Where m is mass, M is molar mass, V is volume.

4. Law of Avogadro. Established 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 the Avogadro constant)

The consequence of this law is that 1 mole of any gas occupies under normal conditions (P 0 \u003d 101.3 kPa and T 0 \u003d 298 K) 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:

6. Gay-Lussac's law

At constant pressure, the change in the volume of a gas is directly proportional to the 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 condition to another:

P 0 , V 0 ,T 0 - volume pressure and temperature under normal conditions: P 0 =760 mm Hg. Art. or 101.3 kPa; T 0 \u003d 273 K (0 0 C)

8. Independent assessment of the value of molecular masses M can be done using the so-called equations of state for an ideal gas or the Clapeyron-Mendeleev equations :

pV=(m/M)*RT=vRT.(1.1)

where R - gas pressure in a closed system, V- volume of the system, t - mass of gas 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 N.C. into equation (1.1):

r = (p V) / (T) \u003d (101.325kPa 22.4 l) / (1 mol 273K) \u003d 8.31J / mol.K)

Examples of problem solving

Example 1 Bringing the volume of gas to normal conditions.

What volume (n.o.) will occupy 0.4×10 -3 m 3 of gas at 50 0 C and a pressure of 0.954×10 5 Pa?

Decision. To bring the volume of gas to normal conditions, use the general formula that combines the laws of Boyle-Mariotte and Gay-Lussac:

pV/T = p 0 V 0 /T 0 .

The volume of gas (n.o.) is , where T 0 = 273 K; p 0 \u003d 1.013 × 10 5 Pa; T = 273 + 50 = 323 K;

m 3 \u003d 0.32 × 10 -3 m 3.

When (n.o.) 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 from hydrogen and air.

Decision. It follows from Avogadro's law that the relative density of one gas over another is equal to the ratio of molecular masses ( M h) of these gases, i.e. D=M 1 /M 2. If a M 1С2Н6 = 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 values ​​of the relative density of these gases with respect to hydrogen.

Decision. 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 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 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 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.

Decision. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:

where m is the mass of gas; M is the molar mass of the 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\u003d 8.3144 × 10 3 J / (kmol × K).

The mass of 1 mole of a substance is called the molar mass. What is the volume of 1 mole of a substance called? Obviously, it is also called the molar volume.

What is the molar volume of water? When we measured 1 mol of water, we did not weigh 18 g of water on the scales - this is inconvenient. We used measuring utensils: a cylinder or a beaker, because we knew that the density of water is 1 g/ml. Therefore, the molar volume of water is 18 ml/mol. For liquids and solids, the molar volume depends on their density (Fig. 52, a). Another thing for gases (Fig. 52, b).

Rice. 52.
Molar volumes (n.a.):
a - liquids and solids; b - gaseous substances

If we take 1 mol of hydrogen H 2 (2 g), 1 mol of oxygen O 2 (32 g), 1 mol of ozone O 3 (48 g), 1 mol of carbon dioxide CO 2 (44 g) and even 1 mol of water vapor H 2 O (18 g) under the same conditions, for example, normal (in chemistry, it is customary to call normal conditions (n.a.) a temperature of 0 ° C and a pressure of 760 mm Hg, or 101.3 kPa), it turns out that 1 mol of any of the gases will occupy the same volume, equal to 22.4 liters, and contain the same number of molecules - 6 × 10 23.

And if we take 44.8 liters of gas, then how much of its substance will be taken? Of course, 2 mol, since the given volume is twice the molar volume. Hence:

where V is the volume of gas. From here

Molar volume is a physical quantity equal to the ratio of the volume of a substance to the amount of a substance.

The molar volume of gaseous substances is expressed in l/mol. Vm - 22.4 l/mol. The volume of one kilomol is called kilomolar and is measured in m 3 / kmol (Vm = 22.4 m 3 / kmol). Accordingly, the millimolar volume is 22.4 ml/mmol.

Task 1. Find the mass of 33.6 m 3 of ammonia NH 3 (n.a.).

Task 2. Find the mass and volume (n.s.) that 18 × 10 20 molecules of hydrogen sulfide H 2 S have.

When solving the problem, let's pay attention to the number of molecules 18 × 10 20 . Since 10 20 is 1000 times smaller than 10 23 , obviously, calculations should be made using mmol, ml/mmol and mg/mmol.

