Using the record of the first law of thermodynamics in differential form (9.2), we obtain an expression for the heat capacity of an arbitrary process:
Imagine full differential internal energy through partial derivatives with respect to parameters and:
Then we rewrite formula (9.6) in the form
Relation (9.7) has an independent meaning, since it determines the heat capacity in any thermodynamic process and for any macroscopic system if the caloric and thermal equations of state are known.
Consider the process at constant pressure and get the general relationship between and.
Based on the formula obtained, one can easily find the relationship between the heat capacities and in an ideal gas. This is what we will do. However, the answer is already known, we actively used it in 7.5.
Robert Mayer's equation
Let us express the partial derivatives on the right-hand side of equation (9.8), using the thermal and caloric equations written for one mole of an ideal gas. Internal energy ideal gas depends only on temperature and does not depend on the volume of gas, therefore
It is easy to obtain from the thermal equation
Substitute (9.9) and (9.10) in (9.8), then
We will finally write down
You hopefully learned (9.11). Yes, of course, this is the Mayer equation. Recall once again that Mayer's equation is valid only for an ideal gas.
9.3. Polytropic processes in ideal gas
As noted above, the first law of thermodynamics can be used to derive the equations of processes occurring in a gas. Big practical use finds a class of processes called polytropic. Polytropic is called a process that takes place at a constant heat capacity .
The process equation is set by the functional relationship of two macroscopic parameters describing the system. On the corresponding coordinate plane, the process equation is clearly represented in the form of a graph - a process curve. The curve depicting the political process is called polytropy. The equation of the polytropic process for any substance can be obtained on the basis of the first law of thermodynamics using its thermal and caloric equations of state. Let us demonstrate how this is done using the example of deriving the process equation for an ideal gas.
Derivation of the equation of a polytropic process in an ideal gas
The requirement for the constancy of the heat capacity in the process makes it possible to write the first law of thermodynamics in the form
Using Mayer's equation (9.11) and the equation of state for an ideal gas, we obtain the following expression for
Dividing equation (9.12) by T and substituting (9.13) into it, we arrive at the expression
Dividing () by, we find
By integrating (9.15), we obtain
This is the polytropic equation in variables
Eliminating from the equation (), using the equality, we obtain the polytropic equation in variables
The parameter is called the polytropic indicator, which can take according to () a variety of values, positive and negative, whole and fractional. There are many processes behind the formula (). The isobaric, isochoric and isothermal processes known to you are special cases of the polytropic.
This class of processes also includes adiabatic or adiabatic process ... Adiabatic is a process that takes place without heat exchange (). This process can be implemented in two ways. The first method assumes that the system has a heat-insulating shell capable of changing its volume. The second is the implementation of such a fast process in which the system does not have time to exchange the amount of heat with environment... The process of sound propagation in gas can be considered adiabatic due to its high speed.
From the definition of heat capacity it follows that in the adiabatic process. According to
where is the adiabatic exponent.
In this case, the polytropic equation takes the form
The equation of the adiabatic process (9.20) is also called the Poisson equation, therefore the parameter is often called the Poisson constant. Constant is an important characteristic of gases. It follows from experience that its values for different gases lie in the range 1.30 ÷ 1.67, therefore, on the diagram of processes, the adiabat "falls" more steeply than the isotherm.
Graphs of polytropic processes for different values are shown in Fig. 9.1.
In fig. 9.1 process diagrams are numbered in accordance with table. 9.1.
This tutorial is a useful addition to the previous topic "".
The ability to do such things is not just a useful thing, it is - necessary... In all areas of mathematics, from school to higher. And in physics too. It is for this reason that tasks of this kind are necessarily present in both the exam and the exam. At all levels - both basic and specialized.
Actually, the entire theoretical part of such tasks is one single phrase. Versatile and simple to the point of disgrace.
We are surprised, but remember:
Any equality with letters, any formula is ALSO an EQUATION!
And where the equation is, there automatically and. So we apply them in a convenient order and - the job is ready.) Have you read the previous lesson? No? However ... Then this link is for you.
Oh, do you know? Fine! Then we apply theoretical knowledge in practice.
Let's start simple.
How do I express one variable through another?
This problem constantly arises when solving systems of equations. For example, there is equality:
3 x - 2 y = 5
Here two variables- x and igrek.
Let's say they ask us to expressxacrossy.
