What is the molar volume? Mol. Avogadro's law. Molar volume of gas

The molar volume of a gas is equal to the ratio of the volume of the gas to the amount of substance of this gas, i.e.

V m = V(X) / n(X),

where V m is the molar volume of gas - a constant value for any gas under given conditions;

V(X) – volume of gas X;

n(X) – amount of gas substance X.

Molar volume of gases at normal conditions (normal pressure pH = 101,325 Pa ≈ 101.3 kPa and temperature Tn = 273.15 K ≈ 273 K) is V m = 22.4 l/mol.

Ideal gas laws

In calculations involving gases, it is often necessary to switch from these conditions to normal ones or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

pV / T = p n V n / T n

Where p is pressure; V - volume; T - temperature on the Kelvin scale; the index “n” indicates normal conditions.

Volume fraction

The composition of gas mixtures is often expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

φ(X) = V(X) / V

where φ(X) is the volume fraction of component X;

V(X) - 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.

Example 1. What volume will ammonia weighing 51 g occupy at a temperature of 20°C and a pressure of 250 kPa?

Example 2. Determine the volume that a gas mixture containing hydrogen, weighing 1.4 g, and nitrogen, weighing 5.6 g, will occupy under normal conditions.

Law of volumetric relations

How to solve a problem using the “Law of Volumetric Relations”?

Law of Volume Ratios: The volumes of gases involved in a reaction are related to each other as small integers equal to the coefficients in the reaction equation.

The coefficients in the reaction equations show the numbers of volumes of reacting and formed gaseous substances.

Example. Calculate the volume of air required to burn 112 liters of acetylene.

1. We compose the reaction equation:

2. Based on the law of volumetric relations, we calculate the volume of oxygen:

112 / 2 = X / 5, from where X = 112 5 / 2 = 280l

3. Determine the volume of air:

V(air) = V(O 2) / φ(O 2)

V(air) = 280 / 0.2 = 1400 l.

P1V1=P2V2, or, which is the same, PV=const (Boyle-Mariotte law). At constant pressure, the ratio of volume to temperature remains constant: V/T=const (Gay-Lussac law). If we fix the volume, then P/T=const (Charles’ law). Combining these three laws gives a universal law that states that PV/T=const. This equation it was found French physicist B. Clapeyron in 1834.

The value of the constant is determined only by the amount of substance gas. DI. Mendeleev derived an equation for one mole in 1874. So it is the value of the universal constant: R=8.314 J/(mol∙K). So PV=RT. In the case of an arbitrary quantity gasνPV=νRT. The amount of a substance itself can be found from mass to molar mass: ν=m/M.

Molar mass is numerically equal to relative molecular mass. The latter can be found from the periodic table; it is indicated in the cell of the element, as a rule, . The molecular weight is equal to the sum of the molecular weights of its constituent elements. In the case of atoms of different valences, an index is required. On at mer, M(N2O)=14∙2+16=28+16=44 g/mol.

Normal conditions for gases at It is commonly assumed that P0 = 1 atm = 101.325 kPa, temperature T0 = 273.15 K = 0°C. Now you can find the volume of one mole gas at normal conditions: Vm=RT/P0=8.314∙273.15/101.325=22.413 l/mol. This table value is the molar volume.

Under normal conditions conditions quantity relative to volume gas to molar volume: ν=V/Vm. For arbitrary conditions you need to use the Mendeleev-Clapeyron equation directly: ν=PV/RT.

Thus, to find the volume gas at normal conditions, you need the amount of substance (number of moles) of this gas multiply by the molar volume equal to 22.4 l/mol. Using the reverse operation, you can find the amount of a substance from a given volume.

To find the volume of one mole of a substance in a solid or liquid state, find its molar mass and divide by its density. One mole of any gas under normal conditions has a volume of 22.4 liters. If conditions change, calculate the volume of one mole using the Clapeyron-Mendeleev equation.

You will need

• Periodic table of Mendeleev, table of density of substances, pressure gauge and thermometer.

Instructions

Determining the volume of one mole or solid
Determine the chemical formula of the solid or liquid you are studying. Then, using the periodic table, find the atomic masses of the elements that are included in the formula. If one is included in the formula more than once, multiply its atomic mass by that number. Add up the atomic masses and get the molecular mass of which it is composed solid or liquid. It will be numerically equal to the molar mass measured in grams per mole.

Using the table of substance densities, find this value for the material of the body or liquid being studied. Then divide the molar mass by the density of this substance, measured in g/cm³ V=M/ρ. The result is the volume of one mole in cm³. If the substance remains unknown, it will be impossible to determine the volume of one mole of it.

: V = n*Vm, where V is the volume of gas (l), n is the amount of substance (mol), Vm is the molar volume of gas (l/mol), at normal (norm) is a standard value and is equal to 22, 4 l/mol. It happens that the condition does not contain the amount of a substance, but there is a mass of a certain substance, then we do this: n = m/M, where m is the mass of the substance (g), M – molar mass substances (g/mol). We find the molar mass using the table D.I. Mendeleev: under each element is its atomic mass, add up all the masses and get what we need. But such tasks are quite rare, usually present in the tasks. The solution to such problems changes slightly. Let's look at an example.

What volume of hydrogen will be released under normal conditions if aluminum weighing 10.8 g is dissolved in excess hydrochloric acid.

If we are dealing with a gas system, then the following formula holds: q(x) = V(x)/V, where q(x)(phi) is the fraction of the component, V(x) is the volume of the component (l), V – system volume (l). To find the volume of a component, we obtain the formula: V(x) = q(x)*V. And if you need to find the volume of the system, then: V = V(x)/q(x).

note

There are other formulas for finding volume, but if you need to find the volume of a gas, only the formulas given in this article are suitable.

Sources:

• "Chemistry Manual", G.P. Khomchenko, 2005.
• how to find the amount of work
• Find the volume of hydrogen during the electrolysis of a ZnSO4 solution

An ideal gas is one in which the interaction between molecules is negligible. In addition to pressure, the state of a gas is characterized by temperature and volume. The relationships between these parameters are shown in gas laws.

Instructions

The pressure of a gas is directly proportional to its temperature, the amount of substance, and inversely proportional to the volume of the container occupied by the gas. The proportionality coefficient is the universal gas constant R, approximately equal to 8.314. It is measured in joules divided by moles and by .

This position forms the mathematical dependence P=νRT/V, where ν is the amount of substance (mol), R=8.314 is the universal gas constant (J/mol K), T is the gas temperature, V is the volume. Pressure is expressed in . It can be expressed by and , with 1 atm = 101.325 kPa.

The considered dependence is a consequence of the Mendeleev-Clapeyron equation PV=(m/M) RT. Here m is the mass of the gas (g), M is its molar mass (g/mol), and the fraction m/M gives the total amount of substance ν, or the number of moles. The Mendeleev-Clapeyron equation is valid for all gases that can be considered. This is the physical gas law.

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 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 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 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).

3.1. When performing measurements of atmospheric air, air working area as well as industrial emissions and hydrocarbons in gas pipelines, 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 lead to standard conditions for comparison of measured air quality parameters in various places and different climatic conditions.

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 (norm): 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

For testing machines, devices 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.