Mole of gas under normal conditions. Molar volume of gas

Lesson 1.

Topic: Amount of substance. Mole

Chemistry is the science of substances. How to measure substances? In what units? In the molecules that make up substances, but this is very difficult to do. In grams, kilograms or milligrams, but this is how mass is measured. What if we combine the mass that is measured on the scale and the number of molecules of the substance, is this possible?

a) H-hydrogen

A n = 1a.u.m.

1a.u.m = 1.66*10 -24 g

Let's take 1g of hydrogen and count the number of hydrogen atoms in this mass (have students do this using a calculator).

N n = 1g / (1.66*10 -24) g = 6.02*10 23

b) O-oxygen

A o = 16 a.u.m = 16 * 1.67 * 10 -24 g

N o = 16 g / (16 * 1.66 * 10 -24) g = 6.02 * 10 23

c) C-carbon

A c = 12a.u.m = 12*1.67*10 -24 g

N c = 12g / (12* 1.66*10 -24) g = 6.02*10 23

Let us conclude: if we take a mass of matter that is equal to atomic mass in size, but taken in grams, then there will always be (for any substance) 6.02 * 10 23 atoms of this substance.

H 2 O - water

18 g / (18 * 1.66 * 10 -24) g = 6.02 * 10 23 water molecules, etc.

N a = 6.02*10 23 - Avogadro’s number or constant.

A mole is the amount of a substance that contains 6.02 * 10 23 molecules, atoms or ions, i.e. structural units.

There are moles of molecules, moles of atoms, moles of ions.

n – number of moles,(the number of moles is often denoted by nu),
N is the number of atoms or molecules,
N a = Avogadro's constant.

Kmol = 10 3 mol, mmol = 10 -3 mol.

Display a portrait of Amedeo Avogadro on a multimedia installation and briefly talk about him, or instruct the student to prepare a short report on the life of the scientist.

Lesson 2.

Topic: “Molar mass of a substance”

What is the mass of 1 mole of a substance? (Students can often draw the conclusion themselves.)

The mass of one mole of a substance is equal to its molecular mass, but expressed in grams. The mass of one mole of a substance is called molar mass and is denoted by M.

Formulas:

M - molar mass,
n - number of moles,
m is the mass of the substance.

The mass of a mole is measured in g/mol, the mass of a kmole is measured in kg/kmol, the mass of a mmol is measured in mg/mol.

Fill out the table (tables are distributed).

Lesson 3.

Topic: Molar volume of gases

Let's solve the problem. Determine the volume of water whose mass at normal conditions 180 g

Given:

Those. volume of liquid and solids We count through density.

But, when calculating the volume of gases, it is not necessary to know the density. Why?

The Italian scientist Avogadro determined that in equal volumes different gases under the same conditions (pressure, temperature) contain same number molecules - this statement is called Avogadro's law.

Those. if at equal conditions V(H 2) =V(O 2), then n(H 2) =n(O 2), and vice versa, if under equal conditions n(H 2) =n(O 2) then the volumes of these gases will be the same. And a mole of a substance always contains the same number of molecules 6.02 * 10 23.

We conclude - under the same conditions, moles of gases should occupy the same volume.

Under normal conditions (t=0, P=101.3 kPa. or 760 mmHg), moles of any gases occupy the same volume. This volume is called molar.

V m =22.4 l/mol

1 kmol occupies a volume of -22.4 m 3 /kmol, 1 mmol occupies a volume of -22.4 ml/mmol.

Example 1.(To be solved on the board):

Example 2.(You can ask students to solve):

Have students fill out the table.

The mass of 1 mole of a substance is called molar. What is the volume of 1 mole of a substance called? Obviously, this is also called molar volume.

What is equal to molar volume water? When we measured 1 mole of water, we did not weigh 18 g of water on the scales - this is inconvenient. We used measuring utensils: a cylinder or a beaker, since we knew that the density of water is 1 g/ml. Therefore, the molar volume of water is 18 ml/mol. In liquids and solids the molar volume depends on their density (Fig. 52, a). It's a different matter for gases (Fig. 52, b).

Rice. 52.
Molar volumes(Well.):
a - liquids and solids; b - gaseous substances

If you take 1 mol of hydrogen H 2 (2 g), 1 mol of oxygen O 2 (32 g), 1 mol of ozone O 3 (48 g), 1 mol carbon dioxide CO 2 (44 g) and even 1 mole of water vapor H 2 O (18 g) under the same conditions, for example normal (in chemistry it is usually called normal conditions (n.s.) temperature 0 ° C and pressure 760 mm Hg. Art. , or 101.3 kPa), then it turns out that 1 mole of any of the gases will occupy the same volume, equal to 22.4 liters, and contain the same number of molecules - 6 × 10 23.

And if you take 44.8 liters of gas, then how much of its substance will be taken? Of course, 2 moles, since the given volume is twice the molar volume. Hence:

where V is the volume of gas. From here

Molar volume is physical quantity, equal to the ratio volume of a substance to the amount of a substance.

The molar volume of gaseous substances is expressed in l/mol. Vm - 22.4 l/mol. The volume of one kilomole is called kilomolar and is measured in m 3 /kmol (Vm = 22.4 m 3 /kmol). Accordingly, the millimolar volume is 22.4 ml/mmol.

Problem 1. Find the mass of 33.6 m 3 of ammonia NH 3 (n.s.).

Problem 2. Find the mass and volume (n.v.) of 18 × 10 20 molecules of hydrogen sulfide H 2 S.