What does this assignment mean? It means that we should get some equality, where there is a pure x on the left. In splendid isolation, without any neighbors and coefficients. And on the right - what will happen.
And how do we get this equality? Very simple! With the help of all the same good old identical transformations! So we use them in a convenient US order, step by step getting to the pure x.
Analyzing the left side of the equation:
3 x – 2 y = 5
Here we get in the way of the three in front of the X and - 2 y... Let's start with - 2y, it will be easier.
We throw - 2y from the left to the right. Changing minus to plus, of course. Those. apply first identical transformation:
3 x = 5 + 2 y
Half the job is done. There is a three left in front of the X. How to get rid of it? Divide both parts into this very three! Those. engage second identical transformation.
So we divide:
That's all. We expressed x through y... On the left - a pure x, and on the right - what happened as a result of the "cleansing" of the x.
It could be at first divide both parts into three, and then transfer. But this would lead to the appearance of fractions in the process of transformations, which is not very convenient. And so, the fraction appeared only at the very end.
Let me remind you that the order of transformations does not matter. How US conveniently, so we do. The most important thing is not the order of application of identical transformations, but their right!
And it is possible from the same equality
3 x – 2 y = 5
express y throughx?
Why not? Can! Everything is the same, only this time we are interested in the pure game on the left. So we clear the game of all unnecessary things.
The first step is to get rid of the expression 3x... We transfer it to the right side:
–2 y = 5 – 3 x
There was a deuce with a minus. We divide both parts by (-2):
And that's all.) We expressedythrough x. Moving on to more serious tasks.
How do I express a variable from a formula?
Not a problem! Similar! If you understand that any formula - the same equation.
For example, a task like this:
From the formula
express variable with.
A formula is also an equation! The task means that through transformations from the proposed formula we need to get some new formula. In which there will be a clean with, and on the right - what happens is what happens ...
However ... How do we have this very with pull out something?
How-how ... Step by step! It is clear what to highlight the clean with straightaway impossible: she sits in a fraction. And the fraction is multiplied by r... So, first of all we clean expression with letter with, i.e. the whole fraction. Here you can divide both sides of the formula by r.
We get:
The next step is to pull out with from the numerator of the fraction. How? Easily! Let's get rid of the fraction. If there is no fraction, there is no numerator.) Multiply both sides of the formula by 2:
Elementary remains. Let's provide the right letter with proud loneliness. For this, the variables a and b move to the left:
That's all, one might say. It remains to rewrite the equality in the usual form, from left to right, and - the answer is ready:
It was not a difficult task. And now the task based on the real version of the exam:
The bathyscaphe locator, steadily sinking vertically downward, emits ultrasonic pulses with a frequency of 749 MHz. The sinking speed of the bathyscaphe is calculated by the formula
where c = 1500 m / s is the speed of sound in water,
f 0 - the frequency of the pulses emitted (in MHz),
fIs the frequency of the signal reflected from the bottom recorded by the receiver (in MHz).
Determine the frequency of the reflected signal in MHz if the sinking speed of the bathyscaphe is 2 m / s.
"A lot of bukaff", yes ... But the letters are lyrics, but the general essence is still the same... The first step is to express this very frequency of the reflected signal (i.e. the letter f) from the formula proposed to us. This is what we will do. We look at the formula:
Directly, naturally, the letter f You cannot pull it out, it is hidden again in a fraction. Moreover, both the numerator and the denominator. Therefore, the most logical step would be to get rid of the fraction. And there - it will be seen. For this we use second transformation - multiply both sides by the denominator.
We get:
And here - another rake. Please pay attention to the brackets of both parts! Often, it is in these very parentheses that errors in such tasks lie. More precisely, not in the parentheses themselves, but in their absence.)
The brackets to the left indicate the letter v multiplies the whole denominator... And not into its separate pieces ...
On the right, after multiplication, the fraction disappeared and there was a lonely numerator. Which, again, the whole wholly multiplied by a letter with... Which is expressed by the parentheses on the right side.)
And now you can expand the brackets:
Fine. The process is underway.) Now the letter f left became common factor ... We take it out of the brackets:
There is nothing left. Divide both parts by a parenthesis (v- c) and - it's in the bag!
