constant) and is equal to 6.02x1023.

This constant is used in cases where calculations require taking into account the microscopic structure of a given substance. Dealing with the number of molecules is difficult, since you have to operate with huge numbers, so the mole is used - a number that determines the number of particles per unit mass.

Formula determining the amount of a substance:

The calculation of the amount of a substance is made in different cases, is used in many formulas and is an important value in molecular physics.

Gas pressure

Gas pressure is an important quantity that has not only theoretical but also practical significance. Let's look at the gas pressure formula used in molecular physics, with explanations necessary for better understanding.

To compile the formula, you will have to make some simplifications. Molecules are complex systems, having a multi-stage structure. For simplicity, we consider gas particles in a certain vessel as elastic homogeneous balls that do not interact with each other (ideal gas).

The speed of movement of minimal particles will also be considered the same. By introducing such simplifications, which do not greatly change the true position, we can derive the following definition: gas pressure is the force exerted by the impacts of gas molecules on the walls of vessels.

At the same time, taking into account the three-dimensionality of space and the presence of two directions of each dimension, it is possible to limit the number of structural units acting on the walls to 1/6.

Thus, bringing together all these conditions and assumptions, we can deduce gas pressure formula under ideal conditions.

The formula looks like this:

where P is gas pressure;

n is the concentration of molecules;

K - Boltzmann constant (1.38×10-23);

Ek - gas molecules.

There is another version of the formula:

P = nkT,

where n is the concentration of molecules;

T - absolute temperature.

Gas volume formula

The volume of a gas is the space that a given amount of gas occupies under certain conditions. Unlike solids, which have a constant volume, practically independent of environmental conditions, gas can change volume depending on pressure or temperature.

The formula for gas volume is the Mendeleev-Clapeyron equation, which looks like this:

PV = nRT

where P is gas pressure;

V - volume of gas;

n is the number of moles of gas;

R - universal gas constant;

T is the gas temperature.

By simple rearrangements we obtain the formula for the volume of gas:

Important! According to Avogadro's law, equal volumes of any gases placed in exactly the same conditions - pressure, temperature - will always contain an equal number of minimal particles.

Crystallization

Crystallization is the phase transition of a substance from a liquid to a solid state, i.e. process is the reverse of melting. The crystallization process occurs with the release of heat, which must be removed from the substance. The temperature coincides with the melting point, the whole process is described by the formula:

Q = λm,

where Q is the amount of heat;

λ - heat of fusion;

This formula describes both crystallization and melting, since they are essentially two sides of the same process. In order for a substance to crystallize, it must be cooled to its melting point, and then remove an amount of heat equal to the product of mass and specific heat of fusion (λ). During crystallization, the temperature does not change.

There is another way to understand this term - crystallization from supersaturated solutions. In this case, the reason for the transition is not only the achievement of a certain temperature, but also the degree of saturation of the solution with a certain substance. At a certain stage, the number of solute particles becomes too large, which causes the formation of small single crystals. They attach molecules from solution, producing layer-by-layer growth. Depending on the growth conditions, the crystals have different shapes.

Number of molecules

The easiest way to determine the number of particles contained in a given mass of a substance is using the following formula:

It follows that the number of molecules is equal to:

That is, it is necessary first of all to determine the amount of substance per certain mass. It is then multiplied by Avogadro's number, resulting in the number of structural units. For compounds, calculations are made by summing the atomic weights of the components. Let's look at a simple example:

Let's determine the number of water molecules in 3 grams. The formula (H2O) contains two atoms and one . The total atomic weight of the minimum particle of water will be: 1+1+16 = 18 g/mol.

Amount of substance in 3 grams of water:

Number of molecules:

1/6 × 6 × 1023 = 1023.

Molecule mass formula

One mole always contains the same number of minimal particles. Therefore, knowing the mass of a mole, we can divide it by the number of molecules (Avogadro’s number), resulting in the mass of a system unit.

