How to derive the Mendeleev clapeyron equation. Equation of state of ideal gas (Mendeleev-Clapeyron equation)

Annotation: traditional presentation of the topic, supplemented by a demonstration on a computer model.

Of the three aggregate states of matter, the simplest is the gaseous state. In gases, the forces acting between molecules are small and under certain conditions they can be neglected.

The gas is called perfect , if:

The size of molecules can be neglected, i.e. molecules can be considered material points;

We can neglect the forces of interaction between molecules (the potential energy of interaction of molecules is much less than their kinetic energy);

The collisions of molecules with each other and with the walls of the vessel can be considered absolutely elastic.

Real gases are close in properties to the ideal at:

Conditions close to normal conditions (t = 0 0 C, p = 1.03 10 5 Pa);

At high temperatures.

The laws that govern the behavior of ideal gases were discovered experimentally quite a long time ago. So, Boyle's law - Mariotte was established in the 17th century. We give the formulations of these laws.

Boyle's Law - Mariotte. Let the gas be under conditions where its temperature is kept constant (such conditions are called isothermal ). Then for a given mass of gas, the product of pressure and volume is a constant value:

This formula is called isotherm equation. Graphically, the dependence of p on V for various temperatures is shown in the figure.

The property of a body to change pressure with a change in volume is called compressibility. If the change in volume occurs at T=const, then the compressibility is characterized by isothermal compressibility factor which is defined as the relative change in volume that causes a change in pressure per unit.

For an ideal gas, it is easy to calculate its value. From the isotherm equation we get:

The minus sign indicates that as the volume increases, the pressure decreases. Thus, the isothermal compressibility of an ideal gas is equal to the reciprocal of its pressure. With increasing pressure, it decreases, because. the greater the pressure, the less the gas has the ability to further compress.

Gay-Lussac law. Let the gas be under conditions where its pressure is maintained constant (such conditions are called isobaric ). They can be carried out by placing gas in a cylinder closed by a movable piston. Then a change in the temperature of the gas will move the piston and change the volume. The pressure of the gas will remain constant. In this case, for a given mass of gas, its volume will be proportional to the temperature:

where V 0 - volume at temperature t = 0 0 C, - volume expansion coefficient gases. It can be represented in a form similar to the compressibility factor:

Graphically, the dependence of V on T for various pressures is shown in the figure.

Moving from temperature in the Celsius scale to absolute temperature, Gay-Lussac's law can be written as:

Charles' Law. If the gas is under conditions where its volume remains constant ( isochoric conditions), then for a given mass of gas, the pressure will be proportional to the temperature:

where p 0 - pressure at temperature t \u003d 0 0 C, - pressure coefficient. It shows the relative increase in gas pressure when it is heated by 10:

Charles' law can also be written as:

Avogadro's law: One mole of any ideal gas at the same temperature and pressure occupies the same volume. Under normal conditions (t = 0 0 C, p = 1.03 10 5 Pa), this volume is equal to m -3 / mol.

The number of particles contained in 1 mole of various substances, called. Avogadro's constant :

It is easy to calculate the number n 0 particles in 1 m 3 under normal conditions:

This number is called Loschmidt number.

Dalton's law: the pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the gases included in it, i.e.

where - partial pressures- the pressure that the components of the mixture would exert if each of them occupied a volume equal to the volume of the mixture at the same temperature.

Equation of Clapeyron - Mendeleev. From the laws of an ideal gas, one can obtain equation of state , linking T, p and V of an ideal gas in a state of equilibrium. This equation was first obtained by the French physicist and engineer B. Clapeyron and Russian scientists D.I. Mendeleev, therefore bears their name.

Let some mass of gas occupies a volume V 1 , has a pressure p 1 and is at a temperature T 1 . The same mass of gas in a different state is characterized by the parameters V 2 , p 2 , T 2 (see figure). The transition from state 1 to state 2 is carried out in the form of two processes: isothermal (1 - 1") and isochoric (1" - 2).

For these processes, one can write down the laws of Boyle - Mariotte and Gay - Lussac:

Eliminating p 1 " from the equations, we get

Since states 1 and 2 were chosen arbitrarily, the last equation can be written as:

This equation is called Clapeyron's equation , in which B is a constant, different for different masses of gases.