In principle, everything is ready. Variable f already expressed... But you can additionally "comb" the resulting expression - take out f 0 for the parenthesis in the numerator and reduce the entire fraction by (-1), thereby getting rid of unnecessary minuses:
Here is an expression. But now you can substitute numerical data. We get:
Answer: 751 MHz
That's all. Hope the general idea is clear.
We make elementary identical transformations in order to secrete the variable of interest to us. The main thing here is not the sequence of actions (it can be anything), but their correctness.
These two lessons only cover two basic identity transformations of equations. They work always... That's why they are basic. In addition to this couple, there are many other transformations that will also be identical, but not always, but only under certain conditions.
For example, squaring both sides of an equation (or formula) (or vice versa, extracting a root from both sides) will be identical transformation if both sides of the equation obviously non-negative.
Or, say, taking the logarithm of both sides of the equation will be the identical transformation if both sides obviously positive. Etc…
Such transformations will be discussed in the relevant topics.
And here and now - examples for training on elementary basic transformations.
A simple task:
From the formula
express the variable a and find its value atS=300, V 0 =20, t=10.
The task is more difficult:
The average speed of a skier (in km / h) over a distance of two circles is calculated by the formula:
whereV 1 andV 2 - average speeds (in km / h) on the first and second circles, respectively. What was the average speed of the skier on the second lap if it is known that the skier ran the first lap at a speed of 15 km / h, and the average speed over the entire distance was 12 km / h?
The task based on the real version of the OGE:
The centripetal acceleration when moving in a circle (in m / s 2) can be calculated by the formulaa= ω 2R, where ω is the angular velocity (in s -1), andRIs the radius of the circle. Using this formula, find the radiusR(in meters), if the angular velocity is 8.5 s -1 and the centripetal acceleration is 289 m / s 2.
A challenge based on a real option profile exam:
To a source with EMF ε = 155 V and internal resistancer= 0.5 Ohm want to connect a load with resistanceROhm. The voltage across this load, expressed in volts, is given by the formula:
At what resistance of the load will the voltage across it be 150 V? Express your answer in ohms.
Answers (in disarray): 4; 15; 2; ten.
And where are the numbers, kilometers per hour, meters, ohms - it's somehow themselves ...)
In each problem in physics, it is required to express the unknown from the formula, the next step is to substitute the numerical values and get the answer, in some cases it is only necessary to express the unknown value. There are many ways to derive the unknown from the formula. If you look at the pages of the Internet, then we will see many recommendations on this matter. This suggests that the scientific community has not yet developed a unified approach to solving this problem, and the methods that are used, as experience in school shows, are all ineffective. Up to 90% of high school students do not know how to correctly express the unknown. Those who know how to do this - perform cumbersome transformations. It is very strange, but physicists, mathematicians, chemists have different approaches, explaining the methods of transferring parameters through the equal sign (they offer the rules of a triangle, cross or proportions, etc.) We can say that they have a different culture of working with formulas. You can imagine what happens to the majority of students who meet with different interpretations of the solution to this problem, consistently attending the lessons of these subjects. This situation is described by a typical conversation on the web:
Learn to express quantities from formulas. Grade 10, I'm ashamed not to know how to make another from one formula.
Don't worry - this is a problem for many of my classmates, even though I'm in grade 9. Teachers show this most often by the triangle method, but it seems to me that it is inconvenient, and it is easy to get confused. I'll show you the easiest way that I use ...
Let's say the formula is given:
Well, a simpler one .... you need to find time from this formula. You take and substitute only different numbers in this formula, based on algebra. Let's say:
and you probably clearly see that to find the time in the algebraic expression 5 you need 45/9, that is, we go to physics: t = s / v
Most students form a psychological block. Students often note that when reading a textbook, difficulties are primarily caused by those fragments of the text in which there are many formulas, that “long conclusions cannot be understood anyway,” but at the same time a feeling of inferiority and disbelief arises.
I propose the following solution to this problem - most students can still solve examples and, therefore, arrange the order of actions. We use this skill.
1... In the part of the formula that contains the variable that needs to be expressed, it is necessary to arrange the order of actions, and we will not do this in monomials that do not contain the desired value.
2. Then, in the reverse sequence of calculations, transfer the elements of the formula to another part of the formula (through the equal sign) with the opposite action ("minus" - "plus", "divide" - "multiply", "squaring" - "square root" ).