Keywords and phrases

1. Molar, millimolar and kilomolar volumes of gases.
2. The molar volume of gases (under normal conditions) is 22.4 l / mol.
3. Normal conditions.

Work with computer

1. Refer to the electronic application. Study the material of the lesson and complete the suggested tasks.
2. Search the Internet for email addresses that can serve as additional sources that reveal the content of the keywords and phrases of the paragraph. Offer the teacher your help in preparing a new lesson - make a report on the key words and phrases of the next paragraph.

Questions and tasks

1. Find the mass and number of molecules at n. y. for: a) 11.2 liters of oxygen; b) 5.6 m 3 nitrogen; c) 22.4 ml of chlorine.
2. Find the volume which, at n. y. will take: a) 3 g of hydrogen; b) 96 kg of ozone; c) 12 × 10 20 nitrogen molecules.
3. Find the densities (mass of 1 liter) of argon, chlorine, oxygen and ozone at n. y. How many molecules of each substance will be contained in 1 liter under the same conditions?
4. Calculate the mass of 5 l (n.a.): a) oxygen; b) ozone; c) carbon dioxide CO 2.
5. Specify which is heavier: a) 5 liters of sulfur dioxide (SO 2) or 5 liters of carbon dioxide (CO 2); b) 2 liters of carbon dioxide (CO 2) or 3 liters of carbon monoxide (CO).

The relationship between pressure and volume of an ideal gas at constant temperature is shown in fig. one.

The pressure and volume of a gas sample are inversely proportional, i.e. their products are constant: pV = const. This relation can be written in a more convenient form for solving problems:

p1 V 1 =p 2 V 2(Boyle-Mariotte law).

Imagine that 50 liters of gas (V 1 ), under a pressure of 2 atm (p 1), compressed to a volume of 25 l (V 2), then its new pressure will be equal to:

The dependence of the properties of ideal gases on temperature is determined by the Gay-Lussac law: the volume of a gas is directly proportional to its absolute temperature (at a constant mass: V = kT, where k- proportionality factor). This relation is usually written in a more convenient form for solving problems:

For example, if 100 liters of gas at a temperature of 300K is heated to 400K without changing the pressure, then at a higher temperature the new volume of gas will be equal to

Recording the combined gas law pV/T== const can be converted to the Mendeleev-Clapeyron equation:

where R- universal gas constant, a is the number of moles of gas.

The Mendeleev-Clapeyron equation allows for a wide variety of calculations. For example, you can determine the number of moles of gas at a pressure of 3 atm and a temperature of 400K, occupying a volume of 70 liters:

One of the consequences of the combined gas law: equal volumes of different gases at the same temperature and pressure contain the same number of molecules. This is Avogadro's law.

In turn, an important consequence also follows from Avogadro's law: the masses of two identical volumes of different gases (of course, at the same pressure and temperature) are related as their molecular weights:

m 1 /m 2 = M 1 /M 2 (m 1 and m 2 are the masses of two gases);

M1 IM 2 is the relative density.

Avogadro's law only applies to ideal gases. Under normal conditions, gases that are difficult to compress (hydrogen, helium, nitrogen, neon, argon) can be considered ideal. For carbon monoxide (IV), ammonia, sulfur oxide (IV), deviations from ideality are already observed under normal conditions and increase with increasing pressure and decreasing temperature.

Example 1. Carbon dioxide with a volume of 1 liter under normal conditions has a mass of 1.977 g. What is the real volume occupied by a mole of this gas (at n.a.)? Explain the answer.

Decision. The molar mass M (CO 2) \u003d 44 g / mol, then the volume of the mole is 44 / 1.977 \u003d 22.12 (l). This value is less than that accepted for ideal gases (22.4 l). The decrease in volume is associated with an increase in the interaction between CO 2 molecules, i.e., a deviation from ideality.

Example 2. Gaseous chlorine weighing 0.01 g, located in a sealed ampoule with a volume of 10 cm 3, is heated from 0 to 273 o C. What is the initial pressure of chlorine at 0 o C and at 273 o C?

Decision. M r (Cl 2)=70.9; hence 0.01 g of chlorine corresponds to 1.4 10 -4 mol. The volume of the ampoule is 0.01 l. Using the Mendeleev-Clapeyron equation pV=vRT, find the initial pressure of chlorine (p 1 ) at 0 o C:

similarly we find the pressure of chlorine (p 2) at 273 o C: p 2 \u003d 0.62 atm.

