Molecular volume of oxygen. How do I find the volume of gas? Molar volume of gas
Where m is mass, M is molar mass, V is volume.
4. Avogadro's law. Installed by the Italian physicist Avogadro in 1811. The same 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 1 mole of any gas occupies under normal conditions (P 0 = 101.3 kPa and T 0 = 298K) a volume equal to 22.4 liters.
5. Boyle-Mariotte's law
At a 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 the combined law of Boyle-Mariotte and Gay-Lussac, which is used to bring gas volumes from one conditions 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 value of molecular masses M can be done using the so-called ideal gas equations of state or the Clapeyron-Mendeleev equation :
pV = (m / M) * RT = vRT.(1.1)
where R - gas pressure in a closed system, V- the 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 the 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 gas volume to normal conditions.
What volume (n.u.) will take up 0.4 × 10 -3 m 3 of gas at 50 0 С and a pressure of 0.954 × 10 5 Pa?
Solution. To bring the volume of gas to normal conditions, a general formula is used that combines the laws of Boyle-Mariotte and Gay-Lussac:
pV / T = p 0 V 0 / T 0.
The gas volume (n.a.) 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 (n.o.), the gas occupies a volume equal to 0.32 × 10 -3 m 3.
Example 2. Calculation of the relative density of a gas by its molecular weight.
Calculate the density of ethane C 2 H 6 in terms of hydrogen and air.
Solution. It follows from Avogadro's law that the relative density of one gas in 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 gas mixture by relative density.
Calculate the average molecular weight of a gas mixture of 80% methane and 20% oxygen (by volume) using the relative hydrogen densities of these gases.
Solution. Calculations are often made according to the mixing rule, which is 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 through 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 gas mixture is 9.6. average molecular weight of a 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 of gas at 13 0 C and a pressure of 1.040 × 10 5 Pa is 0.828 × 10 -3 kg. Calculate the molar mass of the gas.
Solution. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:
where m- gas mass; M- molar mass of gas; R- molar (universal) gas constant, the value of which is determined by the accepted units of measurement.
If the pressure is measured in Pa, and the volume in m 3, then R= 8.3144 × 10 3 J / (kmol × K).
In order to find out the composition of any gaseous substances, it is necessary to be able to operate with such concepts as molar volume, molar mass and density of a substance. In this article, we will look at what molar volume is, and how to calculate it?
Amount of substance
Quantitative calculations are carried out in order to actually carry out a particular process or find out the composition and structure of a certain substance. These calculations are inconvenient to perform with the absolute values of the masses of atoms or molecules due to the fact that they are very small. In most cases, relative atomic masses also cannot be used, since they are not related to generally accepted measures of mass or volume of a substance. Therefore, the concept of the amount 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 the amount of a substance is the mole.
A mole is that amount of a substance that contains as many structural units as there are atoms in 12 g of the carbon isotope.
The mass of 1 atom is 12 amu. e. m., therefore, the number of atoms in 12 g of the carbon isotope is equal to:
Na = 12g / 12 * 1.66057 * 10 in degree-24g = 6.0221 * 10 in degree 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 system Cu, the measurement of the molar volume of gaseous substances is carried out in cubic meters per mole (m 3 / mol)
The molar volume of gaseous substances differs from substances in a liquid and solid state in that a gaseous element in the amount of 1 mol always occupies the same volume (if the same parameters are observed).
The volume of gas depends on temperature and pressure, therefore, 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 is equal to 22.41 dm 3 / mol. This volume is called the ideal gas molar volume. 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.
The table "molar volume of gases"
The following table shows the volume of some gases:
| Gas | Molar volume, l |
| H 2 | 22,432 |
| O 2 | 22,391 |
| Cl 2 | 22,022 |
| CO 2 | 22,263 |
| NH 3 | 22,065 |
| SO 2 | 21,888 |
| Ideal | 22,41383 |
What have we learned?
The molar volume of a gas studied in chemistry (grade 8), along with molar mass and density, are necessary quantities to determine the composition of a chemical substance. A feature of molar gas is that one mole of gas always contains the same volume. This volume is called the molar volume of the gas.
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Before solving problems, you should learn the formulas and rules for how to find the volume of gas. We should remember Avogadro's law. And the gas volume itself can be calculated using several formulas, choosing the appropriate one from them. When choosing 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, in the same volumes of different gases, the same number of molecules will be contained. The number of gas molecules contained in one mole is Avogadro's number. It follows from this law that: 1 Kmole (kilomole) of an ideal gas, and any, at the same pressure and temperature (760 mm Hg and t = 0 * C) always occupies one volume = 22.4136 m3.