It should be noted that this formula applies only to inorganic molecules. Organic molecules are much larger in size, their size or weight have completely different meanings.

Molar mass of gas

Molar mass is mass in kilograms of one mole of a substance. Since one mole contains the same number of structural units, the molar mass formula looks like this:

M = κ × Mr

where k is the proportionality coefficient;

Mr is the atomic mass of the substance.

The molar mass of a gas can be calculated using the Mendeleev-Clapeyron equation:

pV = mRT/M,

from which we can deduce:

M = mRT / pV

Thus, the molar mass of a gas is directly proportional to the product of the mass of the gas and the temperature and the universal gas constant and inversely proportional to the product of the gas pressure and its volume.

Attention! It should be taken into account that the molar mass of a gas as an element may differ from the gas as a substance, for example, the molar mass of the element oxygen (O) is 16 g/mol, and the mass of oxygen as a substance (O2) is 32 g/mol.

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Physics in 5 minutes - molecular physics

Conclusion

Formulas contained in molecular physics and thermodynamics allow one to calculate the quantitative values ​​of all processes occurring with solids and gases. Such calculations are necessary both in theoretical research and in practice, since they contribute to solving practical problems.

One of the basic units in the International System of Units (SI) is The unit of quantity of a substance is the mole.

Mole – this is the amount of a substance that contains as many structural units of a given substance (molecules, atoms, ions, etc.) as there are carbon atoms contained in 0.012 kg (12 g) of a carbon isotope 12 WITH .

Considering that the value of the absolute atomic mass for carbon is equal to m(C) = 1.99 10  26 kg, the number of carbon atoms can be calculated N A, contained in 0.012 kg of carbon.

A mole of any substance contains the same number of particles of this substance (structural units). The number of structural units contained in a substance with an amount of one mole is 6.02 10 23 and is called Avogadro's number (N A ).

For example, one mole of copper contains 6.02 10 23 copper atoms (Cu), and one mole of hydrogen (H 2) contains 6.02 10 23 hydrogen molecules.

Molar mass(M) is the mass of a substance taken in an amount of 1 mole.

Molar mass is designated by the letter M and has the dimension [g/mol]. In physics they use the unit [kg/kmol].

In the general case, the numerical value of the molar mass of a substance numerically coincides with the value of its relative molecular (relative atomic) mass.

For example, the relative molecular weight of water is:

Мr(Н 2 О) = 2Аr (Н) + Аr (O) = 2∙1 + 16 = 18 a.m.u.

The molar mass of water has the same value, but is expressed in g/mol:

M (H 2 O) = 18 g/mol.

Thus, a mole of water containing 6.02 10 23 water molecules (respectively 2 6.02 10 23 hydrogen atoms and 6.02 10 23 oxygen atoms) has a mass of 18 grams. Water, with an amount of substance of 1 mole, contains 2 moles of hydrogen atoms and one mole of oxygen atoms.

1.3.4. The relationship between the mass of a substance and its quantity

Knowing the mass of a substance and its chemical formula, and therefore the value of its molar mass, you can determine the amount of the substance and, conversely, knowing the amount of the substance, you can determine its mass. For such calculations you should use the formulas:

where ν is the amount of substance, [mol]; m– mass of the substance, [g] or [kg]; M – molar mass of the substance, [g/mol] or [kg/kmol].

For example, to find the mass of sodium sulfate (Na 2 SO 4) in an amount of 5 moles, we find:

1) the value of the relative molecular mass of Na 2 SO 4, which is the sum of the rounded values ​​of the relative atomic masses:

Мr(Na 2 SO 4) = 2Аr(Na) + Аr(S) + 4Аr(O) = 142,

2) a numerically equal value of the molar mass of the substance:

M(Na 2 SO 4) = 142 g/mol,

3) and, finally, the mass of 5 mol of sodium sulfate:

m = ν M = 5 mol · 142 g/mol = 710 g.

Answer: 710.

