Finding the molar volume of gases. Ideal gas laws

: 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 quantity 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 is the molar mass of the substance (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 reflected in the 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 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).

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 (norm): 0°C, 101.3 kPa..

Aviation.

The International Civil Aviation Organization (ICAO) defines the International Standard Atmosphere (ISA) as 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.

The gas industry of the Russian Federation, when making payments to consumers, uses atmospheric conditions in accordance with GOST 2939-63: temperature 20 ° C (293.15 K); 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, the following are taken as normal values ​​of climatic factors when testing products (normal climatic test conditions):

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.

Мв = К· Мr (1)

Where: K is the proportionality coefficient equal to 1 g/mol.

In fact, for the carbon isotope 12 6 C Ar = 12, and the molar mass of atoms (by the definition of the concept “mole”) is 12 g/mol. Consequently, the numerical values ​​of the two masses coincide, which means K = 1. It follows that the molar mass of a substance, expressed in grams per mole, has the same numerical value as its relative molecular mass(atomic) weight. Thus, the molar mass of atomic hydrogen is 1.008 g/mol, molecular hydrogen – 2.016 g/mol, molecular oxygen – 31.999 g/mol.

According to Avogadro's law, the same number of molecules of any gas occupies the same volume under the same conditions. On the other hand, 1 mole of any substance contains (by definition) the same number of particles. It follows that at a certain temperature and pressure, 1 mole of any substance in the gaseous state occupies the same volume.

The ratio of the volume occupied by a substance to its quantity is called the molar volume of the substance. Under normal conditions (101.325 kPa; 273 K), the molar volume of any gas is equal to 22,4l/mol(more precisely, Vn = 22.4 l/mol). This statement is true for such a gas, when other types of interaction of its molecules with each other, except for their elastic collision, can be neglected. Such gases are called ideal. For non-ideal gases, called real gases, the molar volumes are different and slightly different from the exact value. However, in most cases the difference is reflected only in the fourth and subsequent significant figures.

Measurements of gas volumes are usually carried out under conditions other than normal. To bring the volume of gas to normal conditions, you can use an equation that combines the Boyle–Mariotte and Gay–Lussac gas laws:

pV / T = p 0 V 0 / T 0

Where: V is the volume of gas at pressure p and temperature T;

V 0 is the volume of gas at normal pressure p 0 (101.325 kPa) and temperature T 0 (273.15 K).

The molar masses of gases can also be calculated using the equation of state of an ideal gas - the Clapeyron - Mendeleev equation:

pV = m B RT / M B ,

Where: p – gas pressure, Pa;

V – its volume, m3;

M B - mass of substance, g;

M B – its molar mass, g/mol;

T – absolute temperature, K;

R is the universal gas constant equal to 8.314 J / (mol K).

If the volume and pressure of a gas are expressed in other units of measurement, then the value of the gas constant in the Clapeyron–Mendeleev equation will take on a different value. It can be calculated using the formula resulting from the unified law of the gas state for a mole of a substance under normal conditions for one mole of gas:

R = (p 0 V 0 / T 0)

Example 1. Express in moles: a) 6.0210 21 CO 2 molecules; b) 1.2010 24 oxygen atoms; c) 2.0010 23 water molecules. What is the molar mass of these substances?

Solution. A mole is the amount of a substance that contains a number of particles of any particular kind equal to Avogadro's constant. Hence, a) 6.0210 21 i.e. 0.01 mol; b) 1.2010 24, i.e. 2 mol; c) 2.0010 23, i.e. 1/3 mol. The mass of a mole of a substance is expressed in kg/mol or g/mol. The molar mass of a substance in grams is numerically equal to its relative molecular (atomic) mass, expressed in atomic mass units (amu)

Since the molecular masses of CO 2 and H 2 O and the atomic mass of oxygen, respectively, are 44; 18 and 16 amu, then their molar masses are equal: a) 44 g/mol; b) 18g/mol; c) 16 g/mol.

Example 2. Calculate the absolute mass of a sulfuric acid molecule in grams.

Solution. A mole of any substance (see example 1) contains Avogadro’s constant N A of structural units (in our example, molecules). The molar mass of H 2 SO 4 is 98.0 g/mol. Therefore, the mass of one molecule is 98/(6.02 10 23) = 1.63 10 -22 g.

Molar volume- the volume of one mole of a substance, the value obtained by dividing the molar mass by the density. Characterizes the packing density of molecules.

Meaning N A = 6.022…×10 23 called Avogadro's number after the Italian chemist Amedeo Avogadro. This is the universal constant for the smallest particles of any substance.

It is this number of molecules that contains 1 mole of oxygen O2, the same number of atoms in 1 mole of iron (Fe), molecules in 1 mole of water H2O, etc.

According to Avogadro's law, 1 mole of an ideal gas at normal conditions has the same volume Vm= 22.413 996(39) l. Under normal conditions, most gases are close to ideal, therefore all reference information on the molar volume of chemical elements refers to their condensed phases, unless otherwise stated

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 was established by the 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 what the solid or liquid is made of. 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. After this, divide the molar mass by the density of the 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.