1. The structure of gaseous, liquid and solid bodies
The molecular kinetic theory makes it possible to understand why a substance can be in gaseous, liquid and solid states.
Gases. In gases, the distance between atoms or molecules is on average many times more sizes the molecules themselves ( fig.8.5). For example, at atmospheric pressure, the volume of a vessel is tens of thousands of times greater than the volume of molecules contained in it.
Gases are easily compressed, while the average distance between molecules decreases, but the shape of the molecule does not change ( fig.8.6).
Molecules with huge speeds - hundreds of meters per second - move in space. Colliding, they bounce off each other in different directions like billiard balls. Weak forces of attraction of gas molecules are not able to keep them near each other. That's why gases can expand indefinitely. They retain neither shape nor volume.
Numerous impacts of molecules on the walls of the vessel create gas pressure.
Liquids. Molecules of a liquid are located almost close to each other ( fig.8.7), so a liquid molecule behaves differently than a gas molecule. In liquids, there is the so-called short-range order, i.e., the ordered arrangement of molecules is preserved at distances equal to several molecular diameters. The molecule oscillates around its equilibrium position, colliding with neighboring molecules. Only from time to time does it make another "jump", falling into a new position of equilibrium. In this equilibrium position, the repulsive force is equal to the attractive force, i.e., the total interaction force of the molecule is zero. Time settled life water molecules, i.e., the time of its oscillations around one specific equilibrium position at room temperature, is on average 10 -11 s. The time of one oscillation is much less (10 -12 -10 -13 s). As the temperature rises, the time of the settled life of molecules decreases.
The nature of molecular motion in liquids, first established by the Soviet physicist Ya.I. Frenkel, makes it possible to understand the basic properties of liquids.
Liquid molecules are located directly next to each other. With a decrease in volume, the repulsive forces become very large. This explains low compressibility of liquids.
As is known, liquids are fluid, i.e. do not retain their shape. It can be explained like this. The external force does not noticeably change the number of molecular jumps per second. But jumps of molecules from one settled position to another occur mainly in the direction of the external force ( fig.8.8). That is why the liquid flows and takes the form of a vessel.
Solids. Atoms or molecules of solids, unlike atoms and molecules of liquids, vibrate around certain equilibrium positions. For this reason, solids retain not only volume but also shape. Potential energy of interaction of molecules solid body much greater than their kinetic energy.
There is another important difference between liquids and solids. A liquid can be compared to a crowd of people, where individual individuals are restlessly jostling in place, and a solid body is like a slender cohort of the same individuals who, although they do not stand at attention, maintain on average certain distances between themselves. If we connect the centers of equilibrium positions of atoms or ions of a solid body, then we get the correct spatial lattice, called crystalline.
Figures 8.9 and 8.10 show the crystal lattices of table salt and diamond. The internal order in the arrangement of crystal atoms leads to regular external geometric shapes.
Figure 8.11 shows Yakutian diamonds.
For a gas, the distance l between molecules is much larger than the dimensions of the molecules r 0:" l>>r 0 .
Liquids and solids have l≈r 0 . The molecules of a liquid are arranged in disorder and from time to time jump from one settled position to another.
In crystalline solids, molecules (or atoms) are arranged in a strictly ordered manner.
2. Ideal gas in molecular kinetic theory
The study of any field of physics always begins with the introduction of a certain model, within the framework of which the study is carried out in the future. For example, when we studied kinematics, the model of the body was a material point, etc. As you may have guessed, the model will never correspond to the actual processes taking place, but often it comes very close to this correspondence.
Molecular physics, and in particular the MKT, is no exception. Many scientists have worked on the problem of describing the model since the eighteenth century: M. Lomonosov, D. Joule, R. Clausius (Fig. 1). The latter, in fact, introduced the ideal gas model in 1857. A qualitative explanation of the basic properties of matter on the basis of molecular kinetic theory is not particularly difficult. However, the theory that establishes quantitative relationships between experimentally measured quantities (pressure, temperature, etc.) and the properties of the molecules themselves, their number and speed of movement, is very complex. In a gas at ordinary pressures, the distance between the molecules is many times greater than their size. In this case, the interaction forces of molecules are negligible and kinetic energy many more molecules potential energy interactions. Gas molecules can be considered as material points or very small hard balls. Instead of real gas, between whose molecules act complex forces interaction, we will consider it model is an ideal gas.
Ideal gas– gas model, in which gas molecules and atoms are represented as very small (disappearing sizes) elastic balls that do not interact with each other (without direct contact), but only collide (see Fig. 2).
It should be noted that rarefied hydrogen (under very low pressure) almost completely satisfies the ideal gas model.
Rice. 2.
Ideal gas is a gas, the interaction between the molecules of which is negligible. Naturally, when molecules of an ideal gas collide, a repulsive force acts on them. Since, according to the model, we can consider gas molecules as material points, we neglect the sizes of molecules, assuming that the volume they occupy is much less than the volume of the vessel.
Recall that in a physical model, only those properties of a real system are taken into account, the consideration of which is absolutely necessary to explain the studied patterns of behavior of this system. No model can convey all the properties of the system. Now we have to solve a rather narrow problem: to calculate, using the molecular-kinetic theory, the pressure of an ideal gas on the walls of a vessel. For this problem, the ideal gas model turns out to be quite satisfactory. It leads to results that are confirmed by experience.