Example 3. What is the volume that is occupied by 10 g of carbon monoxide (II) at a temperature of 15 o C and a pressure of 790 mm Hg. Art.?

Decision.

Tasks

1 . What volume (at N.S.) does 0.5 mol of oxygen occupy?
2 . What volume is occupied by hydrogen containing 18-10 23 molecules (at n.a.)?
3 . What is the molar mass of sulfur oxide (IV) if the hydrogen density of this gas is 32?
4 . What volume is occupied by 68 g of ammonia at a pressure of 2 atm and a temperature of 100 o C?
5 . In a closed vessel with a capacity of 1.5 liters is a mixture of hydrogen sulfide with excess oxygen at a temperature of 27 o C and a pressure of 623.2 mm Hg. Art. Find the total amount of substances in the vessel.
6 . In a large room, the temperature can be measured with a "gas" thermometer. For this purpose, a glass tube having an internal volume of 80 ml was filled with nitrogen at a temperature of 20° C. and a pressure of 101.325 kPa. After that, the tube was slowly and carefully taken out of the room into a warmer room. Due to thermal expansion, the gas escaped from the tube and was collected above the liquid, whose vapor pressure is negligible. The total volume of gas exiting the tube (measured at 20° C. and 101.325 kPa) is 3.5 ml. How many moles of nitrogen did it take to fill the glass tube, and what is the temperature of the warmer room?
7 . A chemist who determined the atomic mass of a new element X in the middle of the 19th century used the following method: he obtained four compounds containing element X (A, B, C and D), and determined the mass fraction of the element (%) in each of them. In a vessel from which air had previously been evacuated, he placed each compound, transferred to a gaseous state at 250 o C, and set the vapor pressure of the substance to 1.013 10 5 Pa. The mass of the gaseous substance was determined from the difference between the masses of empty and full vessels. A similar procedure was carried out with nitrogen. The result was a table like this:

Gas	Total weight, g	Mass fraction () of the element x in the substance,%
N 2	0,652	-
BUT	0,849	97,3
B	2,398	68,9
AT	4,851	85,1
G	3,583	92,2

Determine the probable atomic mass of element X.

8 . In 1826, the French chemist Dumas proposed a method for determining vapor density, applicable to many substances. Using this method, it was possible to find the molecular weights of compounds, using Avogadro's hypothesis that equal amounts of molecules are contained in equal volumes of gases and vapors at equal pressure and temperature. However, experiments with some substances, made according to the Dumas method, contradicted Avogadro's hypothesis and called into question the very possibility of determining the molecular weight by this method. Here is a description of one of these experiments (Fig. 2).

a. At the neck of a vessel a of a known volume, a weighed portion of ammonia b was placed and heated in an oven in up to this temperature t o , at which all ammonia evaporated. The resulting vapors displaced the air from the vessel, some of them stood out in the form of fog. heated up t o the vessel, the pressure in which was equal to atmospheric pressure, was sealed along the constriction r, then cooled and weighed.

Then the vessel was opened, washed from condensed ammonia, dried and weighed again. By difference, the mass m of ammonia was determined.

This mass, when heated to t o had pressure R, equal to atmospheric, in a vessel with a volume v. For vessel a, the pressure and volume of a known mass of hydrogen at room temperature were predetermined. The ratio of the molecular weight of ammonia to the molecular weight of hydrogen was determined by the formula

Got the value M / M (H 2) \u003d 13.4. The ratio calculated from the formula NH 4 Cl was 26.8.

b. The experiment was repeated, but the neck of the vessel was closed with a porous asbestos stopper. d, permeable to gases and vapors. At the same time, we got the relation M/ M (H 2) \u003d 14.2.

in. We repeated experiment b, but increased the initial sample of ammonia by 3 times. The ratio became equal to M/M (H 2) = 16.5.
Explain the results of the described experiment and prove that Avogadro's law was observed in this case.