1.3.5. The relationship between the volume of a substance and its quantity

Under normal conditions (n.s.), i.e. at pressure R , equal to 101325 Pa (760 mm Hg), and temperature T, equal to 273.15 K (0 С), one mole of different gases and vapors occupies the same volume equal to 22.4 l.

The volume occupied by 1 mole of gas or vapor at ground level is called molar volumegas and has the dimension liter per mole.

V mol = 22.4 l/mol.

Knowing the amount of gaseous substance (ν ) And molar volume value (V mol) you can calculate its volume (V) under normal conditions:

V = ν V mol,

where ν is the amount of substance [mol]; V – volume of gaseous substance [l]; V mol = 22.4 l/mol.

And, conversely, knowing the volume ( V) of a gaseous substance under normal conditions, its quantity (ν) can be calculated :

Before solving problems, you should know the formulas and rules of how to find the volume of gas. We should remember Avogadro's law. And the volume of gas itself can be calculated using several formulas, choosing the appropriate one from them. When selecting the required formula, environmental conditions, in particular temperature and pressure, are of great importance.

Avogadro's law

It says that at the same pressure and the same temperature, the same volumes of different gases will contain the same number of molecules. The number of gas molecules contained in one mole is Avogadro's number. From this law it follows that: 1 Kmol (kilomol) of an ideal gas, any gas, at the same pressure and temperature (760 mm Hg and t = 0*C) always occupies one volume = 22.4136 m3.

How to determine gas volume

• The formula V=n*Vm can most often be found in problems. Here the volume of gas in liters is V, Vm is the molar volume of gas (l/mol), which under normal conditions = 22.4 l/mol, and n is the amount of substance in moles. When the conditions do not have the amount of a substance, but there is a mass of the substance, then we proceed this way: n=m/M. Here M is g/mol (molar mass of the substance), and the mass of the substance in grams is m. In the periodic table it is written under each element, as its atomic mass. Let's add up all the masses and get what we are looking for.
• So, how to calculate the volume of gas. Here is the task: dissolve 10 g of aluminum in hydrochloric acid. Question: how much hydrogen can be released at u.? The reaction equation looks like this: 2Al+6HCl(g)=2AlCl3+3H2. At the very beginning, we find the aluminum (quantity) that reacted according to the formula: n(Al)=m(Al)/M(Al). We take the mass of aluminum (molar) from the periodic table M(Al) = 27 g/mol. Let's substitute: n(Al)=10/27=0.37 mol. From the chemical equation it can be seen that 3 moles of hydrogen are formed when 2 moles of aluminum are dissolved. It is necessary to calculate how much hydrogen will be released from 0.4 moles of aluminum: n(H2)=3*0.37/2=0.56mol. Let's substitute the data into the formula and find the volume of this gas. V=n*Vm=0.56*22.4=12.54l.

The volume of 1 mole of a substance is called the Molar volume. Molar mass of 1 mole of water = 18 g/mol 18 g of water occupy a volume of 18 ml. This means the molar volume of water is 18 ml. 18 g of water occupy a volume equal to 18 ml, because the density of water is 1 g/ml CONCLUSION: Molar volume depends on the density of the substance (for liquids and solids).

1 mole of any gas under normal conditions occupies the same volume equal to 22.4 liters. Normal conditions and their designations no. (0 0 C and 760 mmHg; 1 atm.; 101.3 kPa). The volume of a gas with 1 mole of substance is called molar volume and is denoted by – V m

Solving problems Problem 1 Given: V(NH 3) n.s. = 33.6 m 3 Find: m - ? Solution: 1. Calculate the molar mass of ammonia: M(NH 3) = = 17 kg/kmol

CONCLUSIONS 1. The volume of 1 mole of a substance is called the molar volume V m 2. For liquid and solid substances, the molar volume depends on their density 3. V m = 22.4 l/mol 4. Normal conditions (n.s.): and pressure 760 mmHg, or 101.3 kPa 5. The molar volume of gaseous substances is expressed in l/mol, ml/mmol,