3. Gas pressure in molecular kinetic theory
Let the gas be in a closed vessel. Manometer shows gas pressure p0. How does this pressure arise?
Each gas molecule, hitting the wall, acts on it with a certain force for a short period of time. As a result of random impacts on the wall, the pressure changes rapidly with time, approximately as shown in Figure 8.12. However, the effects caused by the impacts of individual molecules are so weak that they are not recorded by the manometer. The manometer records the time-average force acting on each unit area of its surface. sensing element- membranes. Despite small pressure changes, the average pressure p0 in practice, it turns out to be a quite definite value, since there are a lot of impacts on the wall, and the masses of the molecules are very small.
An ideal gas is a model of a real gas. According to this model, gas molecules can be considered as material points, the interaction of which occurs only when they collide. Colliding with the wall, gas molecules exert pressure on it.
4. Micro- and macro-parameters of gas
Now we can begin to describe the parameters of an ideal gas. They are divided into two groups:
Ideal gas parameters
That is, microparameters describe the state of a single particle (microbody), and macroparameters describe the state of the entire gas portion (macrobody). Let us now write the relation connecting some parameters with others, or the basic equation of the MKT:
Here: - the average speed of particles;
Definition. - concentration gas particles - the number of particles per unit volume; ; unit - .
5. Mean value of the squared velocity of molecules
To calculate the average pressure, you need to know the average speed of the molecules (more precisely, the average value of the square of the speed). This is not an easy question. You are used to the fact that each particle has speed. The average speed of molecules depends on the motion of all particles.
Average values. From the very beginning, one must give up trying to follow the movement of all the molecules that make up the gas. There are too many of them, and they move very difficult. We don't need to know how each molecule moves. We must find out what result the movement of all gas molecules leads to.
The nature of the motion of the entire set of gas molecules is known from experience. Molecules participate in random (thermal) motion. This means that the speed of any molecule can be either very large or very small. The direction of movement of molecules constantly changes when they collide with each other.
The velocities of individual molecules can be anything, however the average the value of the modulus of these speeds is quite definite. Similarly, the height of the students in the class is not the same, but its average value is a certain number. To find this number, you need to add the height of individual students and divide this amount by the number of students.
The average value of the square of the speed. In the future, we will need the average value of not the speed itself, but the square of the speed. The average kinetic energy of molecules depends on this value. And the average kinetic energy of molecules, as we will soon see, is of great importance in the entire molecular-kinetic theory.
Let us denote the velocity moduli of individual gas molecules as . The average value of the square of the speed is determined by the following formula:
where N is the number of molecules in the gas.
But the square of the modulus of any vector is equal to the sum of the squares of its projections on the coordinate axes OH, OY, OZ. That's why
The average values of the quantities can be determined using formulas similar to formula (8.9). Between the average value and the average values of the squares of the projections, there is the same relationship as the ratio (8.10):
Indeed, equality (8.10) is valid for each molecule. Adding such equalities for individual molecules and dividing both sides of the resulting equation by the number of molecules N, we arrive at formula (8.11).
Attention! Since the directions of the three axes OH, OY And oz due to the random motion of the molecules, they are equal, the average values of the squares of the velocity projections are equal to each other:
You see, a certain regularity emerges from the chaos. Could you figure it out yourself?
Taking into account relation (8.12), we substitute into formula (8.11) instead of and . Then for the mean square of the velocity projection we get:
i.e. the mean square of the velocity projection is equal to 1/3 of the mean square of the velocity itself. The factor 1/3 appears due to the three-dimensionality of space and, accordingly, the existence of three projections for any vector.
The velocities of molecules vary randomly, but the mean square of the speed is a well-defined value.
6. Basic equation of molecular-kinetic theory
We proceed to the derivation of the basic equation of the molecular-kinetic theory of gases. This equation establishes the dependence of gas pressure on the average kinetic energy of its molecules. After the derivation of this equation in the XIX century. and experimental proof of its validity began fast development quantitative theory, continuing to this day.
The proof of almost any statement in physics, the derivation of any equation can be done with varying degrees of rigor and persuasiveness: very simplified, more or less rigorous, or with the full rigor available modern science.
A rigorous derivation of the equation of the molecular-kinetic theory of gases is rather complicated. Therefore, we restrict ourselves to a highly simplified, schematic derivation of the equation. Despite all the simplifications, the result will be correct.
Derivation of the main equation. Calculate the gas pressure on the wall CD vessel ABCD area S, perpendicular to the coordinate axis OX (fig.8.13).
When a molecule hits a wall, its momentum changes: . Since the modulus of the velocity of molecules does not change upon impact, then 
. According to Newton's second law, the change in the momentum of a molecule is equal to the momentum of the force acting on it from the side of the vessel wall, and according to Newton's third law, the momentum of the force with which the molecule acted on the wall is the same in absolute value. Consequently, as a result of the impact of the molecule, a force acted on the wall, the momentum of which is equal to .
What is the average distance between the molecules of saturated water vapor at 100°C?
Task No. 4.1.65 from the "Collection of tasks for preparing for entrance exams in physics at USPTU"
Given:
\(t=100^\circ\) C, \(l-?\)
The solution of the problem:
Consider water vapor in some arbitrary quantity equal to \(\nu\) mol. To determine the volume \ (V \) occupied by a given amount of water vapor, you need to use the Clapeyron-Mendeleev equation:
In this formula, \(R\) is the universal gas constant, equal to 8.31 J/(mol·K). The pressure of saturated water vapor \(p\) at a temperature of 100 ° C is 100 kPa, this known fact and every student should know.