1. A mole of any gas occupies a volume (at n.a.) of 22.4 liters; 0.5 mol O 2 occupies a volume of 22.40.5 \u003d 11.2 (l).
2. The number of hydrogen molecules equal to 6.02-10 23 (Avogadro's number), at n. y. occupies a volume of 22.4 l (1 mol); then

3. Molar mass of sulfur(IV) oxide: M(SO 2) = 322 = 64 (g/mol).
4. At n. y. 1 mole of NH3, equal to 17 g, occupies a volume of 22.4 liters, 68 g occupies a volume X l ,

From the equation of the gas state p o V o /T o = p 1 V 1 /T 1 we find

mixtures of H 2 S and O 2 .

6 . When filling the tube with nitrogen

In the tube remained (under initial conditions) V 1: 80-3.5 = 76.5 (ml). With an increase in temperature, nitrogen, which occupied a volume of 76.5 ml (V 1) at 20 o C, began to occupy a volume of V 2 = 80 ml. Then, according to Т 1 /Т 2 = = V 1 /V 2 we have

Suppose that at a temperature of 250 ° C, substances A, B, C, D are ideal gases. Then according to Avogadro's law

Mass of element X in 1 mol of substance A, B, C and D (g/mol):

M(A) . 0.973 = 35.45; M(B) . 0.689 = 70.91; M (B). 0.851 = 177.17; M(G) . 0.922= 141.78

Since there must be an integer number of atoms of the element X in the molecule of the substance, it is necessary to find the greatest common divisor of the obtained values. It is 35.44 g / mol, and this number can be considered the probable atomic mass of element X.

8. Any modern chemist can easily explain the results of the experiment. It is well known that the sublimation of ammonia - ammonium chloride - is a reversible process of thermal decomposition of this salt:

NH4Cl		NH3	+ HCl.
53,5		17	36,5

In the gas phase are ammonia and hydrogen chloride, their average relative molecular weight M t

Less clear is the change in the result in the presence of an asbestos plug. However, in the middle of the last century, it was precisely experiments with porous (“borehole”) partitions that showed that ammonia vapor contained two gases. The lighter ammonia passes through the pores more quickly and is easy to spot, either by smell or with wet indicator paper.

A rigorous expression for estimating the relative permeability of gases through porous partitions is given by the molecular-kinetic theory of gases. Average speed of gas molecules
, where R is the gas constant; T - absolute temperature; M - molar mass. According to this formula, ammonia should diffuse faster than hydrogen chloride:

Consequently, when an asbestos stopper is introduced into the neck of the flask, the gas in the flask will have time to somewhat enrich itself with heavy HC1 during the time that the pressure equalizes with atmospheric pressure. The relative density of the gas increases in this case. With an increase in the mass of NH 4 C1, a pressure equal to atmospheric pressure will be established later (the asbestos plug prevents the rapid leakage of vapor from the flask), the gas in the flask will contain more hydrogen chloride than in the previous case; the density of the gas will increase.