Mendeleev combined Clapeyron's equation with Avogadro's law. According to Avogadro's law, 1 mole of any ideal gas at the same p and T occupies the same volume V m, so the constant B will be the same for all gases. This common constant for all gases is denoted R and is called universal gas constant. Then

This equation is ideal gas equation of state , which is also called Clapeyron - Mendeleev equation .

The numerical value of the universal gas constant can be determined by substituting the values ​​of p, T and V m into the Clapeyron - Mendeleev equation under normal conditions:

The Clapeyron - Mendeleev equation can be written for any mass of gas. To do this, recall that the volume of a gas of mass m is related to the volume of one mole by the formula V \u003d (m / M) V m, where M is molar mass of gas. Then the Clapeyron - Mendeleev equation for a gas of mass m will look like:

where is the number of moles.

The equation of state for an ideal gas is often written in terms of Boltzmann's constant :

Based on this, the equation of state can be represented as

where is the concentration of molecules. From the last equation it can be seen that the pressure of an ideal gas is directly proportional to its temperature and concentration of molecules.

Small demo ideal gas laws. After pressing the button "Let's start" You will see the host's comments on what is happening on the screen (black color) and a description of the computer's actions after you press the button "Further"(Brown color). When the computer is "busy" (i.e., experience is in progress), this button is not active. Move on to the next frame only after understanding the result obtained in the current experiment. (If your perception does not match the host's comments, write!)

You can verify the validity of the ideal gas laws on the existing

It is derived on the basis of the combined law of Boyle-Mariotte and Gay-Lussac using Avogadro's law. For one gram-molecule of any substance in an ideal gaseous state, the Mendeleev-Clapeyron equation has the expression:

Or PV=RT (11) .

In the event that there is not one, but n moles of gas, the expression takes the form:

where R- universal gas constant, independent of the nature of the gas.

Since the number of gram-moles of gas, where m- mass of gas, and M- its molecular weight, then expression (12) takes the form:

The numerical value of R depends on the unit of pressure and volume. Its value is expressed in units of energy/mol'deg. To find numeric values R we use equation (11), applying it to 1 mole of an ideal gas under normal conditions,

Substituting into equation (11) the numerical values ​​P=1 atm, T= 273° and V=22.4 l, we obtain

In the international SI system of units, pressure is expressed in newtons per m 2 (N / m 2), and volume in m 3. Then .

Using the Mendeleev-Clapeyron equation, the following calculations can be made: a) finding the physical parameters of the gas state from its molecular weight and other data, b) finding the molecular weight of the gas from data on its physical state (see example 22).

Example 11. How much does nitrogen weigh in a gas tank with a diameter of 3.6 m and a height of 25 m at a temperature of 25ºС and a pressure of 747 mm Hg. Art.?

II example 12. In a flask with a capacity of 500 ml at 25ºС there is 0.615 g of nitric oxide (II). What is the gas pressure in atmospheres, in N / m 2?

Example 13 The mass of a flask with a capacity of 750 cm 3 filled with oxygen at 27°C is 83.35 g. The mass of an empty flask is 82.11 g. Determine the oxygen pressure and mm Hg. on the walls of the flask.

Dalton's Law

This law is formulated as follows: the total pressure of mixtures of gases that do not react with each other is equal to the sum of the partial pressures of the constituent parts (components).

P \u003d p 1 + p 2 + p 3 + ... .. + p n (14)

where P is the total pressure of the gas mixture; p 1 , p 2 , p 3 , …., p n are the partial pressures of the mixture components.

Partial pressure is the pressure exerted by each component of a gas mixture, if we imagine this component occupying a volume equal to the volume of the mixture at the same temperature. In other words, partial pressure is that part of the total pressure of a gas mixture, which is due to a given gas.

From Dalton's law it follows that in the presence of a mixture of gases P in equation (12) is the sum of the number of moles of all components that form a given mixture, and P is the total pressure of the mixture that occupies at a temperature T volume v.

The relationship between partial pressures and the total is expressed by the equations:

where n 1 , n 2 , n 3 is the number of moles of component 1, 2, 3, respectively, in a mixture of gases.

The ratios are called mole fractions of a given component.

If the mole fraction is denoted by N, then the partial pressure of any i-th component of the mixture (where i = 1,2,3,...) will be equal to:

Thus, the partial pressure of each component of the mixture is equal to the product of its mole fraction and the total pressure of the gas mixture.