To determine the number of water vapor molecules \(N\), we use the following formula:
Here \(N_A\) is Avogadro's number, equal to 6.023 10 23 1/mol.
Then for each molecule there is a cube of volume \(V_0\), obviously determined by the formula:
\[(V_0) = \frac(V)(N)\]
\[(V_0) = \frac((\nu RT))((p\nu (N_A))) = \frac((RT))((p(N_A)))\]
Now look at the diagram for the problem. Each molecule is conventionally located in its own cube, the distance between two molecules can vary from 0 to \(2d\), where \(d\) is the length of the edge of the cube. The average distance \(l\) will be equal to the length of the edge of the cube \(d\):
The edge length \(d\) can be found like this:
As a result, we get the following formula:
Let's convert the temperature to the Kelvin scale and calculate the answer:
Answer: 3.72 nm.
If you do not understand the solution and you have some question or you find an error, then feel free to leave a comment below.
Molecules are very small, ordinary molecules cannot be seen even with the strongest optical microscope - but some parameters of molecules can be calculated quite accurately (mass), and some can only be estimated very roughly (size, speed), and it would be nice to understand what “size” is. molecules” and what kind of “molecular speed” we are talking about. So, the mass of a molecule is found as "the mass of one mole" / "the number of molecules in a mole." For example, for a water molecule, m = 0.018/6 1023 = 3 10-26 kg (it can be calculated more precisely - Avogadro's number is known with good accuracy, and the molar mass of any molecule is easy to find).
Estimating the size of a molecule begins with the question of what is considered its size. If only she were a perfectly polished cube! However, it is neither a cube nor a ball, and in general it does not have clearly defined boundaries. How to be in such cases? Let's start from afar. Let's estimate the size of a much more familiar object - a schoolboy. We have all seen the schoolchildren, we will take the mass of the average student equal to 60 kg (and then we will see if this choice greatly affects the result), the density of the schoolchild is about the same as that of water (remember that it is worth taking a deep breath of air, and after that you can “hang” in water, immersed almost completely, and if you exhale, you immediately begin to sink). Now you can find the volume of the student: V \u003d 60/1000 \u003d 0.06 cubic meters. meters. If we now assume that the student has the shape of a cube, then its size is found as the cube root of the volume, i.e. about 0.4 m. This is how the size turned out - less than growth (size "in height"), more thickness (size "in depth"). If we don’t know anything about the shape of the student’s body, then we won’t find anything better than this answer (instead of a cube, you could take a ball, but the answer would be about the same, and it’s more difficult to calculate the diameter of the ball than the edge of the cube). But if we have additional information (from the analysis of photographs, for example), then the answer can be made much more reasonable. Let it become known that the “width” of a schoolchild is, on average, four times less than his height, and his “depth” is three times less. Then H * H / 4 * H / 12 \u003d V, hence H \u003d 1.5 m (it makes no sense to make a more accurate calculation of such a poorly defined value, it is simply illiterate to focus on the capabilities of the calculator in such a “calculation”!). We got a quite reasonable estimate of the height of a schoolchild, if we took a mass of about 100 kg (and there are such schoolchildren!), We get about 1.7 - 1.8 m - also quite reasonable.
Let us now estimate the size of a water molecule. Let's find the volume that falls on one molecule in "liquid water" - in it the molecules are most densely packed (they are more strongly pressed to each other than in a solid, "ice" state). A mole of water has a mass of 18 g and a volume of 18 cu. centimeters. Then one molecule accounts for the volume V= 18 10-6/6 1023 = 3 10-29 m3. If we do not have information about the shape of the water molecule (or - if we do not want to take into account the complex shape of the molecules), the easiest way is to consider it a cube and find the size exactly as we just found the size of a cubic schoolboy: d = (V) 1/3 = 3 10-10 m. That's it! You can evaluate the influence of the shape of rather complex molecules on the result of the calculation, for example, as follows: calculate the size of gasoline molecules, considering the molecules as cubes - and then conduct an experiment by looking at the area of the spot from a drop of gasoline on the water surface. Considering the film as a "liquid surface one molecule thick" and knowing the mass of the drop, we can compare the sizes obtained by these two methods. A very instructive result!
The idea used is also suitable for a completely different calculation. Let us estimate the average distance between neighboring rarefied gas molecules for a specific case - nitrogen at a pressure of 1 atm and a temperature of 300K. To do this, we find the volume that in this gas falls on one molecule, and then everything will turn out simply. So, let's take a mole of nitrogen under these conditions and find the volume of the portion indicated in the condition, and then divide this volume by the number of molecules: V = R T / P NA = 8.3 300/105 6 1023 = 4 10 -26 m3. We will assume that the volume is divided into densely packed cubic cells, and each molecule “on average” sits in the center of its cell. Then the average distance between neighboring (nearest) molecules is equal to the edge of a cubic cell: d = (V)1/3 = 3 10-9 m. occupy a rather small - approximately 1/1000 part - of the volume of the vessel. In this case, too, we carried out the calculation very approximately - it makes no sense to calculate such not too definite values as "the average distance between neighboring molecules" more accurately.