The volume of a gram-molecule of a gas, as well as the mass of a gram-molecule, is a derived unit of measurement and is expressed as the ratio of units of volume - liters or milliliters to a mole. Therefore, the dimension of the gram-molecular volume is l / mol or ml / mol. Since the volume of a gas depends on temperature and pressure, the gram-molecular volume of a gas varies depending on the conditions, but since the gram-molecules of all substances contain the same number of molecules, the gram-molecules of all substances under the same conditions occupy the same volume. under normal conditions. = 22.4 l/mol, or 22400 ml/mol. Recalculation of the gram-molecular volume of gas under normal conditions per volume under given conditions of production. is calculated according to the equation: J- t-tr from which it follows that where Vo is the gram-molecular volume of gas under normal conditions, Umol is the desired gram-molecular volume of gas. Example. Calculate the gram-molecular volume of the gas at 720 mm Hg. Art. and 87°C. Decision. The most important calculations related to the gram-molecular volume of a gas a) Conversion of the volume of gas to the number of moles and the number of moles per volume of gas. Example 1. Calculate how many moles are contained in 500 liters of gas under normal conditions. Decision. Example 2. Calculate the volume of 3 mol of gas at 27 * C 780 mm Hg. Art. Decision. We calculate the gram-molecular volume of gas under the specified conditions: V - ™ ** RP st. - 22.A l / mol. 300 deg \u003d 94 p. -273 vrad 780 mm Hg "ap.--24" ° Calculate the volume of 3 mol GRAM MOLECULAR VOLUME OF GAS V \u003d 24.0 l / mol 3 mol \u003d 72 l b) Conversion of the mass of gas to its volume and volume of a gas per its mass. In the first case, the number of moles of gas is first calculated from its mass, and then the volume of gas is calculated from the found number of moles. In the second case, the number of moles of gas is first calculated from its volume, and then, from the found number of moles, the mass of the gas. Example 1, Calculate the volume (at N.C.) of 5.5 g of carbon dioxide CO * Solution. |icoe ■= 44 g/mol V = 22.4 l/mol 0.125 mol 2.80 l Example 2. Calculate the mass of 800 ml (at n.a.) carbon monoxide CO. Decision. | * w => 28 g / mol m " 28 g / lnm 0.036 did * \u003d" 1.000 g If the mass of the gas is expressed not in grams, but in kilograms or tons, and its volume is expressed not in liters or milliliters, but in cubic meters , then a twofold approach to these calculations is possible: either split higher measures into lower ones, or the calculation of ae with moles is known, and with kilogram-molecules or ton-molecules, using the following ratios: under normal conditions, 1 kilogram-molecule-22,400 l / kmol , 1 ton-molecule - 22,400 m*/tmol. Units: kilogram-molecule - kg/kmol, ton-molecule - t/tmol. Example 1. Calculate the volume of 8.2 tons of oxygen. Decision. 1 ton-molecule Oa » 32 t/tmol. We find the number of ton-molecules of oxygen contained in 8.2 tons of oxygen: 32 t/tmol ** 0.1 Calculate the mass of 1000 -k * ammonia (at n.a.). Decision. We calculate the number of ton-molecules in the specified amount of ammonia: "-stay5JT-0.045 t/mol Calculate the mass of ammonia: 1 ton-molecule NH, 17 t/mol tyv, \u003d 17 t/mol 0.045 t/mol * 0.765 t General principle of calculation, related to gas mixtures, is that the calculations related to the individual components are performed separately, and then the results are summed.Example 1. Calculate what volume a gas mixture consisting of 140 g of nitrogen and 30 e of hydrogen will occupy under normal conditions. Solution Calculate the number of moles of nitrogen and hydrogen contained in the mixture (No. "= 28 u/mol; cn, = 2 g/mol): 140 £ 30 in 28 g/mol W Total 20 mol. GRAM MOLECULAR VOLUME OF GAS Calculate the volume of the mixture : Ueden in 22 "4 AlnoAb 20 mol " 448 l Example 2. Calculate the mass of 114 mixture (at n.a.) of carbon monoxide and carbon dioxide, the volume composition of which is expressed by the ratio: /lso: /iso, = 8:3. Decision. According to the indicated composition, we find the volumes of each gas by the method of proportional division, after which we calculate the corresponding number of moles: t / II l "8 Q" "11 J 8 Q Ksoe 8 + 3 8 * Va> "a & + & * VCQM grfc - 0 "36 ^- grfc "" 0.134 jas * Calculating the mass of each of the gases from the found number of moles of each of them. 1 "co 28 g / mol; jico. \u003d 44 g / mol moo" 28 e! mol 0.36 mol "South tco. \u003d 44 e / zham" - 0.134 "au> - 5.9 g By adding the found masses of each of the components, we find the mass of the mixture: gas by gram-molecular volume Above was considered the method of calculating the molecular weight of a gas by relative density.Now we will consider the method of calculating the molecular weight of a gas by gram-molecular volume.In the calculation, it is assumed that the mass and volume of the gas are directly proportional to each other.It follows "that the volume of a gas and its mass are related to each other as the gram-molecular volume of a gas is to its gram-molecular mass, which in mathematical which form is expressed as follows: V_ Ushts / i (x where Un * "- gram-molecular volume, p - gram-molecular weight. Hence _ Huiol t p? Let's consider the calculation technique on a specific example. "Example. The mass of 34 $ ju gas at 740 mm Hg, spi and 21 ° C is 0.604 g. Calculate the molecular weight of the gas. Decision. To solve, you need to know the gram-molecular volume of the gas. Therefore, before proceeding to the calculations, it is necessary to dwell on some specific gram-molecular volume of the gas. You can use the standard gram-molecular volume of gas, which is equal to 22.4 l / mol. Then the volume of gas specified in the condition of the problem must be brought to normal conditions. But it is possible, on the contrary, to calculate the gram-molecular volume of a gas under the conditions specified in the problem. With the first method of calculation, the following design is obtained: at 740 * mrt.st .. 340 ml - 273 deg ^ Q ^ 0 760 mm Hg. Art. 294 deg ™ 1 l.1 - 22.4 l / mol 0.604 in _ s, ypya. -m-8 \u003d 44 g, M0Ab In the second method, we find: V - 22»4 A! mol No. mm Hg. st.-29A deg 0A77 l1ylv. Uiol 273 vrad 740 mmHg Art. ~ R * 0 ** In both cases, we calculate the mass of the gram molecule, but since the gram molecule is numerically equal to the molecular mass, we thereby find the molecular mass.