How to determine the volume of gas
- 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 there is no amount of matter under the conditions, but at the same time there is a mass of matter, then we do this: n = m / M. Here M is g / mol (molar mass of a substance), and the mass of a substance in grams is m. In the periodic table, it is written under each element, as its atomic mass. Let's add all the masses and get the desired one.
- So how to calculate the gas volume. Here's the problem: dissolve 10 g of aluminum in hydrochloric acid. Question: how much hydrogen can be released during n. at.? The reaction equation looks like this: 2Al + 6HCl (ex) = 2AlCl3 + 3H2. At the very beginning, we find aluminum (amount), which has reacted according to the formula: n (Al) = m (Al) / M (Al). We take the mass of aluminum (molar) from the periodic table M (Al) = 27g / mol. Substitute: n (Al) = 10/27 = 0.37 mol. It can be seen from the chemical equation that 3 moles of hydrogen were formed by dissolving 2 moles of aluminum. It is necessary to calculate how much hydrogen will be released from 0.4 mol of aluminum: n (H2) = 3 * 0.37 / 2 = 0.56 mol. Let's substitute the data into the formula and find the volume of this gas. V = n * Vm = 0.56 * 22.4 = 12.54 liters.
From the provisions that one mole of any substance includes the number of particles of this substance equal to Avogadro's number, and that equal numbers of particles of different gases under the same physical conditions are contained in equal volumes of these gases, the consequence follows:
equal amounts of any gaseous substances under the same physical conditions occupy equal volumes
For example, the volume of one mole of any gas has (at p, T = const) the same value. Consequently, the equation of the reaction proceeding with the participation of gases sets not only the ratio of their quantities and masses, but also their volumes.
the molar volume of a gas (V M) is the volume of a gas that contains 1 mole of this gas
V M = V / n
The unit of the molar volume of gas in SI is cubic meter per mole (m 3 / mol), but fractional units are more often used - liter (cubic decimeter) per mole (l / mol, dm 3 / mol) and mlliliter (cubic centimeter) per mole ( cm 3 / mol).
In accordance with the definition of the molar volume for any gas, the ratio of its volume V to the quantity n will be the same provided that it is an ideal gas.
Under normal conditions (n.a.) - 101.3 kPa, 0 ° С - the molar volume of an ideal gas is
V M = 2.241381 · 10 -2 m 3 / mol ≈ 22.4 L / mol
In chemical calculations, a rounded value of 22.4 L / mol is used, since the exact value refers to an ideal gas, and most real gases differ in properties from it. Real gases with a very low equilibrium condensation temperature (H 2, O 2, N 2) under normal conditions have a volume almost equal to 22.4 l / mol, and gases condensing at high temperatures have a slightly lower molar volume at n. u .: for CO 2 - 22.26 l / mol, for NH 3 - 22.08 l / mol.
Knowing the volume of a certain gas under given conditions, one can determine the amount of substances in this volume, and vice versa, by the amount of substance in a given portion of gas, one can find the volume of this portion:
n = V / V M; V = V M * n
Molar volume of gas at normal level is a fundamental physical constant that is widely used in chemical calculations. It allows you to use the volume of a gas instead of its mass, which is very convenient in analytical chemistry (gas analyzers based on the measurement of volume), since it is easier to measure the volume of a gas than its mass.
The value of the molar volume of gas at normal conditions. is the coefficient of proportionality between the Avogadro and Loschmidt constants:
V M = N A / N L = 6.022 10 23 (mol -1) / 2.24 10 4 (cm 3 / mol) = 2.69 10 19 (cm -3)
Using the values of the molar volume and the molar mass of the gas, the density of the gas can be determined:
ρ = M / V M
In calculations based on the law of equivalents for gaseous substances (reagents, products), instead of the equivalent mass, it is more convenient to use the equivalent volume, which is the ratio of the volume of a portion of a given gas to the equivalent amount of a substance in this portion:
V eq = V / n eq = V / zn = V M / z; (p, T = const)
The unit of equivalent volume is the same as the unit of molar volume. The value of the equivalent volume of gas is a constant of a given gas only in a specific reaction, since it depends on the equivalence factor f eq.
