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 - volume and temperature pressure 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 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 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 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).
For theoretical material, see the page "Molar volume of gas".
Basic formulas and concepts:
From Avogadro's law, for example, it follows that under the same conditions, 1 liter of hydrogen and 1 liter of oxygen contain the same number of molecules, although their sizes vary greatly.
First corollary of Avogadro's law:
The volume occupied by 1 mole of any gas under normal conditions (n.s.) is 22.4 liters and is called molar volume of gas(Vm).
V m =V/ν (m 3 /mol)
What are called normal conditions (n.s.):
- normal temperature = 0°C or 273 K;
- normal pressure = 1 atm or 760 mm Hg. or 101.3 kPa
From the first corollary of Avogadro’s law it follows that, for example, 1 mole of hydrogen (2 g) and 1 mole of oxygen (32 g) occupy the same volume, equal to 22.4 liters at ground level.
Knowing V m, you can find the volume of any quantity (ν) and any mass (m) of gas:
V=V m ·ν V=V m ·(m/M)
Typical problem 1: What is the volume at no. occupies 10 moles of gas?
V=V m ·ν=22.4·10=224 (l/mol)
Typical problem 2: What is the volume at no. takes up 16 g of oxygen?
V(O 2)=V m ·(m/M) M r (O 2)=32; M(O 2)=32 g/mol V(O 2)=22.4·(16/32)=11.2 l
Second corollary of Avogadro's law:
Knowing the gas density (ρ=m/V) at normal conditions, we can calculate the molar mass of this gas: M=22.4·ρ
Density (D) of one gas is otherwise called the ratio of the mass of a certain volume of the first gas to the mass of a similar volume of the second gas, taken under the same conditions.
Typical task 3: Determine the relative density of carbon dioxide compared to hydrogen and air.
D hydrogen (CO 2) = M r (CO 2)/M r (H 2) = 44/2 = 22 D air = 44/29 = 1.5
- one volume of hydrogen and one volume of chlorine give two volumes of hydrogen chloride: H 2 +Cl 2 =2HCl
- two volumes of hydrogen and one volume of oxygen give two volumes of water vapor: 2H 2 + O 2 = 2H 2 O
Task 1. How many moles and molecules are contained in 44 g of carbon dioxide?
Solution:
M(CO 2) = 12+16 2 = 44 g/mol ν = m/M = 44/44 = 1 mol N(CO 2) = ν N A = 1 6.02 10 23 = 6.02 ·10 23
Task 2. Calculate the mass of one molecule of ozone and an argon atom.
Solution:
M(O 3) = 16 3 = 48 g m(O 3) = M(O 3)/N A = 48/(6.02 10 23) = 7.97 10 -23 g M(Ar) = 40 g m(Ar) = M(Ar)/N A = 40/(6.02 10 23) = 6.65 10 -23 g
Task 3. What is the volume at standard conditions? occupies 2 moles of methane.
Solution:
ν = V/22.4 V(CH 4) = ν 22.4 = 2 22.4 = 44.8 l
Task 4. Determine the density and relative density of carbon monoxide (IV) from hydrogen, methane and air.
Solution:
M r (CO 2)=12+16·2=44; M(CO 2)=44 g/mol M r (CH 4)=12+1·4=16; M(CH 4)=16 g/mol M r (H 2)=1·2=2; M(H 2)=2 g/mol M r (air)=29; M(air)=29 g/mol ρ=m/V ρ(CO 2)=44/22.4=1.96 g/mol D(CH 4)=M(CO 2)/M(CH 4)= 44/16=2.75 D(H 2)=M(CO 2)/M(H 2)=44/2=22 D(air)=M(CO 2)/M(air)=44/24= 1.52
Task 5. Determine the mass of the gas mixture, which includes 2.8 cubic meters of methane and 1.12 cubic meters of carbon monoxide.
Solution:
M r (CO 2)=12+16·2=44; M(CO 2)=44 g/mol M r (CH 4)=12+1·4=16; M(CH 4) = 16 g/mol 22.4 cubic meters CH 4 = 16 kg 2.8 cubic meters CH 4 = x m(CH 4) = x = 2.8 16/22.4 = 2 kg 22.4 cubic meters CO 2 = 28 kg 1.12 cubic meters CO 2 = x m(CO 2)=x=1.12·28/22.4=1.4 kg m(CH 4)+m(CO 2)=2+1, 4=3.4 kg
Task 6. Determine the volumes of oxygen and air required to burn 112 cubic meters of divalent carbon monoxide when it contains non-combustible impurities in a volume fraction of 0.50.