Gas laws and foundations of the MKT.
If the gas is sufficiently rarefied (and this is a common thing, we most often have to deal with rarefied gases), then almost any calculation is done using a formula that relates pressure P, volume V, amount of gas ν and temperature T - this is the famous “equation state of an ideal gas» P·V= ν·R·T. How to find one of these quantities, if all the others are given, is quite simple and understandable. But it is possible to formulate the problem in such a way that the question will be about some other quantity - for example, about the density of a gas. So, the task is to find the density of nitrogen at a temperature of 300K and a pressure of 0.2 atm. Let's solve it. Judging by the condition, the gas is rather rarefied (air, consisting of 80% of nitrogen and at a much higher pressure, can be considered rarefied, we breathe it freely and easily pass through it), and if this were not the case, we would still have other formulas no - use this, beloved. The condition does not specify the volume of any portion of the gas, we will set it ourselves. Let's take 1 cubic meter of nitrogen and find the amount of gas in this volume. Knowing the molar mass of nitrogen M = 0.028 kg / mol, we find the mass of this portion - and the problem is solved. The amount of gas ν= P V/R T, the mass m = ν M = M P V/R T, hence the density ρ= m/V = M P/R T = 0.028 20000/( 8.3 300) ≈ 0.2 kg/m3. The volume we chose was never included in the answer, we chose it for specificity - it’s easier to reason this way, because you don’t have to immediately realize that the volume can be anything, but the density will turn out to be the same. However, you can figure it out - "taking a volume, say, five times more, we will increase exactly five times the amount of gas, therefore, no matter what volume we take, the density will be the same." You could simply rewrite your favorite formula, substituting into it the expression for the amount of gas through the mass of a portion of gas and its molar mass: ν \u003d m / M, then the ratio m / V \u003d M P / R T is immediately expressed, and this is the density . It was possible to take a mole of gas and find the volume occupied by it, after which the density is immediately found, because the mass of the mole is known. In general, the simpler the task, the more equal and beautiful ways to solve it ...
Here is another problem where the question may seem unexpected: find the difference in air pressure at a height of 20 m and at a height of 50 m above ground level. Temperature 00С, pressure 1 atm. Solution: if we find the air density ρ under these conditions, then the pressure difference ∆P = ρ·g·∆H. We find the density in the same way as in the previous problem, the only difficulty is that air is a mixture of gases. Assuming that it consists of 80% nitrogen and 20% oxygen, we find the mass of a mole of the mixture: m = 0.8 0.028 + 0.2 0.032 ≈ 0.029 kg. The volume occupied by this mole is V= R·T/P and the density is found as the ratio of these two quantities. Then everything is clear, the answer will be approximately 35 Pa.
The density of the gas will also have to be calculated when finding, for example, the lifting force of a balloon of a given volume, when calculating the amount of air in scuba cylinders necessary for breathing underwater for a known time, when calculating the number of donkeys needed to transport a given amount of mercury vapor through the desert, and in many other cases.
But the task is more complicated: an electric kettle boils noisily on the table, the power consumption is 1000 W, efficiency. heater 75% (the rest "leaves" into the surrounding space). From the nozzle - the area of \u200b\u200bthe "nose" is 1 cm2 - a jet of steam flies out, estimate the gas velocity in this jet. All the necessary data is taken from the tables.
Solution. We will assume that saturated steam is formed in the kettle above the water, then a jet of saturated water vapor flies out of the spout at +1000C. The pressure of such a vapor is 1 atm, it is easy to find its density. Knowing the power used for evaporation P = 0.75 P0 = 750 W and the specific heat of vaporization (evaporation) r = 2300 kJ / kg, we find the mass of steam generated over time τ: m = 0.75 P0 τ / r. We know the density, then it is easy to find the volume of this amount of steam. The rest is already clear - let's imagine this volume as a column with a cross-sectional area of 1 cm2, the length of this column, divided by τ, will give us the departure speed (such a length flies out in a second). So, the jet departure velocity from the teapot spout V = m/(ρ S τ) = 0.75P0 τ/(r ρ S τ) = 0.75P0 R T/(r P M S) = 750 8.3 373/(2.3 106 1 105 0.018 1 10-4) ≈ 5 m/s.
(c) Zilberman A. R.
liquids, amorphous and crystalline bodies
gases and liquids
gases, liquids and crystalline bodies
approximately equal to the diameter of the molecule
less than the diameter of the molecule
about 10 times the diameter of the molecule
depends on gas temperature
liquids
crystalline bodies
amorphous bodies
only gas structure models
only models of the structure of amorphous bodies
models of the structure of gases and liquids
models of the structure of gases, liquids and solids
the distance between molecules increases
molecules begin to attract each other
increasing order in the arrangement of molecules
the distance between molecules decreases
hasn't changed
increased 5 times
decreased by 5 times
increased by the root of five times
The distances between molecules are comparable to the sizes of molecules (under normal conditions) for
In gases under normal conditions, the average distance between molecules
The least order in the arrangement of particles is typical for
The distance between adjacent particles of a substance, on average, is many times greater than the size of the particles themselves. This statement is consistent with the model
During the transition of water from a liquid to a crystalline state
At constant pressure, the concentration of gas molecules increased by 5 times, and its mass did not change. Average kinetic energy forward movement gas molecules
The table shows the melting and boiling points of some substances:
substance | Boiling temperature | substance | Melting temperature |
naphthalene |
Choose the correct statement.