Molar volume of gas
Molar volume of gas From the provisions that one mole of any substance includes the number of particles of this substance equal to Avogadro's number, and that equal numbers of particles of different gases with the same
Gas volume under normal conditions
Topic 1
LESSON 7
Topic. Molar volume of gases. Calculation of gas volume under normal conditions
Lesson objectives: to acquaint students with the concept of "molar volume"; to reveal the features of using the concept of "molar volume" for gaseous substances; teach students to use the knowledge gained to calculate the volume of gases under normal conditions.
Lesson type: combined.
Forms of work: teacher's story, guided practice.
Equipment: Periodic table of chemical elements of D. I. Mendeleev, cards with tasks, a cube with a volume of 22.4 liters (with a side of 28.2 cm).
II. Checking homework, updating basic knowledge
Pupils take homework on sheets of paper for review.
1) What is “amount of substance”?
2) A unit for measuring the amount of a substance.
3) How many particles are there in 1 mole of a substance?
4) What is the relationship between the amount of a substance and the state of aggregation in which this substance is located?
5) How many water molecules are there in 1 mole of ice?
6) What about 1 mole of liquid water?
7) In 1 mole of water vapor?
8) What mass will they have:
III. Learning new material
Creation and solution of a problem situation Problematic issue. How much will it take:
We cannot answer these questions right away, because the volume of a substance depends on the density of the substance. And according to the formula V = m / ρ, the volume will be different. 1 mole of vapor takes up more volume than 1 mole of water or ice.
Because in liquid and gaseous substances, the distance between water molecules is different.
Many scientists have studied gaseous substances. A significant contribution to the study of this issue was made by the French chemist Joseph Louis Gay-Lussac and the English physicist Robert Boyle, who formulated a number of physical laws describing the state of gases.
Do you know of these patterns?
All gases are equally compressed and have the same thermal expansion coefficient. The volumes of gases do not depend on the size of individual molecules, but on the distance between the molecules. Distances between molecules depend on their speed of movement, energy and, accordingly, temperature.
Based on these laws and his research, the Italian scientist Amedeo Avogadro formulated the law:
Equal volumes of different gases contain the same number of molecules.
Under normal conditions, gaseous substances have a molecular structure. Gas molecules are very small compared to the distance between them. Therefore, the volume of a gas is determined not by the size of particles (molecules), but by the distance between them, which is approximately the same for any gas.
A. Avogadro concluded that if we take 1 mole, i.e. 6.02 · 1023 molecules of any gases, then they will occupy the same volume. But at the same time, this volume is measured under the same conditions, that is, at the same temperature and pressure.
The conditions under which such calculations are carried out were called normal conditions.
Normal conditions (n.c.):
T = 273 K or t = 0 ° C
P = 101.3 kPa or P = 1 atm. = 760 mm Hg. Art.
The volume of 1 mol of a substance is called the molar volume (Vm). For gases under normal conditions it is 22.4 l / mol.
A cube with a volume of 22.4 liters is on display.
Such a cube contains 6.02-1023 molecules of any gases, for example, oxygen, hydrogen, ammonia (NH 3), methane (CH4).
Under what conditions?
At a temperature of 0 ° C and a pressure of 760 mm Hg. Art.
It follows from Avogadro's law that
where Vm = 22.4 l / mol of any gas at n. v.
So, knowing the volume of the gas, you can calculate the amount of matter, and vice versa.
IV. Formation of skills and abilities
Practice by example
Calculate what volume 3 moles of oxygen will occupy at n. v.
Calculate the number of carbon (IV) oxide molecules in a volume of 44.8 liters (n.v).
2) Let's calculate the number of С O 2 molecules according to the formulas:
N (CO 2) = 2 mol 6.02 1023 molecules / mol = 12.04 1023 molecules.
Answer: 12.04 1023 molecules.
Calculate the volume of nitrogen with a mass of 112 g (n.v.).
V (N 2) = 4 mol · 22.4 L / mol = 89.6 L.
V. Homework
Work through the relevant paragraph of the textbook, answer the questions.
Creative assignment (home practice). Solve problems 2, 4, 6 on your own from the map.
Card-task for lesson 7
Calculate the volume of 7 mol of nitrogen N 2 (n.v.).
Calculate the number of 112 L hydrogen molecules.
(Answer: 30.1 1023 molecules)
Calculate the volume of 340 g of hydrogen sulfide.