Solution:
- determine the volume of pure CO in the mixture: V(CO)=112·0.5=66 cubic meters
- determine the volume of oxygen required to burn 66 cubic meters of CO: 2CO+O 2 =2CO 2 2mol+1mol 66m 3 +X m 3 V(CO)=2·22.4 = 44.8 m 3 V(O 2)=22 .4 m 3 66/44.8 = X/22.4 X = 66 22.4/44.8 = 33 m 3 or 2V(CO)/V(O 2) = V 0 (CO)/V 0 (O 2) V - molar volumes V 0 - calculated volumes V 0 (O 2) = V(O 2)·(V 0 (CO)/2V(CO))
Task 7. How will the pressure change in a vessel filled with hydrogen and chlorine gases after they react? Is it the same for hydrogen and oxygen?
Solution:
- H 2 +Cl 2 =2HCl - as a result of the interaction of 1 mole of hydrogen and 1 mole of chlorine, 2 moles of hydrogen chloride are obtained: 1 (mol) + 1 (mol) = 2 (mol), therefore, the pressure will not change, since the resulting volume of the gas mixture is equal to the sum of the volumes of the components that reacted.
- 2H 2 + O 2 = 2H 2 O - 2 (mol) + 1 (mol) = 2 (mol) - the pressure in the vessel will decrease by one and a half times, since from 3 volumes of the components that reacted, 2 volumes of the gas mixture are obtained.
Task 8. 12 liters of a gas mixture of ammonia and tetravalent carbon monoxide at no. have a mass of 18 g. How much of each gas is in the mixture?
Solution:
V(NH 3)=x l V(CO 2)=y l M(NH 3)=14+1 3=17 g/mol M(CO 2)=12+16 2=44 g/mol m( NH 3)=x/(22.4·17) g m(CO 2)=y/(22.4·44) g System of equations volume of mixture: x+y=12 mass of mixture: x/(22.4· 17)+y/(22.4·44)=18 After solving we get: x=4.62 l y=7.38 l
Task 9. What amount of water will be obtained as a result of the reaction of 2 g of hydrogen and 24 g of oxygen?
Solution:
2H 2 +O 2 =2H 2 O
From the reaction equation it is clear that the number of reactants does not correspond to the ratio of the stoichiometric coefficients in the equation. In such cases, calculations are carried out using a substance that is less abundant, i.e., this substance will end up first during the reaction. To determine which of the components is deficient, you need to pay attention to the coefficient in the reaction equation.
Amounts of starting components ν(H 2)=4/2=2 (mol) ν(O 2)=48/32=1.5 (mol)
However, there is no need to rush. In our case, for a reaction with 1.5 moles of oxygen, 3 moles of hydrogen (1.5 2) are needed, but we only have 2 moles, i.e., 1 mole of hydrogen is missing for all one and a half moles of oxygen to react. Therefore, we will calculate the amount of water using hydrogen:
ν(H 2 O)=ν(H 2)=2 mol m(H 2 O) = 2 18=36 g
Problem 10. At a temperature of 400 K and a pressure of 3 atmospheres, the gas occupies a volume of 1 liter. What volume will this gas occupy at zero level?
Solution:
From the Clapeyron equation:
P·V/T = Pn ·Vn/Tn Vn = (PVT n)/(Pn T) Vn = (3·1·273)/(1·400) = 2.05 l
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 Greek letter v (nude) 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 contained in 12 g of a 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 international system 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 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 |
What have we learned?
The molar volume of a gas studied in chemistry (grade 8), along with molar mass and density, are necessary quantities for determining the composition of a particular chemical substance. A feature of a molar gas is that one mole of gas always contains the same volume. This volume is called the molar volume of the gas.
Test on the topic
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In chemistry, they do not use the absolute masses of molecules, but use the relative molecular mass. It shows how many times the mass of a molecule is greater than 1/12 the mass of a carbon atom. This quantity is denoted by Mr.