When solving the problem, let's pay attention to the number of molecules 18 × 10 20. Since 10 20 is 1000 times less than 10 23, obviously, calculations should be carried out using mmol, ml/mmol and mg/mmol.

Key words and phrases

1. Molar, millimolar and kilomolar volumes of gases.
2. The molar volume of gases (under normal conditions) is 22.4 l/mol.
3. Normal conditions.

Work with computer

1. Talk to electronic application. Study the lesson material and complete the assigned tasks.
2. Search on the Internet email addresses, which can serve additional sources, revealing the content of keywords and phrases in the paragraph. Offer your help to the teacher in preparing a new lesson - send a message by keywords and phrases in the next paragraph.

Questions and tasks

1. Find the mass and number of molecules at n. u. for: a) 11.2 liters of oxygen; b) 5.6 m 3 nitrogen; c) 22.4 ml of chlorine.
2. Find the volume that at n. u. will take: a) 3 g of hydrogen; b) 96 kg of ozone; c) 12 × 10 20 nitrogen molecules.
3. Find the densities (mass 1 liter) of argon, chlorine, oxygen and ozone at room temperature. u. How many molecules of each substance will be contained in 1 liter under the same conditions?
4. Calculate the mass of 5 liters (n.s.): a) oxygen; b) ozone; c) carbon dioxide CO 2.
5. Indicate which is heavier: a) 5 l sulfur dioxide(SO 2) or 5 liters of carbon dioxide (CO 2); b) 2 l of carbon dioxide (CO 2) or 3 l carbon monoxide(SO).

Where m-mass, M-molar mass, V-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 volume given quantity 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 equations of state ideal gas or Clapeyron-Mendeleev equations :

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

Where R - gas pressure in 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 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, average molecular mass 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 is that the ratio of the volumes of gases in a two-component gas mixture inversely proportional to the differences between the density of the mixture and the densities of the gases that make up this mixture. Let's denote relative density gas mixture by hydrogen through D H2. She will be more density methane, but less density 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 m 3, then R=8.3144×10 3 J/(kmol×K).

3.1. When performing measurements of atmospheric air, work area air, as well as industrial emissions and hydrocarbons in gas lines, there is a problem of bringing the volumes of measured air to normal (standard) conditions. Often in practice, when air quality measurements are taken, the measured concentrations are not recalculated to normal conditions, resulting in unreliable results.

Here is an excerpt from the Standard:

“Measurements lead to standard conditions using the following formula:

C 0 = C 1 * P 0 T 1 / P 1 T 0

where: C 0 - result expressed in units of mass per unit volume of air, kg / cubic meter. m, or the amount of substance per unit volume of air, mol/cubic. m, at standard temperature and pressure;

C 1 - result expressed in units of mass per unit volume of air, kg / cubic meter. m, or the amount of substance per unit volume

air, mol/cub. m, at temperature T 1, K, and pressure P 1, kPa.”

The formula for reduction to normal conditions in a simplified form has the form (2)

C 1 = C 0 * f, where f = P 1 T 0 / P 0 T 1

standard conversion factor for normalization. The parameters of air and impurities are measured at different values ​​of temperature, pressure and humidity. The results provide standard conditions for comparing measured air quality parameters in different locations and different climates.

3.2. Industry normal conditions

Normal conditions are standard physical conditions with which the properties of substances are usually related (Standard temperature and pressure, STP). Normal conditions are defined by IUPAC (International Union of Practical and Applied Chemistry) as follows: Atmospheric pressure 101325 Pa = 760 mm Hg. Air temperature 273.15 K = 0° C.

Standard conditions (Standard Ambient Temperature and Pressure, SATP) are normal ambient temperature and pressure: pressure 1 Bar = 10 5 Pa = 750.06 mm T. Art.; temperature 298.15 K = 25 °C.

Other areas.

Air quality measurements.

The results of measuring the concentrations of harmful substances in the air of the working area lead to the following conditions: temperature 293 K (20 ° C) and pressure 101.3 kPa (760 mm Hg).

Aerodynamic parameters of pollutant emissions must be measured in accordance with current government standards. The volumes of exhaust gases obtained from the results of instrumental measurements must be reduced to normal conditions (n.s.): 0°C, 101.3 kPa..

Aviation.

International organization civil aviation(ICAO) defines the International Standard Atmosphere (ISA) at sea level with a temperature of 15 °C, an atmospheric pressure of 101325 Pa and a relative humidity of 0%. These parameters are used when calculating the movement of aircraft.

Gas industry.

Gas industry Russian Federation when making payments to consumers, it uses atmospheric conditions in accordance with GOST 2939-63: temperature 20°C (293.15K); pressure 760 mm Hg. Art. (101325 N/m²); humidity is 0. Thus, the mass of a cubic meter of gas according to GOST 2939-63 is slightly less than under “chemical” normal conditions.

Tests

To test machines, instruments and other technical products for normal values climatic factors When testing products (normal climatic test conditions), the following are accepted:

Temperature - plus 25°±10°С; Relative humidity – 45-80%

Atmospheric pressure 84-106 kPa (630-800 mmHg)

Verification of measuring instruments

The nominal values ​​of the most common normal influencing quantities are selected as follows: Temperature - 293 K (20 ° C), atmospheric pressure - 101.3 kPa (760 mm Hg).

Rationing

The guidelines regarding the establishment of air quality standards indicate that maximum permissible concentrations in atmospheric air are established under normal indoor conditions, i.e. 20 C and 760 mm. rt. Art.

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 m 3, then R=8.3144×10 3 J/(kmol×K).