Change in pressure at constant volume. 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. Storage and transportation of gases
Ideal gas law.
Experimental:
The main parameters of gas are temperature, pressure and volume. The volume of gas depends significantly on the pressure and temperature of the gas. Therefore, it is necessary to find the relationship between the volume, pressure and temperature of the gas. This ratio is called equation of state.
It was experimentally discovered that for a given amount of gas the following relation holds to a good approximation: at constant temperature, the volume of gas is inversely proportional to the pressure applied to it (Fig. 1):
V~1/P , at T=const.
For example, if the pressure acting on a gas doubles, the volume will decrease to half its original volume. This relationship is known as Boyle's law (1627-1691)-Mariotte (1620-1684), it can be written like this:
This means that when one of the quantities changes, the other will also change, and in such a way that their product remains constant.
The dependence of volume on temperature (Fig. 2) was discovered by J. Gay-Lussac. He discovered that at constant pressure, the volume of a given amount of gas is directly proportional to the temperature:
V~T, at Р =const.
The graph of this dependence passes through the origin of coordinates and, accordingly, at 0K its volume will become equal to zero, which obviously has no physical meaning. This has led to the assumption that -273 0 C is the minimum temperature that can be achieved.
The third gas law, known as Charles's law named after Jacques Charles (1746-1823). This law states: at constant volume, gas pressure is directly proportional to absolute temperature (Fig. 3):
P ~T, at V=const.
A well-known example of this law is an aerosol can that explodes in a fire. This occurs due to a sharp increase in temperature at a constant volume.
These three laws are experimental, valid in real gases only as long as the pressure and density are not very high and the temperature is not too close to the condensation temperature of the gas, so the word "law" is not very suitable for these properties of gases, but it has become generally accepted.
The gas laws of Boyle-Mariotte, Charles and Gay-Lussac can be combined into one more general relationship between volume, pressure and temperature, which is valid for a certain amount of gas:
This shows that when one of the quantities P, V or T changes, the other two quantities will also change. This expression turns into these three laws when one value is taken as constant.
Now we should take into account one more quantity, which until now we considered constant - the amount of this gas. It has been experimentally confirmed that: at constant temperature and pressure, the closed volume of a gas increases in direct proportion to the mass of this gas:
This dependence connects all the main quantities of gas. If we introduce a proportionality coefficient into this proportionality, we get equality. However, experiments show that this coefficient is different in different gases, so instead of mass m, the amount of substance n (number of moles) is introduced.
As a result we get:
Where n is the number of moles, and R is the proportionality coefficient. The quantity R is called universal gas constant. To date, the most accurate value of this value is:
R=8.31441 ± 0.00026 J/mol
Equality (1) is called equation of state of an ideal gas or ideal gas law.
Avogadro's number; ideal gas law at the molecular level:
That the constant R has the same value for all gases is a magnificent reflection of the simplicity of nature. This was first realized, albeit in a slightly different form, by the Italian Amedeo Avogadro (1776-1856). He experimentally found that Equal volumes of gas at the same pressure and temperature contain the same number of molecules. Firstly: from equation (1) it is clear that if different gases contain an equal number of moles, have the same pressures and temperatures, then, provided R is constant, they occupy equal volumes. Secondly: the number of molecules in one mole is the same for all gases, which directly follows from the definition of a mole. Therefore, we can say that the value of R is constant for all gases.
The number of molecules in one mole is called Avogadro's numberN A. It is currently established that Avogadro's number is equal to:
N A =(6.022045 ± 0.000031) 10 -23 mol -1
Since the total number of molecules N of a gas is equal to the number of molecules in one mole multiplied by the number of moles (N = nN A), the ideal gas law can be rewritten as follows:
Where k is called Boltzmann constant and has the same value:
k= R/N A =(1.380662 ± 0.000044) 10 -23 J/K
Directory of compressor equipment
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.
According to Boyle's law V1: V2 = P2: P1 at constant temperature
According to Gay-Lusac's law V1: V2 = T1: T2 at constant pressure
P1: P2 = T1: T2 at constant volume
From the formulas presented above, you can see that two of the three quantities can be considered as variables if the third is constant. There is no state in which pressure, volume and temperature can all be considered variables.