The melting point of mercury is greater than the boiling point of ether
The boiling point of alcohol is less than the melting point of mercury
The boiling point of alcohol is greater than the melting point of naphthalene
The boiling point of ether is less than the melting point of naphthalene
The temperature of the solid body dropped by 17 ºС. On the absolute temperature scale, this change was
1) 290 K 2) 256 K 3) 17 K 4) 0 K
9. In a vessel of constant volume there is an ideal gas in the amount of 2 mol. How should the absolute temperature of a vessel with gas be changed when 1 mol of gas is released from the vessel so that the pressure of the gas on the walls of the vessel increases by 2 times?
1) increase by 2 times 3) increase by 4 times
2) decrease by 2 times 4) decrease by 4 times
10. At temperature T and pressure p, one mole of an ideal gas occupies a volume V. What is the volume of the same gas, taken in an amount of 2 mol, at a pressure of 2p and a temperature of 2T?
1) 4V 2) 2V 3) V 4) 8V
11. The temperature of hydrogen, taken in an amount of 3 mol, in a vessel is equal to T. What is the temperature of oxygen, taken in an amount of 3 mol, in a vessel of the same volume and at the same pressure?
1) T 2) 8T 3) 24 T 4) T/8
12. In a vessel closed by a piston, there is an ideal gas. A graph of the dependence of gas pressure on temperature with changes in its state is shown in the figure. Which state of the gas corresponds to the smallest value of volume?
1) A 2) B 3) C 4) D
13. In a vessel of constant volume there is an ideal gas, the mass of which is changed. The diagram shows the process of changing the state of the gas. At which point on the diagram is the mass of gas the greatest?
1) A 2) B 3) C 4) D
14. At the same temperature, saturated steam in a closed vessel differs from unsaturated steam in the same vessel
1) pressure
2) the speed of movement of molecules
3) the average energy of the chaotic movement of molecules
4) no admixture of foreign gases
15. Which point on the diagram corresponds to the maximum gas pressure?
can't give an exact answer
17. Balloon with a volume of 2500 cubic meters and a shell mass of 400 kg, it has an opening at the bottom through which the air in the ball is heated by a burner. To what minimum temperature must the air in the balloon be heated in order for the balloon to take off with a load (basket and aeronaut) weighing 200 kg? The ambient temperature is 7ºС, its density is 1.2 kg per cubic meter. The shell of the sphere is assumed to be inextensible.
MKT and thermodynamics
MKT and thermodynamics
For this section, each option included five tasks with a choice
response, of which 4 are basic and 1 is advanced. Based on exam results
The following elements of content were learned:
Application of the Mendeleev–Clapeyron equation;
Dependence of gas pressure on the concentration of molecules and temperature;
The amount of heat during heating and cooling (calculation);
Features of heat transfer;
Relative air humidity (calculation);
Work in thermodynamics (graph);
Application of the equation of state of a gas.
Among the tasks of the basic level of difficulty, the following questions were raised:
1) Change in internal energy in various isoprocesses (for example, when
isochoric increase in pressure) - 50% of completion.
2) Graphs of isoprocesses - 56%.
Example 5
The constant mass of an ideal gas is involved in the process shown
on the image. The highest gas pressure in the process is reached
1) at point 1
2) on the entire segment 1–2
3) at point 3
4) on the entire segment 2–3
Answer: 1
3) Determination of air humidity - 50%. These assignments included a photo
psychrometer, according to which it was necessary to take readings of dry and wet
thermometers, and then determine the humidity of the air using part
psychrometric table given in the task.
4) Application of the first law of thermodynamics. These tasks were the most
difficult among the tasks of the basic level in this section - 45%. Here
it was necessary to use the graph, determine the type of isoprocess
(either isotherms or isochores were used) and in accordance with this
determine one of the parameters given the other.
Among the tasks of an advanced level, there were presented computational tasks for
application of the equation of state of gas, with which an average of 54% coped
students, as well as previously used tasks to determine the change
parameters of an ideal gas in an arbitrary process. Dealing with them successfully
only a group of strong graduates, and the average percentage of completion was 45%.
One of these tasks is shown below.
Example 6
An ideal gas is contained in a vessel closed by a piston. Process
the change in the state of the gas is shown in the diagram (see figure). How
did the volume of the gas change during its transition from state A to state B?
1) increased all the time
2) decreased all the time
3) first increased, then decreased
4) first decreased, then increased
Answer: 1
Activities Quantity
jobs %
photos2 10-12 25.0-30.0
4. PHYSICS
4.1. Characteristics of control measuring materials in physics
2007
Examination paper for unified state exam in 2007 had
the same structure as in the previous two years. It consisted of 40 tasks,
differing in form of presentation and level of complexity. In the first part of the work
30 tasks with a choice of answers were included, where each task was given
four possible answers, of which only one was correct. The second part contained 4
short answer questions. They were computational problems, after solving
which required the answer to be given as a number. The third part of the exam
work - these are 6 calculation tasks, to which it was necessary to bring a complete
expanded solution. The total time to complete the work was 210 minutes.
