Volume formula under normal conditions. Molar volume

The volume of gas can be determined using several formulas. It is necessary to choose the appropriate one based on the data in the condition of the quantities problem. A major role in selecting the desired formula is played by these media, and in particular: pressure and temperature.

Instructions

1. The formula that is especially often encountered in problems is: V = n*Vm, where V is the volume of gas (l), n is the number of substance (mol), Vm is the molar volume of gas (l/mol), under typical conditions (n.s.) is a standard value and is equal to 22.4 l/mol. It happens that the condition does not contain the number of a substance, but there is a mass of a certain substance, then we do this: n = m/M, where m is the mass of the substance (g), M is the molar mass of the substance (g/mol). We find the molar mass using the table D.I. Mendeleev: under each element its nuclear mass is written, we add up all the masses and get the one we need. But such problems are quite rare; usually the problem contains a reaction equation. The solution to such problems changes a little. Let's look at an example.

2. What volume of hydrogen will be released under typical conditions if aluminum weighing 10.8 g is dissolved in excess hydrochloric acid. We write the reaction equation: 2Al + 6HCl(ex.) = 2AlCl3 + 3H2. Solve the problem about this equation. Find the number of aluminum substances that reacted: n(Al) = m(Al)/M(Al). In order to substitute the data into this formula, we need to calculate the molar mass of aluminum: M(Al) = 27 g/mol. We substitute: n(Al) = 10.8/27 = 0.4 mol. From the equation we see that when 2 moles of aluminum are dissolved, 3 moles of hydrogen are formed. We calculate what amount of hydrogen substance is formed from 0.4 mol of aluminum: n(H2) = 3 * 0.4/2 = 0.6 mol. After this, we substitute the data into the formula for finding the volume of hydrogen: V = n*Vm = 0.6*22.4 = 13.44 liters. So we got the result.

3. If we are dealing with a gas system, then the following formula holds: q(x) = V(x)/V, where q(x)(phi) is the volume fraction of the component, V(x) is the volume of the component (l), V – system volume (l). To find the volume of a component, we obtain the formula: V(x) = q(x)*V. And if you need to find the volume of the system, then: V = V(x)/q(x).

A gas in which the interaction between molecules is negligibly small is considered impeccable. In addition to pressure, the state of a gas is characterized by temperature and volume. The relationships between these parameters are reflected in the gas laws.

Instructions

1. The pressure of a gas is directly proportional to its temperature, the amount of substance, and inversely proportional to the volume of the vessel occupied by the gas. The proportionality indicator is the universal gas continuous R, approximately equal to 8.314. It is measured in joules divided by moles and kelvins.

2. This arrangement forms the mathematical connection P=?RT/V, where? – number of substance (mol), R=8.314 – universal gas continuous (J/mol K), T – gas temperature, V – volume. Pressure is expressed in pascals. It can also be expressed in atmospheres, with 1 atm = 101.325 kPa.

3. The considered connectivity is a consequence of the Mendeleev-Clapeyron equation PV=(m/M) RT. Here m is the mass of the gas (g), M is its molar mass (g/mol), and the fraction m/M results in the number of substance?, or the number of moles. The Mendeleev-Clapeyron equation is objective for all gases that can be considered impeccable. This is a fundamental physical and chemical gas law.

4. When monitoring the behavior of an ideal gas, we talk about so-called typical conditions - environmental conditions that we often deal with in reality. Thus, typical data (n.s.) assume a temperature of 0 degrees Celsius (or 273.15 degrees on the Kelvin scale) and a pressure of 101.325 kPa (1 atm). A value has been discovered that equals the volume of one mole of an ideal gas under the following conditions: Vm = 22.413 l/mol. This volume is called molar. Molar volume is one of the main chemical constants used in solving problems.

5. The main thing to understand is that with continuous pressure and temperature, the volume of the gas also does not change. This fascinating postulate is formulated in Avogadro's law, which states that the volume of a gas is directly proportional to the number of moles.

Video on the topic

Note!
There are other formulas for finding volume, but if you need to find the volume of a gas, only the formulas given in this article are suitable.

One of the basic units in the International System of Units (SI) is The unit of quantity of a substance is the mole.

Mole – this is the amount of a substance that contains as many structural units of a given substance (molecules, atoms, ions, etc.) as there are carbon atoms contained in 0.012 kg (12 g) of a carbon isotope 12 WITH .

Considering that the value of the absolute atomic mass for carbon is equal to m(C) = 1.99 10  26 kg, the number of carbon atoms can be calculated N A, contained in 0.012 kg of carbon.

A mole of any substance contains the same number of particles of this substance (structural units). The number of structural units contained in a substance with an amount of one mole is 6.02 10 23 and is called Avogadro's number (N A ).