2.1. Relative gas density d equal to the ratio of densities (ρ 1 and ρ 2) of gases (at the same pressure and temperature):

d \u003d ρ 1: ρ 2 ≈ M 1: M 2 (2.1)

where M 1 and M 2 are the molecular weights of gases.

Relative gas density:

relative to air: d ≈ M/29
with respect to hydrogen: d ≈ M/2

where М, 29 and 2 are the corresponding molecular weights of the given gas, air and hydrogen.

2.2. Weight quantity a (in g) gas in a given volume V (in dm 3):

• a \u003d M * 1.293 * p * 273 * V / 28.98 (273 + t) * 760 \u003d 0.01605 * p * M * V / 273 + t (2.2)

where M is the molecular weight of the gas, p is the gas pressure, mm Hg, t is the gas temperature, 0 C.

The amount of gas in g per 1 dm 3 under normal conditions

where d is the relative density of the gas relative to air.

2.3.Volume V occupied by a given weight quantity a of gas :

V \u003d a * 22.4 * 760 * (273 + t) / M * p (2.4)

2.5. Gas mixtures

The mass (in g) of a mixture of n shaped components having volumes V 1, V 2 ... V n and molecular weights M 1, M 2 ... M n is equal to

Where 22.4 is the volume of 1 mol of a substance in the gaseous state at 273 K and 101.32 kPa (0 ° C and 760 mm Hg)

Since the volume of the mixture V \u003d V 1 + V 2 + ... + V n, then 1 dm 3 of it has a mass:

The average molecular weight M of the gas mixture (with additivity of its properties) is equal to:

The concentration of the components of gas mixtures is most often expressed as a percentage by volume. Volume concentration (V 1 /V·100) numerically coincides with the fraction of the partial pressure of the component (р 1 /р·100) and with its molar concentration (M 1 /M·100).

The proportions of individual components i in the gas mixture are equal, %

massive voluminous

where q i is the mass content of the i-th component in the mixture.

Equal volumes of different gases under the same conditions contain the same number of molecules, so

p 1: p 2: ... = V 1: V 2: ... = M 1: M 2: ...

where M is the number of moles.

Number of moles of the component:

If the gas is under the same conditions(P, T) and it is necessary to determine its volume or mass under other conditions (P´, T´), then the following formulas are used:

for volume conversion

for mass conversion

At T = const partial pressure P us saturated steam in a gas mixture, regardless of the total pressure, is constant. At 101.32 kPa and T K, 1 mole of gas or vapor occupies a volume of 22.4 (T / 273) dm 3. If the vapor pressure at this temperature is P us, then the volume of 1 mol is:

Thus, the mass of 1m 3 pair of molecular weight M at temperature T and pressure P is equal to us, in g / m 3

Knowing the mass content of saturated steam in 1 m 3 of the mixture, we can calculate its pressure:

The volume of dry gas is calculated by the formula:

where P sat., T is the pressure of saturated water vapor at temperature T.

Bringing the volumes of dry V (T, P) dry. and wet V (T, P) vl. gases to normal conditions (n.o.) (273 K and 101.32 kPa) are produced according to the formulas:

Formula

are used to recalculate the volume of wet gas at P and T to other P´, T´, provided that the equilibrium pressure of water vapor also changes with temperature. Expressions for recalculating gas volumes under different conditions are similar:

If the water vapor pressure of saturated steam at any temperature is P sat. , but it is necessary to calculate G n.o.s. - its content in 1 m 3 of gas at n.o., then equation (1.2) is used, but in this case T is not the saturation temperature, but is equal to 273 K.

From this it follows that:

G n.o.s. = 4.396 10 -7 Mr sat. .

The pressure of saturated water vapor, if its content is known in 1 m 3 at n.o. calculated according to the formula.