Gay-Lussac's Law: At constant pressure, the volume of a gas changes in direct proportion to the absolute temperature.
Boyle-Marriott's Law: At constant temperature, the pressure produced by a given mass of gas is inversely proportional to the volume of the gas.
Gas laws
The study of the properties of gaseous substances and chemical reactions involving gases played such an important role in the development of atomic-molecular theory that gas laws deserve special consideration.
Experimental studies on the study of chemical reactions between gaseous substances led J.-L. Gay Lussac (1805) for opening law of volumetric relations: at constant temperature and pressure, the volumes of reacting gases relate to each other and to the volumes of gaseous reaction products as small integers . Thus, during the formation of hydrogen chloride from simple substances (H 2 + Cl 2 = 2HCl), the volumes of reacting and resulting substances relate to each other as 1: 1: 2, and during the synthesis of H 2 O from simple substances (2H 2 + O 2 = 2H 2 O) this ratio is 2:1:2.
These proportions are explained in Avogadro's law: equal volumes of different gases under the same conditions (temperature and pressure) contain an equal number of molecules. Molecules of simple gaseous substances, such as hydrogen, oxygen, chlorine, etc., consist of two atoms.
Avogadro's law has two important consequences:
The molecular mass (n.m.) of a gas or vapor (M 1) is equal to the product of its relative density (D) to any other gas by the molecular mass of the latter (M 2)
M 1 = D ∙ M 2;
D = M 1 / M 2 – the ratio of the mass of a given gas to the mass of another gas taken in the same volume, at the same temperature and the same pressure.
For example, nitrogen is 7 times heavier than helium, since the density of nitrogen relative to helium is:
D He (N 2) = M(N 2) / M(Not) = 28/4 =7
- moles of any gas under normal conditions (P 0 = 1 atm or 101.325 kPa or 760 mm Hg and temperature T 0 = 273.15 K or 0 ° C) occupies a volume of 22.4 liters.
The gaseous state of a substance of a given mass is characterized by three parameters: pressure R, volume V and temperature T. The following relationships were experimentally established between these quantities.
P 2 / P 1 = V 1 / V 2 , or PV= const.
V 1 / T 1 = V 2 / T 2 , or V/T= const.
P 1 / T 1 = R 2 / T 2 , or R/T= const.
These three laws can be combined into one universal gas law:
P 1 V 1 / T 1 = P 2 V 2 / T 2 , or РV/Т= const.
This equation was established by B. Clapeyron (1834). The value of the constant in the equation depends only on the amount of gas substance. The equation for one mole of gas was derived by D.I. Mendeleev (1874). For one mole of gas the constant is called universal gas constant and is designated R= 8.314 J/(mol∙TO)= 0.0821 l∙atm/(mol∙K)
РV=RT,
For an arbitrary amount of gas ν the right side of this equation must be multiplied by ν :
РV= νRT or РV= (t/M)RT ,
which is called Clapeyron-Mendeleev equation. This equation is valid for all gases in any quantities and for all values P, V And T, at which gases can be considered ideal.
Absolute temperature scale.
If we continue the isochore into the region of negative temperatures, then at the point of intersection with the x-axis we have
P = P0(1 + t) = 0. (21)
Hence the temperature at which the pressure of an ideal gas vanishes, t = –273°С(more precisely, –273.16°С). This temperature was chosen as the starting point for the thermodynamic temperature scale, which was proposed by the English scientist Kelvin. This temperature is called Kelvin zero (or absolute zero).
The temperature measured on the thermodynamic temperature scale is designated T. They call her thermodynamic temperature. Since the melting point of ice at normal atmospheric pressure, taken as 0 ° WITH, equal to 273.16 K –1 , That
T = 273.16 + t. (22)
Clayperon equation.
Let us obtain another form of equations describing isobaric and isochoric processes by replacing in equations (18) and (20) the temperature measured on the Celsius scale with thermodynamic temperature:
V = V0(1 + ·t) = V0() = V0
Designating gas volumes at temperatures T1 And T2, How V1 And V2, let's write
V1 = V0, V2 = V0.
Dividing these equalities term by term, we obtain the Gay-Lussac law in the form V1/V2 = T1/T2
= const. (23)
Similarly, we obtain a new form of Charles’s law:
Charles's and Gay-Lussac's laws can be combined into one general law connecting the parameters P, V And T at constant gas mass.