However, there are cases when all quantities are variable and one factor is unknown. In practical cases, such problems can be solved by analogy with the examples below:
Gas at a temperature of 20 o C occupies a volume of 0.98 m 3 in a cylinder with a diameter of 50 mm; a force of 980 N is applied to the piston. What will be the displacement of the piston if the force applied to the piston is doubled and the temperature is increased to 50 o C?
Piston displacement is easy to determine by specifying volume changes. However, in the problem only one volume value is specified (0.98 m 3), and the other is unknown.
To establish the relationships between all parameters that are variable, volume changes must be considered separately in the two phases.
Case A 1st phase
The gas is heated from a temperature t = 20 o C, which corresponds to the absolute temperature T1 = 20 + 273 = 293 o K, to a temperature of 50 o C, which corresponds to T2 = (50 + 273) = 323 o K. If the pressure on the piston remains constant with a load of 980N, an increase in gas volume will occur. According to Gay-Lusac's law V1: V2 = T1: T2
Vх = (0.98 323)/293 = 1.08 dm 3 (intermediate value)
2nd phase
The gas, having reached a volume Vx = 1.08 dm 3 as a result of an increase in temperature to T2 (323 o K), now receives an additional effect - the force applied to the piston has increased. As a result, it increases to P2 = 980 2 = 1960 N, and the volume decreases as the air is compressed by the piston. According to Boyle's law Vx: V2 = P2: P1 (Vx P1 = V2 P2)
Substituting the given values:
V2 = (1.08 980)/1960 = 0.54 dm 3 (final value)
Note that the parameters P1 and P2 were presented as symbols of applied force rather than units of pressure. This is not an error as force relates directly to pressure in this example since the diameter of the piston does not change.
This is confirmed by the following calculations.
I. Piston surface area in cm2 (3.14 D2)/4
Diameter = 50 mm = 5 cm S = (3.14 52)/4 = 19.6 cm 2
The pressure at each stage can now be calculated.
II. Initial pressure P1=Initial force/Surface area = 980N/19.6cm2 = 50N/cm2 =5kg/cm2
Final pressure P2= Final force/Surface area = (980 2)/19.6 =100N/cm=10kg/cm2
If the surface areas of the piston are equal, doubling the applied force will double the pressure.
Substituting the given values:
Vх P1 = V2 P2
V2 = (1.08 dm 3 50 N/cm 2)/100 N/cm 2 = (1.08 dm 3 5 kg/cm 2)/10 kg/cm 2 = 0.54 dm 3
The same result was obtained in the previous calculation.
You can get the result by directly using the following expression, which is a combination of two initial formulas:
(P 1 V1)/T1 = (P2 V2)/T2
In the example, volume V2 is required in order to calculate the displacement of the piston
V2 = (P1 V1 T2)/(T1 P2) = (5 0.98 323)/(293 10) = 0.54 dm 2
Using both volumes, the change in piston position can be calculated using geometry:
Volume = surface area height Height in cm = volume in cm 2 / area in cm 2
Initial height = 980cm 3 /19.6cm 2 =50cm. Final height = 540cm 3 /19.6cm 2 =27.5cm
Piston displacement = 50-27.5 = 22.5 cm In this problem, it was assumed that the heating of the gas occurred as a result of an increase in the temperature of the external environment.
If you remember the experiment with a bicycle pump, when the air is compressed and does not have the opportunity to expand, heat is released, that is, the air temperature increases and this heat is transferred to the outer surfaces of the pump. The reverse process occurs when the gas expands.
If the gas has the opportunity to expand, its temperature will decrease.
Changes in air temperature give rise to:
I. The emergence of heat during the compression stage.
II. Heat absorption during the expansion stage.
Temperature changes can be calculated as shown using the values from the previous example.
The amount of gas at a temperature of 293°K occupies a volume V1 = 0.98 dm 3 at a pressure of 5 bar. If the pressure is increased to 10 bar, the volume decreases to V2 = 0.54 dm 3.
What will the temperature of the gas be? It is important to remember that Boyle's law only works when the temperature is constant. Therefore, at 293°K, an increase in pressure from P1 to P2 leads to a decrease in gas volume from V1 to Vx: V1: Vx = P2: P1 that is. V1 P1 = Vx P2
Substituting known values: Vx = (0.98 5)/10 = 0.49 dm 3
Using the Gay-Lusac law and considering pressure as a constant value P2 (to which the volume Vx is already assigned), we can write:
Vx: V2 = T1: T2 that is, Vx T2 = V2 T1
Substituting the known values: T2 = (0.54 293)/0.49 = 323°K This value is equal to the value given in the initial example.