Education Content Elements Codifier and Specification
examination work were compiled on the basis of the Mandatory Minimum
1999 No. 56) and took into account the Federal component of the state standard
secondary (complete) education in physics, profile level (Order of the Ministry of Defense dated 5
March 2004 No. 1089). The content element codifier has not changed since
compared with 2006 and included only those elements that are simultaneously
are present as in the Federal component of the state standard
(profile level, 2004), and in the Mandatory minimum maintenance
Education 1999
Compared to the 2006 control measuring materials in the options
The 2007 USE has been amended in two ways. The first of these was to redistribute
assignments in the first part of the work on a thematic basis. Regardless of the difficulty
(basic or advanced levels), first all tasks in mechanics followed, then
in MKT and thermodynamics, electrodynamics and, finally, in quantum physics. Second
the change concerned the purposeful introduction of tasks that check
formation of methodological skills. In 2007, A30 tasks tested skills
analyze results experimental studies, expressed as
tables or graphs, as well as build graphs based on the results of the experiment. Selection
tasks for the A30 line was carried out based on the need for verification in this
series of variants of one type of activity and, accordingly, regardless of
thematic affiliation of a particular task.
In the examination paper, tasks of basic, advanced
and high levels of difficulty. The tasks of the basic level tested the assimilation of the most
important physical concepts and laws. Elevated tasks supervised
the ability to use these concepts and laws to analyze more complex processes or
the ability to solve problems for the application of one or two laws (formulas) for any of
topics of school physics course. Tasks high level Difficulties are calculated
tasks that reflect the level of requirements for university entrance exams and
require the application of knowledge from two or three sections of physics at once in a modified or
new situation.
KIM 2007 included assignments for all major content
sections of the physics course:
1) "Mechanics" (kinematics, dynamics, statics, conservation laws in mechanics,
mechanical vibrations and waves);
2) “Molecular physics. Thermodynamics";
3) "Electrodynamics" (electrostatics, D.C., a magnetic field,
electromagnetic induction, electromagnetic oscillations and waves, optics);
4) " The quantum physics» (elements of SRT, wave-particle duality, physics
atom, nuclear physics).
Table 4.1 shows the distribution of tasks by content blocks in each
part of the examination paper.
Table 4.1
depending on the type of tasks
All work
(with choice
(with brief
Jobs % No.
Jobs % No.
jobs %
1 Mechanics 11-131 27.5-32.5 9-10 22.5-25.0 1 2.5 1-2 2.5-5.0
2 MKT and thermodynamics 8-10 20.0-25.0 6-7 15.0-17.5 1 2.5 1-2 2.5-5.0
3 Electrodynamics 12-14 30.0-35.5 9-10 22.5-15.0 2 5.0 2-3 5.0-7.5
4 Quantum physics and
STO 6-8 15.0-20.0 5-6 12.5-15.0 – – 1-2 2.5-5.0
Table 4.2 shows the distribution of tasks by content blocks in
depending on the level of difficulty.
table4.2
Distribution of tasks by sections of the course of physics
depending on the level of difficulty
All work
A basic level of
(with choice
elevated
(with choice of answer
and brief
High level
(with extended
Answer section)
Jobs % No.
Jobs % No.
Jobs % No.
jobs %
1 Mechanics 11-13 27.5-32.5 7-8 17.5-20.0 3 7.5 1-2 2.5-5.0
2 MKT and thermodynamics 8-10 20.0-25.0 5-6 12.5-15.0 2 5.0 1-2 2.5-5.0
3 Electrodynamics 12-14 30.0-35.5 7-8 17.5-20.0 4 10.0 2-3 5.0-7.5
4 Quantum physics and
STO 6-8 15.0-20.0 4-5 10.0-12.5 1 2.5 1-2 2.5-5.0
When developing the content of the examination paper, it was taken into account
the need to check the mastery of various activities. Wherein
tasks of each of the series of options were selected taking into account the distribution by type
activities presented in table 4.3.
1 The change in the number of tasks for each of the topics is associated with different topics of complex tasks C6 and
tasks A30, testing methodological skills on the material of different sections of physics, in
different series of options.
table4.3
Distribution of tasks by types of activity
Activities Quantity
jobs %
1 Understand the physical meaning of models, concepts, quantities 4-5 10.0-12.5
2 Explain physical phenomena, to distinguish the influence of different
factors on the course of phenomena, manifestations of phenomena in nature or
their use in technical devices and everyday life
3 Apply the laws of physics (formulas) to analyze processes on
quality level 6-8 15.0-20.0
4 Apply the laws of physics (formulas) to analyze processes on
calculated level 10-12 25.0-30.0
5 Analyze the results of experimental studies 1-2 2.5-5.0
6 Analyze information obtained from graphs, tables, diagrams,
photos2 10-12 25.0-30.0
7 Solve problems of various levels of complexity 13-14 32.5-35.0
All tasks of the first and second parts of the examination paper were evaluated at 1
primary score. Solutions to the problems of the third part (С1-С6) were checked by two experts in
in accordance with the generalized evaluation criteria, taking into account the correctness and
completeness of the answer. The maximum score for all tasks with a detailed answer was 3
points. The task was considered solved if the student scored at least 2 points for it.