In addition to the partial pressure in gas mixtures, the partial volume of each of the gases is distinguished v 1 , v 2 , v 3 etc.

The partial volume is called the volume that would be occupied by a separate ideal gas, which is part of an ideal mixture of gases, if, with the same amount, it had the pressure and temperature of the mixture.

The sum of the partial volumes of all components of the gas mixture is equal to the total volume of the mixture

V = v 1 ,+v2 + v 3 + ... + v n (16) .

The ratio, etc., is called the volume fraction of the first, second, etc. components of the gas mixture. For ideal gases, the mole fraction is equal to the volume fraction. Therefore, the partial pressure of each component of the mixture is also equal to the product of its volume fraction and the total pressure of the mixture.

; ; p i = r i´ P (17).

Partial pressure is usually found from the value of the total pressure, taking into account the composition of the gas mixture. The composition of the gas mixture is expressed in weight percent, volume percent and mole percent.

The volume percentage is the volume fraction increased by 100 times (the number of volume units of a given gas contained in 100 volume units of the mixture)

mole percent q called the mole fraction, increased by 100 times.

The weight percentage of a given gas is the number of mass units of it contained in 100 mass units of the gas mixture.

where m 1 , m 2 are the masses of the individual components of the gas mixture; m- the total mass of the mixture.

To switch from volume percent to weight percent, which is necessary in practical calculations, use the formula:

where r i (%) - volume percentage i-th gas mixture component; M i is the molecular weight of this gas; M cf - the average molecular weight of a mixture of gases, which is calculated by the formula

M cf = M 1 ´r 1 + M 2 ´r 2 + M 3 ´r 3 + ….. + M i ´r i (19)

where M 1 , M 2 , M 3 , M i are the molecular weights of individual gases.

If the composition of the gas mixture is expressed by the number of masses of individual components, then the average molecular weight of the mixture can be expressed by the formula

where G 1 , G 2 , G 3 , G i are the mass fractions of gases in the mixture: ; ; etc.

Example 14 5 liters of nitrogen at a pressure of 2 atm, 2 liters of oxygen at a pressure of 2.5 atm and 3 liters of carbon dioxide at a pressure of 5 atm are mixed, and the volume provided to the mixture is 15 liters. Calculate the pressure under which the mixture is and the partial pressures of each gas.

Nitrogen, which occupied a volume of 5 liters at a pressure P 1 = 2 atm, after mixing with other gases, spread in a volume V 2 = 15 liters. Partial pressure of nitrogen p N 2\u003d P 2 we find from the Boyle-Mariotte law (P 1 V 1 \u003d P 2 V 2). Where

The partial pressures of oxygen and carbon dioxide are found in a similar way:

The total pressure of the mixture is .

Example 15 A mixture consisting of 2 moles of hydrogen, some moles of oxygen and 1 mole of nitrogen at 20°C and a pressure of 4 atm occupies a volume of 40 liters. Calculate the number of moles of oxygen in the mixture and the partial pressures of each of the gases.

From the equation (12) Mendeleev-Clapeyron we find the total number of moles of all gases that make up the mixture

The number of moles of oxygen in the mixture is

The partial pressures of each of the gases are calculated using equations (15a):

Example 17. The composition of benzene hydrocarbon vapors over absorbent oil in benzene scrubbers, expressed in units of mass, is characterized by the following values: benzene C 6 H 6 - 73%, toluene C 6 H 5 CH 3 - 21%, xylene C 6 H 4 (CH 3) 2 - 4%, trimethylbenzene C 6 H 3 (CH 3) 3 - 2%. Calculate the content of each component by volume and the partial vapor pressures of each substance if the total pressure of the mixture is 200 mm Hg. Art.

To calculate the content of each component of the vapor mixture by volume, we use the formula (18)

Therefore, it is necessary to know M cf, which can be calculated from formula (20):

The partial pressures of each component in the mixture are calculated using equation (17)

p benzene= 0.7678´200 = 153.56 mmHg ; p toluene= 0.1875´200 = 37.50 mmHg ;

p xylene= 0.0310´200 = 6.20 mmHg ; p trimethylbenzene= 0.0137´200 = 2.74 mmHg

Similar information.

The ideal gas model is used to explain the properties of matter in the gaseous state.