Indeed, let us assume that the initial state of the gas at m = const characterized by parameters V1, P1, T1, and the final one – accordingly V2, P2, T2. Let the transition from the initial state to the final state occur through two processes: isothermal and isobaric. During the first process, we change the pressure from P1 on P2. The volume that the gas will occupy after this transition is denoted by V, then according to the Boyle–Mariotte law, P1V1 = P2V, where
(25)
At the second stage, reduce the temperature from T1 before T2, and the volume will change from the value V before V2; therefore, according to Charles's law 
where
(26)
In equations (25) and (26), the left sides are equal; therefore the right ones are equal, then 
, or
, (27)
i.e. we can write that
. (28)
Expression (28) is called Clapeyron's equation or the unified gas law.
The equation of state of an ideal gas is the Mendeleev-Clapeyron equation.
The value of the constant included in equation (28), which is denoted as R, for one mole of any gas is the same, so this constant is called universal gas constant.
Let's find the numerical value R in SI, for which we take into account that, as follows from Avogadro’s law, one mole of any gas at the same pressure and the same temperature occupies the same volume. In particular when T0 = 273K and pressure P0 = 105 Pa the volume of one mole of gas is equal to V0 = 22.4·10–³ m³. Then R == 8.31 J/(mol K).
Equation (28) for one mole of gas can be written as
From equation (29) it is easy to obtain an equation for any mass of gas. Gas mass m will take up volume V = V0(m/M), Where M– mass 1 mol, m/M– number of moles of gas. Multiplying both sides of equation (29) by m/M, we get 
.
Because 
, then we finally get
. (30)
Equation (30) is called the Mendeleev–Clapeyron equation and is the basic equation relating the parameters of a gas in a state of thermal equilibrium. That's why they call him equation of state of an ideal gas.
2. Isochoric process. V is constant. P and T change. Gas obeys Charles's law . Pressure, at constant volume, is directly proportional to absolute temperature
3. Isothermal process. T is constant. P and V change. In this case, the gas obeys the Boyle-Mariotte law . The pressure of a given mass of gas at a constant temperature is inversely proportional to the volume of the gas.
4. From a large number of processes in gas, when all parameters change, we single out a process that obeys the unified gas law. For a given mass of gas, the product of pressure and volume divided by absolute temperature is a constant.
This law is applicable for a large number of processes in gas, when gas parameters do not change very quickly.
All of the listed laws for real gases are approximate. Errors increase with increasing gas pressure and density.
Work order:
1. part of the work.
1. Lower the glass ball hose into a vessel with water at room temperature (Fig. 1 in the appendix). Then we heat the ball (with our hands, with warm water). Assuming the gas pressure is constant, write how the volume of the gas depends on the temperature
Conclusion:………………..
2. Connect a cylindrical vessel with a millimanometer with a hose (Fig. 2). Let's heat the metal vessel and the air in it using a lighter. Assuming the gas volume is constant, write how the gas pressure depends on temperature.
Conclusion:………………..
3. Squeeze the cylindrical vessel connected to the millimanometer with your hands, reducing its volume (Fig. 3). Assuming the gas temperature is constant, write how the gas pressure depends on the volume.
Conclusion:……………….
4. Connect the pump to the ball chamber and pump in several portions of air (Fig. 4). How did the pressure, volume and temperature of the air pumped into the chamber change?
Conclusion:………………..
5. Pour about 2 cm 3 of alcohol into the bottle, close it with a stopper with a hose (Fig. 5) attached to the injection pump. Let's make a few pumps until the cork leaves the bottle. How do the pressure, volume and temperature of air (and alcohol vapor) change after the cork is removed?
Conclusion:………………..
Part of work.
Checking the Gay-Lussac law.
1. Take the heated glass tube out of the hot water and lower the open end into a small vessel with water.
2. Hold the handset vertically.
3. As the air in the tube cools, water from the vessel enters the tube (Figure 6).
4. Find and
Length of tube and air column (at the beginning of the experiment)
The volume of warm air in the tube,
The cross-sectional area of the tube.
The height of the column of water that entered the tube when the air in the tube cooled.
Length of the cold air column in the tube
The volume of cold air in the tube.
Based on Gay-Lussac's law, we have for two states of air
Or (2) (3)
Temperature of hot water in the bucket
Room temperature
We need to check equation (3) and therefore the Gay–Lussac law.
5. Let's calculate
6. Find the relative measurement error when measuring length, taking Dl=0.5 cm.
7. Find the absolute error of the ratio
=……………………..
8. Record the result of the reading
………..…..
9. Find the relative measurement error T, taking
10. Find the absolute calculation error
11. Write down the result of the calculation
12. If the interval for determining the temperature ratio (at least partially) coincides with the interval for determining the ratio of the lengths of air columns in the tube, then equation (2) is valid and the air in the tube obeys the Gay-Lussac law.