Gas volume under normal conditions
Molar volume of gases. Calculation of the volume of gas under normal conditions - AMOUNT OF SUBSTANCE. CHEMICAL FORMULAS CALCULATIONS - ALL CHEMISTRY LESSONS - Grade 8 - Lesson Notes - Chemistry Lessons - Lesson Plan - Lesson Synopsis - Lesson Plans - Chemistry Lesson Development - CHEMISTRY - Standard and Academic Level School Program - All Chemistry Lessons for Eighth Grade 12 Year Old schools
Gas laws. Avogadro's law. Molar volume of gas
French scientist J.L. Gay Lussac established the law volumetric relationship:
For instance, 1 l of chlorine connects with 1 liter of hydrogen , forming 2 l of hydrogen chloride ; 2 l of sulfur oxide (IV) connect with 1 liter of oxygen, forming 1 liter of sulfur oxide (VI).
This law allowed the Italian scientist A. Avogadro assume that molecules of simple gases ( hydrogen, oxygen, nitrogen, chlorine, etc.
) consist of two identical atoms
... When hydrogen combines with chlorine, their molecules break down into atoms, and the latter form hydrogen chloride molecules. But since two molecules of hydrogen chloride are formed from one hydrogen molecule and one chlorine molecule, the volume of the latter must be equal to the sum of the volumes of the initial gases.
Thus, the volume ratios are easily explained if we proceed from the concept of the diatomic nature of molecules of simple gases ( H2, Cl2, O2, N2, etc.
) - This, in turn, serves as proof of the diatomic nature of the molecules of these substances.
The study of the properties of gases allowed A. Avogadro to formulate a hypothesis, which was subsequently confirmed by experimental data, and therefore began to be called Avogadro's law:
An important point follows from Avogadro's law consequence: under the same conditions, 1 mole of any gas occupies the same volume.
This volume can be calculated if the mass is known 1 l gas. Under normal conditions, (n.o.) i.e. temperature 273K (O ° C) and pressure 101 325 Pa (760 mm Hg) , the mass of 1 liter of hydrogen is 0.09 g, its molar mass is 1.008 2 = 2.016 g / mol. Then the volume occupied by 1 mole of hydrogen under normal conditions is equal to 22.4 l
Under the same conditions, the mass 1L oxygen 1,492g ; molar 32g / mol ... Then the volume of oxygen at (n.a.) is also equal to 22.4 mol.
The molar volume of a gas is the ratio of the volume of a substance to the amount of this substance:
where V m - molar volume of gas (dimension l / mol ); V is the volume of the system substance; n - the amount of matter in the system. Recording example: V m gas (Well.) = 22.4 l / mol.
Based on Avogadro's law, the molar masses of gaseous substances are determined. The greater the mass of gas molecules, the greater the mass of one and the same volume of gas. Equal volumes of gases under the same conditions contain the same number of molecules, and, consequently, the moles of gases. The ratio of masses of equal volumes of gases is equal to the ratio of their molar masses:
where m 1 - the mass of a certain volume of the first gas; m 2 - the mass of the same volume of the second gas; M 1 and M 2 - molar masses of the first and second gases.
Usually, the density of a gas is determined in relation to the lightest gas - hydrogen (denoted D H2 ). The molar mass of hydrogen is 2g / mol ... Therefore, we get.
The molecular weight of a substance in a gaseous state is equal to its double density for hydrogen.
Often the density of a gas is measured in relation to air. (D B ) ... Although air is a mixture of gases, one still speaks of its average molar mass. It is equal to 29g / mol. In this case, the molar mass is determined by the expression M = 29D B .
Determination of molecular weights showed that molecules of simple gases consist of two atoms (H2, F2, Cl2, O2 N2) , and molecules of inert gases - from one atom (He, Ne, Ar, Kr, Xe, Rn). For noble gases, "molecule" and "atom" are equivalent.
Boyle's Law - Mariotte:
at constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is.From here pV = const
,
where R
- pressure, V
- gas volume.
Gay Lussac's Law:
at constant pressure and the change in gas volume is directly proportional to temperature, i.e.
V / T = const,
where T
- temperature on a scale TO
(kelvin)
The combined gas law of Boyle - Mariotte and Gay-Lussac:
pV / T = const.
This formula is usually used to calculate the volume of a gas under given conditions, if its volume is known under different conditions. If a transition is made from normal conditions (or to normal conditions), then this formula is written as follows:
pV / T = p
V
/ T
,
where R
, V
, T
- pressure, gas volume and temperature under normal conditions ( R
= 101 325 Pa
, T
= 273 K
V
= 22.4 l / mol)
.