Ideal gas name a gas for which the size of molecules and the forces of molecular interaction can be neglected; Collisions of molecules in such a gas occur according to the law of collision of elastic balls.

real gases behave like an ideal one when the average distance between the molecules is many times greater than their sizes, i.e., at sufficiently large rarefaction.

The state of the gas is described by three parameters V, P, T, between which there is an unambiguous relationship, called the Mendeleev-Clapeyron equation.

R - molar gas constant, determines the work that 1 mole of gas does when it is heated isobarically by 1 K.

This name of this equation is due to the fact that it was first obtained by D.I. Mendeleev (1874) on the basis of a generalization of the results previously obtained by the French scientist B.P. Clapeyron.

A number of important consequences follow from the equation of state of an ideal gas:

At the same temperatures and pressures, equal volumes of any ideal gases contain the same number of molecules(Avagadro's law).

The pressure of a mixture of chemically non-interacting ideal gases is equal to the sum of the partial pressures of these gases(Dalton's law ).

The ratio of the product of pressure and volume of an ideal gas to its absolute temperature is a constant value for a given mass of a given gas(combined gas law)

Any change in the state of a gas is called a thermodynamic process.

During the transition of a given mass of gas from one state to another, in the general case, all gas parameters can change: volume, pressure and temperature. However, sometimes any two of these parameters change, while the third remains unchanged. The processes in which one of the parameters of the state of the gas remains constant, while the other two change, are called isoprocesses .

§ 9.2.1Isothermal process (T=const). Boyle-Mariotte law.

The process that takes place in a gas in which the temperature remains constant is called isothermal ("izos" - "same"; "terme" - "warmth").

In practice, this process can be realized by slowly decreasing or increasing the volume of gas. With slow compression and expansion, conditions are created to maintain a constant gas temperature due to heat exchange with the environment.

If the volume V is increased at a constant temperature, the pressure P decreases; when the volume V decreases, the pressure P increases, and the product of P and V is preserved.

pV = const (9.11)

This law is called Boyle-Mariotte law, since it was opened almost simultaneously in the 17th century. French scientist E. Mariotte and English scientist R. Boyle.

Boyle-Mariotte law is formulated like this: The product of gas pressure and volume for a given mass of gas is a constant value:

The graphical dependence of the gas pressure P on the volume V is depicted as a curve (hyperbola), which is called isotherms(fig.9.8). Different temperatures correspond to different isotherms. The isotherm corresponding to the higher temperature lies above the isotherm corresponding to the lower temperature. And in the VT (volume - temperature) and PT (pressure - temperature) coordinates, the isotherms are straight lines perpendicular to the temperature axis (Fig.).

§ 9.2.2Isobaric process (P= const). Gay-Lussac's law

The process that takes place in a gas in which the pressure remains constant is called isobaric ("baros" - "gravity"). The simplest example of an isobaric process is the expansion of a heated gas in a cylinder with a free piston. The expansion of the gas observed in this case is called thermal expansion.

Experiments conducted in 1802 by the French physicist and chemist Gay-Lussac showed that The volume of gas of a given mass at constant pressure lhoarfrostincreases with temperature(Gay-Lussac's law) :

V = V 0 (1 + αt) (9.12)

The value α is called temperature coefficient of volume expansion(for all gases)

If we replace the temperature measured on the Celsius scale with the thermodynamic temperature, we get the Gay-Lussac law in the following formulation: at constant pressure, the ratio of the volume given by the mass of an ideal gas to its absolute temperature is a constant value, those.

Graphically, this dependence in the coordinates Vt is depicted as a straight line emerging from the point t=-273°C. This line is called isobar(Fig. 9.9). Different pressures correspond to different isobars. Since the volume of a gas decreases with increasing pressure at constant temperature, the isobar corresponding to a higher pressure lies below the isobar corresponding to a lower pressure. In PV and PT coordinates, isobars are straight lines perpendicular to the pressure axis. At low temperatures, close to the temperature of liquefaction (condensation) of gases, the Gay-Lussac law is not fulfilled, so the red line on the graph is replaced by a white one.

§ 9.2.3Isochoric process (V= const). Charles' law

The process that takes place in a gas, in which the volume remains constant, is called isochoric ("horema" - capacity). For the implementation of the isochoric process, the gas is placed in a hermetic vessel that does not change its volume.