For example, one mole of copper contains 6.02 10 23 copper atoms (Cu), and one mole of hydrogen (H 2) contains 6.02 10 23 hydrogen molecules.

Molar mass(M) is the mass of a substance taken in an amount of 1 mole.

Molar mass is designated by the letter M and has the dimension [g/mol]. In physics they use the unit [kg/kmol].

In the general case, the numerical value of the molar mass of a substance numerically coincides with the value of its relative molecular (relative atomic) mass.

For example, the relative molecular weight of water is:

Мr(Н 2 О) = 2Аr (Н) + Аr (O) = 2∙1 + 16 = 18 a.m.u.

The molar mass of water has the same value, but is expressed in g/mol:

M (H 2 O) = 18 g/mol.

Thus, a mole of water containing 6.02 10 23 water molecules (respectively 2 6.02 10 23 hydrogen atoms and 6.02 10 23 oxygen atoms) has a mass of 18 grams. Water, with an amount of substance of 1 mole, contains 2 moles of hydrogen atoms and one mole of oxygen atoms.

1.3.4. The relationship between the mass of a substance and its quantity

Knowing the mass of a substance and its chemical formula, and therefore the value of its molar mass, you can determine the amount of the substance and, conversely, knowing the amount of the substance, you can determine its mass. For such calculations you should use the formulas:

where ν is the amount of substance, [mol]; m– mass of the substance, [g] or [kg]; M – molar mass of the substance, [g/mol] or [kg/kmol].

For example, to find the mass of sodium sulfate (Na 2 SO 4) in an amount of 5 moles, we find:

1) the value of the relative molecular mass of Na 2 SO 4, which is the sum of the rounded values ​​of the relative atomic masses:

Мr(Na 2 SO 4) = 2Аr(Na) + Аr(S) + 4Аr(O) = 142,

2) a numerically equal value of the molar mass of the substance:

M(Na 2 SO 4) = 142 g/mol,

3) and, finally, the mass of 5 mol of sodium sulfate:

m = ν M = 5 mol · 142 g/mol = 710 g.

Answer: 710.

1.3.5. The relationship between the volume of a substance and its quantity

Under normal conditions (n.s.), i.e. at pressure R , equal to 101325 Pa (760 mm Hg), and temperature T, equal to 273.15 K (0 С), one mole of different gases and vapors occupies the same volume equal to 22.4 l.

The volume occupied by 1 mole of gas or vapor at ground level is called molar volumegas and has the dimension liter per mole.

V mol = 22.4 l/mol.

Knowing the amount of gaseous substance (ν ) And molar volume value (V mol) you can calculate its volume (V) under normal conditions:

V = ν V mol,

where ν is the amount of substance [mol]; V – volume of gaseous substance [l]; V mol = 22.4 l/mol.

And, conversely, knowing the volume ( V) of a gaseous substance under normal conditions, its quantity (ν) can be calculated :

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

What is the molar volume of water? When we measured 1 mole 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, since 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). It's a different matter for gases (Fig. 52, b).

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

If you take 1 mole of hydrogen H2 (2 g), 1 mole of oxygen O2 (32 g), 1 mole of ozone O3 (48 g), 1 mole of carbon dioxide CO2 (44 g) and even 1 mole of water vapor H2 O (18 g) under the same conditions, for example normal (in chemistry it is customary to call normal conditions (n.s.) a temperature of 0 ° C and a pressure of 760 mm Hg, or 101.3 kPa), then 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 you take 44.8 liters of gas, then how much of its substance will be taken? Of course, 2 moles, 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 substance.

The molar volume of gaseous substances is expressed in l/mol. Vm - 22.4 l/mol. The volume of one kilomole 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.

Problem 1. Find the mass of 33.6 m 3 of ammonia NH 3 (n.s.).

Problem 2. Find the mass and volume (n.v.) of 18 × 10 20 molecules of hydrogen sulfide H 2 S.

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

Key words 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.

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Questions and tasks

1. Find the mass and number of molecules at n. u. for: a) 11.2 liters of oxygen; b) 5.6 m 3 nitrogen; c) 22.4 ml of chlorine.
2. Find the volume that at n. u. will take: a) 3 g of hydrogen; b) 96 kg of ozone; c) 12 × 10 20 nitrogen molecules.
3. Find the densities (mass 1 liter) of argon, chlorine, oxygen and ozone at room temperature. u. How many molecules of each substance will be contained in 1 liter under the same conditions?
4. Calculate the mass of 5 liters (n.s.): a) oxygen; b) ozone; c) carbon dioxide CO 2.
5. Indicate 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 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?

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.

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.