Based on the points assigned for the completion of all tasks of the examination
work was translated into "test" scores on a 100-point scale and into marks
on a five-point scale. Table 4.4 reflects the relationship between primary,
test marks on a five-point system over the past three years.
table4.4
Primary score ratio, test scores and school grades
Years, points 2 3 4 5
2007 primary 0-11 12-22 23-35 36-52
test 0-32 33-51 52-68 69-100
2006 primary 0-9 10-19 20-33 34-52
test 0-34 35-51 52-69 70-100
2005 primary 0-10 11-20 21-35 36-52
test 0-33 34-50 51-67 68-100
Comparison of the boundaries of primary scores shows that this year the conditions
the corresponding marks were more stringent than in 2006, but
approximately corresponded to the conditions of 2005. This was due to the fact that in the past
year single exam not only those who were going to enter universities passed in physics
in the relevant profile, but also almost 20% of students (of the total number of applicants),
who studied physics basic level(for them, this exam was by decision
region is required).
In total, 40 options were prepared for the exam in 2007,
which were five series of 8 options, created according to different plans.
The series of variants differed in controlled content elements and types.
activities for the same line of tasks, but in general they all had approximately
2 In this case, we mean the form of presentation of information in the text of the task or distractors,
so the same job can check two activities.
the same average level of difficulty and corresponded to the plan of the examination
of the work given in Appendix 4.1.
4.2. Characteristics of USE participants in physics2007 of the year
The number of participants in the USE in physics this year amounted to 70,052 people, which
significantly lower than in the previous year, and approximately in line with the indicators
2005 (see table 4.5). The number of regions in which graduates took the USE in
physics, increased to 65. The number of graduates who chose physics in the format
USE, differs significantly for different regions: from 5316 people. in the Republic
Tatarstan up to 51 people in the Nenets autonomous region. As a percentage of
the total number of graduates, the number of participants in the USE in physics ranges from
0.34% in Moscow to 19.1% in the Samara region.
table4.5
Number of Exam Participants
Year Number Girls Boys
regions
participants Number % Number %
2005 54 68 916 18 006 26,1 50 910 73,9
2006 61 90 3893 29 266 32,4 61 123 67,6
2007 65 70 052 17 076 24,4 52 976 75,6
The physics exam is chosen predominantly by young men, and only a quarter of
of the total number of participants are girls who chose to continue
education universities of physical and technical profile.
The distribution of exam participants by
types of settlements (see table 4.6). Nearly half of the graduates who took
USE in physics, lives in major cities and only 20% are students who have completed
rural schools.
table4.6
Distribution of exam participants by types of settlements, in which
their educational institutions are located
Number of examinees Percentage
Type locality examinees
Settlement of rural type (village,
village, farm, etc.) 13,767 18,107 14,281 20.0 20.0 20.4
Urban settlement
(working settlement, urban settlement
type, etc.)
4 780 8 325 4 805 6,9 9,2 6,9
City with a population of less than 50 thousand people 7,427 10,810 7,965 10.8 12.0 11.4
City with a population of 50-100 thousand people 6,063 8,757 7,088 8.8 9.7 10.1
City with a population of 100-450 thousand people 16,195 17,673 14,630 23.5 19.5 20.9
City with a population of 450-680 thousand people 7,679 11,799 7,210 11.1 13.1 10.3
A city with a population of over 680,000.
people 13,005 14,283 13,807 18.9 15.8 19.7
St. Petersburg - 72 7 - 0.1 0.01
Moscow - 224 259 - 0.2 0.3
No data – 339 – – 0.4 –
Total 68,916 90,389 70,052 100% 100% 100%
3 In 2006, in one of the regions entrance exams in universities in physics were held only in
USE format. This led to such a significant increase in the number of participants in the exam.
The composition of exam participants by types of educational institutions practically does not change.
institutions (see table 4.7). Like last year, the vast majority
test takers were finishing educational institutions, and only about 2%
graduates came to the exam from educational institutions of primary or
middle vocational education.
table4.7
Distribution of exam participants by types of educational institutions
Number
examinees
Percent
Type educational institution examinees
2006 G. 2007 G. 2006 G. 2007 G.
General education institutions 86,331 66,849 95.5 95.4
Evening (shift) general education
institutions 487 369 0.5 0.5
General education boarding school,
cadet school, boarding school with
initial flight training
1 144 1 369 1,3 2,0
Educational institutions of primary and
secondary vocational education 1,469 1,333 1.7 1.9
No data 958 132 1.0 0.2
Total: 90,389 70,052 100% 100%
4.3. The main results of the examination work in physics
In general, the results of the examination work in 2007 were
slightly higher than last year, but about the same level as
figures for the previous year. Table 4.8 shows the results of the USE in physics in 2007.
on a five-point scale, and in table 4.9 and in fig. 4.1 - on test scores in 100-
point scale. For clarity of comparison, the results are presented in comparison with
the previous two years.
table4.8
Distribution of exam participants by level
training(percentage of the total)
Years "2" Marks "n3o" 5 points "b4n" on the scale "5"
2005 10,5% 40,7% 38,1% 10,7%
2006 16,0% 41,4% 31,1% 11,5%
2007 12,3% 43,2% 32,5% 12,0%
table4.9
Distribution of exam participants
based on test scores2005-2007 gg.