The French physicist J. Charles established: the pressure of a gas of a given mass at constant volume increases linearly with increasingtemperature(Charles law):

Р = Р 0 (1 + γt) (9.14)

(p - gas pressure at temperature t, ° C; p 0 - its pressure at 0 ° C].

The quantity γ is called pressure temperature coefficient. Its value does not depend on the nature of the gas: for all gases.

If we replace the temperature measured on the Celsius scale with the thermodynamic temperature, we get Charles's law in the following formulation: at a constant volume, the ratio of the pressure of a given mass of an ideal gas to its absolute temperature is a constant value, those.

Graphically, this dependence in the coordinates Pt is depicted as a straight line coming out of the point t=-273°C. This line is called isochore(Fig. 9.10). Different volumes correspond to different isochores. Since with an increase in the volume of a gas at a constant temperature, its pressure decreases, the isochore corresponding to a larger volume lies below the isochore corresponding to a smaller volume. In PV and VT coordinates, isochores are straight lines that are perpendicular to the volume axis. In the region of low temperatures close to the temperature of liquefaction (condensation) of gases, Charles's law, as well as the Gay-Lussac law, is not fulfilled.

The unit of temperature on the thermodynamic scale is the kelvin (K); corresponds to 1°C.

The temperature measured on the thermodynamic temperature scale is called thermodynamic temperature. Since the melting point of ice at normal atmospheric pressure, taken as 0 ° C, is 273.16 K -1, then

The Mendeleev-Clapeyron equation is the equation of state for an ideal gas, referred to 1 mole of gas. In 1874, D. I. Mendeleev, based on the Clapeyron equation, combining it with Avogadro's law, using the molar volume V m and referring it to 1 mole, derived the equation of state for 1 mole of an ideal gas:

pV=RT, where R is the universal gas constant,

R = 8.31 J / (mol. K)

The Clapeyron-Mendeleev equation shows that for a given mass of gas, it is possible to simultaneously change three parameters characterizing the state of an ideal gas. For an arbitrary mass of gas M, the molar mass of which is m: pV = (M/m) . RT. or pV = N A kT,

where N A is Avogadro's number, k is Boltzmann's constant.

Derivation of the equation:

Using the equation of state of an ideal gas, one can investigate processes in which the mass of the gas and one of the parameters - pressure, volume or temperature - remain constant, and only the other two change, and theoretically obtain gas laws for these conditions for changing the state of the gas.

isobaric process- the process of changing the state of the system at constant pressure. For a gas of a given mass, the ratio of the volume of the gas to its temperature remains constant if the pressure of the gas does not change. This Gay-Lussac's law. According to Gay-Lussac's law, the volume of a gas is directly proportional to its temperature: V/T=const. Graphically, this dependence in V-T coordinates is depicted as a straight line coming out of the point Т=0. This line is called an isobar. Different pressures correspond to different isobars. Gay-Lussac's law is not respected at low temperatures close to the temperature of liquefaction (condensation) of gases.

Isochoric process- the process of changing the state of the system at a constant volume. For a given mass of gas, the ratio of the pressure of a gas to its temperature remains constant if the volume of the gas does not change. This is Charles' gas law. According to Charles' law, the pressure of a gas is directly proportional to its temperature: P/T=const. Graphically, this dependence in P-T coordinates is depicted as a straight line coming out of the point T=0. This line is called an isochore. Different volumes correspond to different isochores. Charles's law is not respected in the region of low temperatures, close to the temperature of liquefaction (condensation) of gases.

So, from the law pV \u003d (M / m) . RT derives the following laws:

T = const=> PV = const- Boyle's Law - Mariotte.

p = const => V/T = const- Gay-Lussac's law.

If an ideal gas is a mixture of several gases, then according to Dalton's law, the pressure of a mixture of ideal gases is equal to the sum of the partial pressures of its constituent gases. Partial pressure is the pressure that a gas would produce if it alone occupied the entire volume equal to the volume of the mixture.

Some may be interested in the question, how did you manage to determine the Avogadro constant N A \u003d 6.02 10 23? The value of the Avogadro number was experimentally established only at the end of the 19th - beginning of the 20th century. Let us describe one of these experiments.