Relative molecular mass is equal to the sum of the relative atomic masses of its constituent atoms. Let's calculate the relative molecular mass of water.
You know that a water molecule contains two hydrogen atoms and one oxygen atom. Then its relative molecular mass will be equal to the sum of the products of the relative atomic mass everyone chemical element by the number of its atoms in a water molecule:
Knowing the relative molecular weights gaseous substances, you can compare their densities, i.e. calculate the relative density of one gas from another - D(A/B). The relative density of gas A to gas B is equal to the ratio of their relative molecular masses:
Let's calculate the relative density of carbon dioxide to hydrogen:
Now we calculate the relative density of carbon dioxide to hydrogen:
D(arc/hydr) = Mr(arc) : Mr(hydr) = 44:2 = 22.
Thus, carbon dioxide 22 times heavier than hydrogen.
As you know, Avogadro's law applies only to gaseous substances. But chemists need to have an idea of the number of molecules and in portions of liquid or solids. Therefore, to compare the number of molecules in substances, chemists introduced the value - molar mass .
Molar mass is denoted M, it is numerically equal to the relative molecular weight.
The ratio of the mass of a substance to its molar mass is called amount of substance .
The amount of substance is indicated n. This is a quantitative characteristic of a portion of a substance, along with mass and volume. The amount of a substance is measured in moles.
The word "mole" comes from the word "molecule". The number of molecules in equal amounts of a substance is the same.
It has been experimentally established that 1 mole of a substance contains particles (for example, molecules). This number is called Avogadro's number. And if we add a unit of measurement to it - 1/mol, then it will be physical quantity- Avogadro's constant, which is denoted by N A.
Molar mass is measured in g/mol. Physical meaning molar mass is that this mass is 1 mole of a substance.
According to Avogadro's law, 1 mole of any gas will occupy the same volume. The volume of one mole of gas is called molar volume and is denoted Vn.
Under normal conditions (which is 0 °C and normal pressure - 1 atm. or 760 mm Hg or 101.3 kPa), the molar volume is 22.4 l/mol.
Then the amount of gas substance at ground level is can be calculated as the ratio of gas volume to molar volume.
TASK 1. What amount of substance corresponds to 180 g of water?
TASK 2. Let us calculate the volume at zero level that will be occupied by carbon dioxide in an amount of 6 mol.
Bibliography
- Collection of problems and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry, 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p. 29-34)
- Ushakova O.V. Workbook in chemistry: 8th grade: to the textbook P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 27-32)
- Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§§ 12, 13)
- Chemistry: inorg. chemistry: textbook. for 8th grade. general education institution / G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009. (§§ 10, 17)
- Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.
- Unified collection of digital educational resources ().
- Electronic version of the journal “Chemistry and Life” ().
- Chemistry tests (online) ().
Homework
1.p.69 No. 3; p.73 No. 1, 2, 4 from the textbook “Chemistry: 8th grade” (P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005).
2. №№ 65, 66, 71, 72 from the Collection of problems and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry, 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006.
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.
The molar volume of gases under normal conditions (normal pressure p n = 101,325 Pa ≈ 101.3 kPa and temperature T n = 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?
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1. Determine the amount of ammonia substance: n(NH 3) = m(NH 3) / M(NH 3) = 51 / 17 = 3 mol. 2. The volume of ammonia under normal conditions is: V(NH 3) = V m n(NH 3) = 22.4 3 = 67.2 l. 3. Using formula (3), we reduce the volume of ammonia to these conditions (temperature T = (273 + 20) K = 293 K): V(NH 3) = pn Vn (NH 3) / pT n = 101.3 293 67.2 / 250 273 = 29.2 l. Answer: V(NH 3) = 29.2 l. |
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.
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1. Find the amounts of hydrogen and nitrogen substances: n(N 2) = m(N 2) / M(N 2) = 5.6 / 28 = 0.2 mol n(H 2) = m(H 2) / M(H 2) = 1.4 / 2 = 0.7 mol 2. Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of the gases, i.e. V(mixtures) = V(N 2) + V(H 2) = V m n(N 2) + V m n(H2) = 22.4 0.2 + 22.4 0.7 = 20.16 l. Answer: V(mixture) = 20.16 l. |
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.