Year Test score scale interval
0-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100
2005 0,09% 0,57% 6,69% 19,62% 24,27% 24,44% 16,45% 6,34% 1,03% 0,50% 68 916
2006 0,10% 0,19% 6,91% 23,65% 23,28% 19,98% 15,74% 7,21% 2,26% 0,68% 90 389
2007 0,07% 1,09% 7,80% 19,13% 27,44% 20,60% 14,82% 6,76% 1,74% 0,55% 70 052
0-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100
Test score
Percentage of students who received
corresponding test score
Rice. 4.1 Distribution of exam participants by test scores received
Table 4.10 compares the scale in test scores in a 100-point
scale with the results of tasks examination option in primary
table4.10
Comparison of intervals of primary and test scores in2007 year
Scale interval
test scores 0-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100
Scale interval
primary scores 0-3 4-6 7-10 11-15 16-22 23-29 30-37 38-44 45-48 49-52
To obtain 35 points (score 3, primary score - 13) the test-taker
it was enough to correctly answer the 13 simplest questions of the first part
work. To score 65 points (grade 4, primary score - 34), the graduate must
was, for example, correctly answer 25 tasks with a choice of answers, solve three out of four
short answer problems and two more high-level problems
difficulties. Those who received 85 points (score 5, primary score 46) practically
performed the first and second parts of the work perfectly and solved at least four tasks
third part.
The best of the best (range from 91 to 100 points) need not only
freely navigate in all matters of the school course of physics, but also in practice
avoid even technical errors. So, to get 94 points (primary score
– 49) it was possible to “not get” only 3 primary points, allowing, for example,
arithmetic errors in solving one of the problems of a high level of complexity
and make a mistake in the answer to any two questions with a choice of answers.
Unfortunately, this year there was no increase in the number of graduates who scored
on USE results in physics, the highest possible score. Table 4.11
the number of 100 points for the last four years is given.
table4.11
Number of people tested, scored on the exam results100 points
Year 2004 2005 2006 2007
Number of students 6 23 33 28
The leaders of this year are 27 boys and only one girl (Romanova A.I. from
Novovoronezh secondary school No. 1). Like last year, among the graduates of Lyceum No. 153
Ufa - two students at once who scored 100 points each. The same results (two 100-
ballnik) and achieved the gymnasium No. 4 named after. A.S. Pushkin in Yoshkar-Ola.
Let us consider how the projection of the resulting force of interaction between them on the straight line connecting the centers of the molecules changes depending on the distance between the molecules. If the molecules are located at distances exceeding their sizes by several times, then the forces of interaction between them practically do not affect. Interaction forces between molecules are short-range.
At distances exceeding 2-3 molecular diameters, the repulsive force is practically zero. Only the force of attraction is noticeable. As the distance decreases, the attractive force increases and at the same time the repulsive force begins to affect. This force increases very rapidly when the electron shells of the molecules begin to overlap.
Figure 2.10 graphically shows the dependence of the projection F r interaction forces of molecules on the distance between their centers. On distance r 0 , approximately equal to the sum of the radii of the molecules, F r = 0 , since the force of attraction is equal in absolute value to the force of repulsion. At r > r 0 there is an attractive force between molecules. The projection of the force acting on the right molecule is negative. At r < r 0 there is a repulsive force with a positive projection value F r .
Origin of elastic forces
The dependence of the interaction forces of molecules on the distance between them explains the appearance of an elastic force during compression and tension of bodies. If you try to bring the molecules closer to a distance less than r0, then a force begins to act that prevents the approach. On the contrary, when the molecules move away from each other, an attractive force acts, returning the molecules to their original positions after the cessation of external influence.
With a small displacement of molecules from equilibrium positions, the forces of attraction or repulsion grow linearly with increasing displacement. In a small section, the curve can be considered a straight line segment (the thickened section of the curve in Fig. 2.10). That is why, at small deformations, Hooke's law turns out to be valid, according to which the elastic force is proportional to the deformation. At large displacements of molecules, Hooke's law is no longer valid.
Since the distances between all molecules change when the body is deformed, the neighboring layers of molecules account for an insignificant part of the total deformation. Therefore, Hooke's law is fulfilled at deformations that are millions of times greater than the size of the molecules.
Atomic force microscope
The device of the atomic force microscope (AFM) is based on the action of repulsive forces between atoms and molecules at small distances. This microscope, in contrast to the tunnel microscope, allows you to obtain images of non-conductive surfaces. Instead of a tungsten tip, AFM uses a small piece of diamond sharpened to atomic dimensions. This fragment is fixed on a thin metal holder. When the tip approaches the surface under study, the electron clouds of diamond atoms and the surface begin to overlap and repulsive forces arise. These forces deflect the tip of the diamond point. Deviation is recorded using a laser beam reflected from a mirror mounted on a holder. The reflected beam drives a piezoelectric arm similar to that of a tunneling microscope. The feedback mechanism ensures that the height of the diamond needle above the surface is such that the curvature of the holder plate remains unchanged.
In Figure 2.11 you see an AFM image of the polymer chains of the amino acid alanine. Each bump represents one amino acid molecule.
At present, atomic microscopes have been designed, the device of which is based on the action of molecular forces of attraction at distances several times greater than the size of an atom. These forces are approximately 1000 times smaller than the repulsive forces in the AFM. Therefore, a more complex sensitive system is used to register forces.
Atoms and molecules are made up of electrically charged particles. Due to the action of electric forces at short distances, the molecules attract, but begin to repel when the electron shells of the atoms overlap.
- In contact with 0
- Google+ 0
- OK 0
- Facebook 0