Conclusion:……………………………………………………………………………………………………
Report requirement:
1. Title and purpose of the work.
2. List of equipment.
3. Draw pictures from the application and draw conclusions for experiments 1, 2, 3, 4.
4. Write the content, purpose, calculations of the second part of the laboratory work.
5. Write a conclusion on the second part of the laboratory work.
6. Construct graphs of isoprocesses (for experiments 1,2,3) in the axes: ; ; .
7. Solve problems:
1. Determine the density of oxygen if its pressure is 152 kPa and the root mean square speed of its molecules is 545 m/s.
2. A certain mass of gas at a pressure of 126 kPa and a temperature of 295 K occupies a volume of 500 liters. Find the volume of gas under normal conditions.
3. Find the mass of carbon dioxide in a cylinder with a capacity of 40 liters at a temperature of 288 K and a pressure of 5.07 MPa.
Application
Let's consider how gas pressure depends on temperature when its mass and volume remain constant.
Let's take a closed vessel with gas and heat it (Fig. 4.2). We will determine the gas temperature using a thermometer, and the pressure using a pressure gauge M.
First, we will place the vessel in melting snow and designate the gas pressure at 0 ° C, and then we will gradually heat the outer vessel and record the values for the gas. It turns out that the graph of the dependence on, constructed on the basis of such experience, looks like a straight line (Fig. 4.3, a). If we continue this graph to the left, it will intersect with the x-axis at point A, corresponding to zero gas pressure.
From the similarity of triangles in Fig. 4.3, but you can write:
If we denote the constant by y, we get
In essence, the proportionality coefficient y in the experiments described should express the dependence of the change in gas pressure on its type.
The quantity characterizing the dependence of the change in gas pressure on its type during the process of temperature change at a constant volume and constant mass of the gas is called the temperature coefficient of pressure. The temperature coefficient of pressure shows by what part of the pressure of a gas taken at 0 ° C its pressure changes when heated by
Let us derive the unit of temperature coefficient y in SI:
By repeating the described experiment for different gases at different masses, it can be established that, within the experimental errors, point A for all graphs is obtained in the same place (Fig. 4.3, b). In this case, the length of the segment OA is equal to Thus, for all cases, the temperature at which the gas pressure should become zero is the same and is equal to and the temperature coefficient of pressure Note that the exact value of y is When solving problems, they usually use an approximate value of y equal to
From experiments, the value of y was first determined by the French physicist J. Charles, who in 1787 established the following law: the temperature coefficient of pressure does not depend on the type of gas and is equal to Note that this is true only for gases with low density and with small changes in temperature ; at high pressures or low temperatures, y depends on the type of gas. Only an ideal gas strictly obeys Charles's law.
The French physicist Charles discovered a law (in 1787) that expresses the dependence of the change in gas pressure on temperature at constant volume.
Experience shows that when a gas is heated at a constant volume, the gas pressure increases. A scalar quantity measured by the change in unit pressure of a gas taken at 0 0 C from a change in its temperature by 1 0 C is called the thermal pressure coefficient γ.
According to the definition, thermal pressure coefficient?
where p 0 is the gas pressure at 0°C, p- gas pressure after heating to t°. Let's do the following experiment (Fig. 13, a). Place vessel A in water with ice with taps 1 and 2 open. When the vessel:: and the air contained in it cool to 0°С, close valve 2. Initial state of air in the vessel: t° = 0°C, p 0 = 1 at. Without changing the volume of air, place the vessel in hot water. The air in the vessel heats up, its pressure increases at temperature t° 1 = 40°C it becomes p 1 = 1.15 at. Thermal pressure coefficient
Through more accurate experiments, having determined the thermal pressure coefficient for various gases, Charles discovered that at constant volume all gases have the same thermal pressure coefficient 
From the formula for the thermal pressure coefficient
We will replace t° = T-273°. Then
Replacing we get
hence, р = р 0 γТ.
If the gas pressure at temperature T 1 is designated p 1, and at temperature T 2 - p 2, That р 1 = γр 0 Т 1 And р 2 = γр 0 Т 2. Comparing the pressures, we obtain the formula for Charles’s law:
For a given mass of gas at constant volume, the gas pressure changes in direct proportion to the change in the absolute temperature of the gas. This is the formulation of Charles's law. The process of changing the state of a gas at a constant volume is called isochoric. The formula of Charles's law is the equation of the isochoric state of a gas. The higher the gas temperature, the greater the average kinetic energy of the molecules, and therefore, the greater their speed. In this regard, the number of impacts of molecules on the walls of the vessel increases, i.e. pressure. In Fig. 13, b shows a graph of Charles's law.