If the mass and amount of gas are known, but it is necessary to calculate its volume, or vice versa, use Mendeleev-Cliperon equation:
where n - the amount of gas substance, mol; m - weight, g; M - molar mass of gas, g / yol ; R is a universal gas constant. R = 8.31 J / (mol * K)
Gas laws
Gas laws. Avogadro's law. Molar volume of gas French scientist J.L. Gay-Lussac established the law of volumetric relationships: For example, 1 liter of chlorine combines with 1 liter of hydrogen, forming 2
Where m is mass, M is molar mass, V is volume.
4. Avogadro's law. Installed by the Italian physicist Avogadro in 1811. The same 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 1 mole of any gas occupies under normal conditions (P 0 = 101.3 kPa and T 0 = 298K) a volume equal to 22.4 liters.
5. Boyle-Mariotte's law
At a 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 the combined law of Boyle-Mariotte and Gay-Lussac, which is used to bring gas volumes from one conditions 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 value of molecular masses M can be done using the so-called ideal gas equations of state or the Clapeyron-Mendeleev equation :
pV = (m / M) * RT = vRT.(1.1)
where R - gas pressure in a closed system, V- the 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 the 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 gas volume to normal conditions.
What volume (n.u.) will take up 0.4 × 10 -3 m 3 of gas at 50 0 С and a pressure of 0.954 × 10 5 Pa?
Solution. To bring the volume of gas to normal conditions, a general formula is used that combines the laws of Boyle-Mariotte and Gay-Lussac:
pV / T = p 0 V 0 / T 0.
The gas volume (n.a.) 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 (n.o.), the gas occupies a volume equal to 0.32 × 10 -3 m 3.
Example 2. Calculation of the relative density of a gas by its molecular weight.
Calculate the density of ethane C 2 H 6 in terms of hydrogen and air.
Solution. It follows from Avogadro's law that the relative density of one gas in 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 gas mixture by relative density.
Calculate the average molecular weight of a gas mixture of 80% methane and 20% oxygen (by volume) using the relative hydrogen densities of these gases.
Solution. Calculations are often made according to the mixing rule, which is 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 through 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 gas mixture is 9.6. average molecular weight of a 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 of gas at 13 0 C and a pressure of 1.040 × 10 5 Pa is 0.828 × 10 -3 kg. Calculate the molar mass of the gas.
Solution. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:
where m- gas mass; M- molar mass of gas; R- molar (universal) gas constant, the value of which is determined by the accepted units of measurement.
If the pressure is measured in Pa, and the volume in m 3, then R= 8.3144 × 10 3 J / (kmol × K).
3.1. When performing measurements of atmospheric air, air in the working area, 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 carrying out measurements of air quality, the conversion of measured concentrations to normal conditions is not used, as a result of which unreliable results are obtained.
Here is an excerpt from the Standard:
“The measurements are brought to standard conditions using the following formula:
C 0 = C 1 * P 0 T 1 / P 1 T 0
where: С 0 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume of air, mol / cu. m, at standard temperature and pressure;
С 1 - the result, expressed in units of mass per unit volume of air, kg / cu. m, or the amount of substance per unit volume
air, mol / cu. 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 comparing measured air quality parameters in different locations and different climates.
3.2 Industry normal conditions
Normal conditions are the standard physical conditions to which the properties of substances (Standard temperature and pressure, STP) are usually related. Reference 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. st .; 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 applicable government standards. The volumes of waste gases obtained from the results of instrumental measurements should be brought to normal conditions (n.o.): 0 ° С, 101.3 kPa ..
Aviation.
The International Civil Aviation Organization (ICAO) defines an 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 facilities.
The gas industry of the Russian Federation, when making settlements with 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.
Testing
To test machines, instruments and other technical products, the following are taken for normal values of climatic factors during product testing (normal climatic test conditions):
Temperature - plus 25 ° ± 10 ° С; Relative humidity - 45-80%
Atmospheric pressure 84-106 kPa (630-800 mm Hg)
Verification of measuring instruments
The nominal values of the most common normal influencing quantities are chosen as follows: Temperature - 293 K (20 ° C), atmospheric pressure - 101.3 kPa (760 mm Hg).
Rationing
In the guidelines for the establishment of air quality standards, it is indicated that MPCs in the ambient air are set under normal indoor conditions, i.e. 20 C and 760 mm. rt. Art.