Topics of the Unified State Examination codifier: isoprocesses - isothermal, isochoric, isobaric processes.
Throughout this paper we will stick to the following assumption: the mass and chemical composition of the gas remain unchanged. In other words, we believe that:
That is, there is no gas leakage from the vessel or, conversely, gas inflow into the vessel;
That is, the gas particles do not experience any changes (say, there is no dissociation - the breakdown of molecules into atoms).
These two conditions are satisfied in many physically interesting situations (for example, in simple models of heat engines) and therefore deserve separate consideration.
If the mass of a gas and its molar mass are fixed, then the state of the gas is determined three macroscopic parameters: pressure, volume And temperature. These parameters are related to each other by the equation of state (Mendeleev-Clapeyron equation).
Thermodynamic process(or simply process) is a change in the state of a gas over time. During the thermodynamic process, the values of macroscopic parameters - pressure, volume and temperature - change.
Of particular interest are isoprocesses- thermodynamic processes in which the value of one of the macroscopic parameters remains unchanged. By fixing each of the three parameters in turn, we obtain three types of isoprocesses.
1. Isothermal process runs at a constant gas temperature: .
2. Isobaric process runs at constant gas pressure: .
3. Isochoric process occurs at a constant volume of gas: .
Isoprocesses are described by very simple laws of Boyle - Mariotte, Gay-Lussac and Charles. Let's move on to studying them.
Isothermal process
Let an ideal gas undergo an isothermal process at temperature . During the process, only the gas pressure and its volume change.
Let us consider two arbitrary states of the gas: in one of them the values of macroscopic parameters are equal, and in the second - . These values are related by the Mendeleev-Clapeyron equation:
As we said from the beginning, mass and molar mass are assumed to be constant.
Therefore, the right sides of the written equations are equal. Therefore, the left sides are also equal:
(1)
Since the two states of the gas were chosen arbitrarily, we can conclude that During an isothermal process, the product of gas pressure and its volume remains constant:
(2)
This statement is called Boyle-Mariotte law.
Having written the Boyle-Mariotte law in the form
(3)
You can also give this formulation: in an isothermal process, the gas pressure is inversely proportional to its volume. If, for example, during isothermal expansion of a gas its volume increases three times, then the gas pressure decreases three times.
How to explain the inverse relationship between pressure and volume from a physical point of view? At a constant temperature, the average kinetic energy of gas molecules remains unchanged, that is, simply put, the force of impacts of molecules on the walls of the vessel does not change. As the volume increases, the concentration of molecules decreases, and accordingly the number of impacts of molecules per unit time per unit wall area decreases - the gas pressure drops. On the contrary, as the volume decreases, the concentration of molecules increases, their impacts occur more frequently and the gas pressure increases.
Isothermal process graphs
In general, graphs of thermodynamic processes are usually depicted in the following coordinate systems:
-diagram: abscissa axis, ordinate axis;
-diagram: abscissa axis, ordinate axis.
The graph of an isothermal process is called isotherm.
An isotherm on an -diagram is a graph of an inversely proportional relationship.
Such a graph is a hyperbola (remember algebra - the graph of a function). The hyperbola isotherm is shown in Fig. 1 .
Rice. 1. Isotherm on -diagram
Each isotherm corresponds to a certain fixed temperature value. It turns out that the higher the temperature, the higher the corresponding isotherm lies on -diagram.
In fact, let us consider two isothermal processes performed by the same gas (Fig. 2). The first process occurs at temperature, the second - at temperature.
Rice. 2. The higher the temperature, the higher the isotherm
We fix a certain volume value. On the first isotherm it corresponds to pressure, on the second - class="tex" alt="p_2 > p_1"> . Но при фиксированном объёме давление тем больше, чем выше температура (молекулы начинают сильнее бить по стенкам). Значит, class="tex" alt="T_2 > T_1"> .!}
In the remaining two coordinate systems, the isotherm looks very simple: it is a straight line perpendicular to the axis (Fig. 3):
Rice. 3. Isotherms on and -diagrams
Isobaric process
Let us recall once again that an isobaric process is a process taking place at constant pressure. During the isobaric process, only the volume of the gas and its temperature change.