In a vessel evacuated to a deep vacuum with a volume of V = 30 ml, a sample of the radium element weighing 0.5 g was placed and kept there for one year. It was known that 1 g of radium emitted 3.7 x 10 10 alpha particles per second. These particles are helium nuclei, which immediately accept electrons from the walls of the vessel and turn into helium atoms. During the year, the pressure in the vessel increased to 7.95·10 -4 atm (at a temperature of 27°C). The change in the mass of radium over the year can be neglected. So, what is N A equal to?

First, let's find how many alpha particles (that is, helium atoms) were formed in one year. Let's denote this number as N atoms:

N = 3.7 10 10 0.5 g 60 sec 60 min 24 hours 365 days = 5.83 10 17 atoms.

We write the Clapeyron-Mendeleev equation PV = n RT and note that the number of moles of helium n= N/N A . From here:

N A = NRT = 5,83 . 10 17 . 0,0821 . 300 = 6,02 . 10 23

PV 7.95. 10 -4 . 3 . 10-2

At the beginning of the 20th century, this method of determining the Avogadro constant was the most accurate. But why did the experiment last so long (within a year)? The fact is that radium is very difficult to extract. With its small amount (0.5 g), the radioactive decay of this element produces very little helium. And the less gas in a closed vessel, the less pressure it will create and the greater the measurement error. It is clear that an appreciable amount of helium can be formed from radium only over a sufficiently long time.

Mendeleev's Clapeyron equation originates from the French engineer Clapeyron B. who lived from 1799 to 1864. Since the parameters of the state of an ideal gas have a connection, he connected the available experimental laws of gases and revealed a connection in the parameters.

pW/T = const

And Mendeleev D.I. our Russian scientist, who lived from 1834 to 1907, combined it with Avogadro's law. From this law it follows that if P and T are the same, then a mole of any gas occupies an equal molar volume. Wm=22.4l. From which Mendeleev's conclusion follows - the constant value on the right side of the equation is the same for any gas. The designation is written as R, and is called the universal gas constant.

The digital expression R is calculated by substitution. Mendeleev's Clapeyron equation looks like:

PW=nRT

in him:
R- gas pressure, W- liter volume, T- temperature, measured in kelvins, n- number of moles, R- UGP.

For instance: Oxygen is in a container of 2.6 liters, under a pressure of 2.3 atm and 26 degrees C. It is not known how many moles of O 2 are contained in the container?

According to the gas law, we find how many moles n

n \u003d PW / RT from which: n \u003d (2.3 atm * 2.6 l) / (0.0821 l * atm / mol * K * 299K) \u003d 0.24 mol O 2

The temperature must be converted to kelvins (273 0 С + 26 0 С) = 299K. In order to avoid errors when solving equations, it is necessary to pay attention to the quantities in which data are given for Mendeleev-Clapeyron equations The pressure can be in mm Hg - we translate into atmospheres (1 atm \u003d 760 mm r / s). If in pascals when converting to atmospheres, it is important to remember that 101325 Pa = 1 atm.

If you make calculations where the units are in m 3 and Pa. Here you need to use R \u003d 8.314 J / K * mol (constant gas).

Let's look at an example:
Given: Helium volume 16.5 liters, temperature - 78 0 С, pressure 45.6 atm. What will be its volume under normal conditions? Number of moles? We can quickly find out how many moles n it contains using the Mendeleev-Clapeyron Equation, but what if the value of R is forgotten. Under normal conditions, 1 mole (1atm and 273K) fills 22.4 liters. I.e

PW \u003d nRT, it follows from this, R \u003d PW / nT \u003d (1 atm * 22.4 l) / (1 mol * 273 K) \u003d 0.082

If you make it so that R would be reduced. We get the following solution.
Initial data: P 1 \u003d 45.6 atm, W 1 \u003d 16.5 l, T 1 \u003d 351K.
Final data: P 2 \u003d 1atm, W 2 \u003d?, T 2 \u003d 273K.

We see that the equation is exactly true for both initial and final data
P 1 W 1 = nRT 1
P 2 W 2 = nRT 2

In order to find out the volume of gas, we divide the values ​​\u200b\u200bin the equation
P 1 W 1 / P 2 W 2 \u003d T 1 / T 2,
insert the values ​​we know
W 2 \u003d 45.6 * 16.5 * 273 / 351 \u003d 585 liters

This means that under normal conditions, the volume of helium will be 585 liters. We divide 585 by the molar gas volume in norms. conditions (22.4 l / * mol) we get how many moles in helium 585 / 22.4 \u003d 26.1 m.

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