In order to find out the composition of any gaseous substances, you must be able to operate with concepts such as molar volume, molar mass and density of the substance. In this article, we will look at what molar volume is and how to calculate it?
Quantity of substance
Quantitative calculations are carried out in order to actually carry out a particular process or to find out the composition and structure of a certain substance. These calculations are inconvenient to perform with absolute values of the mass of atoms or molecules due to the fact that they are very small. Relative atomic masses also cannot be used in most cases, since they are not related to generally accepted measures of mass or volume of a substance. Therefore, the concept of quantity of a substance was introduced, which is denoted by the Greek letter v (nu) or n. The amount of a substance is proportional to the number of structural units (molecules, atomic particles) contained in the substance.
The unit of quantity of a substance is the mole.
A mole is an amount of substance that contains the same number of structural units as there are atoms in 12 g of carbon isotope.
The mass of 1 atom is 12 a. e.m., therefore the number of atoms in 12 g of carbon isotope is equal to:
Na= 12g/12*1.66057*10 to the power-24g=6.0221*10 to the power of 23
The physical quantity Na is called Avogadro's constant. One mole of any substance contains 6.02 * 10 to the power of 23 particles.
Rice. 1. Avogadro's law.
Molar volume of gas
The molar volume of a gas is the ratio of the volume of a substance to the amount of that substance. This value is calculated by dividing the molar mass of a substance by its density using the following formula:
where Vm is the molar volume, M is the molar mass, and p is the density of the substance.
Rice. 2. Molar volume formula.
In the international C system, the molar volume of gaseous substances is measured in cubic meters per mole (m 3 /mol)
The molar volume of gaseous substances differs from substances in liquid and solid states in that a gaseous element with an amount of 1 mole always occupies the same volume (if the same parameters are met).
The volume of gas depends on temperature and pressure, so when calculating, you should take the volume of gas under normal conditions. Normal conditions are considered to be a temperature of 0 degrees and a pressure of 101.325 kPa. The molar volume of 1 mole of gas under normal conditions is always the same and equal to 22.41 dm 3 /mol. This volume is called the molar volume of an ideal gas. That is, in 1 mole of any gas (oxygen, hydrogen, air) the volume is 22.41 dm 3 /m.
Rice. 3. Molar volume of gas under normal conditions.
Table "molar volume of gases"
The following table shows the volume of some gases:
| Gas | Molar volume, l |
| H 2 | 22,432 |
| O2 | 22,391 |
| Cl2 | 22,022 |
| CO2 | 22,263 |
| NH 3 | 22,065 |
| SO 2 | 21,888 |
| Ideal | 22,41383 |
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 the desired one.
- 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.
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 combining the gas laws of Boyle - Mariotte and Gay - Lussac:
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) 6.0210 21 CO 2 molecules; b) 1.2010 24 oxygen atoms; c) 2.0010 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) 6.0210 21 i.e. 0.01 mol; b) 1.2010 24, i.e. 2 mol; c) 2.0010 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
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 (no.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 (293.15K); 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, 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.
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,