A typical example of an isobaric process: gas is located under a massive piston that can move freely. If the mass of the piston and the cross section of the piston are , then the gas pressure is constant all the time and equal to
where is atmospheric pressure.
Let an ideal gas undergo an isobaric process at pressure . Consider again two arbitrary states of the gas; this time the values of the macroscopic parameters will be equal to and .
Let us write down the equations of state:
Dividing them by each other, we get:
In principle, this could already be enough, but we will go a little further. Let's rewrite the resulting relationship so that in one part only the parameters of the first state appear, and in the other part - only the parameters of the second state (in other words, we “spread the indices” across different parts):
(4)
And from here now - due to the arbitrariness of the choice of states! - we get Gay-Lussac's law:
(5)
In other words, at constant gas pressure, its volume is directly proportional to temperature:
(6)
Why does volume increase with increasing temperature? As the temperature rises, the molecules begin to beat harder and lift the piston. At the same time, the concentration of molecules drops, the impacts become less frequent, so that in the end the pressure remains the same.
Isobaric process graphs
The graph of an isobaric process is called isobar. On the -diagram, the isobar is a straight line (Fig. 4):
Rice. 4. Isobar on the -diagram
The dotted section of the graph means that in the case of a real gas at sufficiently low temperatures, the ideal gas model (and with it the Gay-Lussac law) stops working. In fact, as the temperature decreases, gas particles move more and more slowly, and the forces of intermolecular interaction have an increasingly significant influence on their movement (analogy: a slow ball is easier to catch than a fast one). Well, at very low temperatures, gases completely turn into liquids.
Let us now understand how the position of the isobar changes with pressure changes. It turns out that the greater the pressure, the lower the isobar goes on -diagram.
To verify this, consider two isobars with pressures and (Fig. 5):
Rice. 5. The lower the isobar, the greater the pressure
Let us fix a certain temperature value. We see that . But at a fixed temperature, the greater the pressure, the smaller the volume (Boyle-Mariotte law!).
Therefore, class="tex" alt="p_2 > p_1"> .!}
In the remaining two coordinate systems, the isobar is a straight line perpendicular to the axis (Fig. 6):
Rice. 6. Isobars on and -diagrams
Isochoric process
An isochoric process, recall, is a process that takes place at a constant volume. In an isochoric process, only the gas pressure and its temperature change.
It is very simple to imagine an isochoric process: it is a process taking place in a rigid vessel of a fixed volume (or in a cylinder under a piston when the piston is fixed).
Let an ideal gas undergo an isochoric process in a vessel with a volume of . Again, consider two arbitrary gas states with parameters and . We have:
Divide these equations by each other:
As in the derivation of Gay-Lussac’s law, we “split” the indices into different parts:
(7)
Due to the arbitrariness of the choice of states, we arrive at Charles' law:
(8)
In other words, at a constant volume of gas, its pressure is directly proportional to temperature:
(9)
An increase in the pressure of a gas of a fixed volume when it is heated is a completely obvious thing from a physical point of view. You can easily explain this yourself.
Graphs of an isochoric process
The graph of an isochoric process is called isochore. On the -diagram, the isochore is a straight line (Fig. 7):
Rice. 7. Isochore on the -diagram
The meaning of the dotted section is the same: the inadequacy of the ideal gas model at low temperatures.
Rice. 8. The lower the isochore, the greater the volume
The proof is similar to the previous one. We fix the temperature and see that . But at a fixed temperature, the lower the pressure, the larger the volume (again, the Boyle-Mariotte law). Therefore, class="tex" alt="V_2 > V_1"> .!}
In the remaining two coordinate systems, an isochore is a straight line perpendicular to the axis (Fig. 9):
Rice. 9. Isochores on and -diagrams
Boyle's laws - Mariotte, Gay-Lussac and Charles's laws are also called gas laws.
We derived gas laws from the Mendeleev-Clapeyron equation. But historically, everything was the other way around: gas laws were established experimentally, and much earlier. The equation of state subsequently appeared as their generalization.
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°C, close tap 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 precise 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.