That is, we find the last action in the expression and transfer the monomial or polynomial performing this action through the equal sign first, but with the opposite action. Thus, sequentially, finding the last action in the expression, transfer all known quantities from one part of the equality to the other. In conclusion, let's rewrite the formula so that the unknown variable is on the left.
We get a clear algorithm of work, we know exactly how many transformations need to be performed. We can use already known formulas for training, we can invent our own. To start working on mastering this algorithm, a presentation was created.
Experience with students shows that this method is well accepted by them. The teachers' reaction to my performance at the Specialized School Teacher Festival also speaks of the positive grain inherent in this work.
In order to derive a formula for the complex, it is necessary, first of all, by analysis to establish what elements the substance consists of and in what weight ratios the elements included in it are connected to each other. Usually the composition of a complex is expressed as a percentage, but it can be expressed in any other numbers indicating the ratio the ratio between the weight amounts of elements that form a given substance. For example, the composition of alumina containing 52.94% of aluminum and 47.06% of oxygen will be quite definite if we say that they are combined in a weight ratio of 9: 8, that is, by 9 wt. including aluminum accounts for 8 weight. including oxygen. It is clear that the ratio of 9: 8 should be equal to the ratio of 52.94: 47.06.
Knowing the weight composition of a complex and the atomic weights of the elements that form it, it is easy to find the relative number of atoms of each element in the molecule of the substance taken and thus establish its simplest formula.
Suppose, for example, that you want to derive the formula for calcium chloride containing 36% calcium and 64% chlorine. The atomic weight of calcium is 40, chlorine is 35.5.
Let us denote the number of calcium atoms in the calcium chloride molecule through NS, and the number of chlorine atoms through at. Since the calcium atom weighs 40, and the chlorine atom weighs 35.5 oxygen units, the total weight of the calcium atoms that make up the calcium chloride molecule will be 40 NS, and the weight of chlorine atoms is 35.5 at. The ratio of these numbers, obviously, should be equal to the ratio of the weight amounts of calcium and chlorine in any amount of calcium chloride. But the last ratio is 36: 64.
Equating both relations, we get:
40x: 35.5y = 36:64
Then we get rid of the coefficients for unknowns NS and at by dividing the first terms of the proportion by 40, and the second by 35.5:
The numbers 0.9 and 1.8 express the relative number of atoms in the calcium chloride molecule, but they are fractional, while the molecule can contain only an integer number of atoms. To express an attitude NS:at two integers, we divide both terms ^ of the second relation by the smallest of them. We get
NS: at = 1:2
Consequently, in a molecule of calcium chloride, there are two chlorine atoms per calcium atom. This condition is satisfied by a number of formulas: CaCl 2, Ca 2 Cl 4, Ca 3 Cl 6, etc. Since we have no data to judge which of the written formulas corresponds to the actual atomic composition of the calcium chloride molecule, we will focus on the simplest of which CaCl 2, indicating the smallest possible number of atoms in the calcium chloride molecule.
However, the arbitrariness in the choice of the formula disappears if, along with the weight composition of the substance, its molecular the weight. In this case, it is easy to derive a formula that expresses the true composition of the molecule. Let's give an example.
By analysis, it was found that glucose contains 4.5 wt. including carbon 0.75 wt. including hydrogen and 6 wt. including oxygen. Its molecular weight was found to be 180. It is required to derive the glucose formula.
As in the previous case, we first find the ratio between the number of carbon atoms (atomic weight 12), hydrogen and oxygen in the glucose molecule. Denoting the number of carbon atoms through NS, hydrogen through at and oxygen through z, we make up the proportion:
2x : at: 16z = 4.5: 0.75: 6
where
Dividing all three terms of the second half of the equality by 0.375, we get:
NS : at:z = 1: 2: 1
Consequently, the simplest formula for glucose will be CH 2 O. But calculated from it would be equal to 30, while in reality glucose is 180, that is, six times more. Obviously, for glucose, you need to take the formula C 6 H 12 O 6.
Formulas based, in addition to analysis data, also on the determination of molecular weight and indicating the real number of atoms in a molecule, are called true or molecular formulas; formulas derived only from analysis data are called the simplest or empirical.
Acquainted with the conclusion chemical formulas, ”It is easy to understand how precise molecular weights are set. As we have already mentioned, the existing methods for determining molecular weights in most cases do not give completely accurate results. But, knowing at least the approximate and percentage composition of a substance, you can establish its formula, which expresses the atomic composition of a molecule. Since the weight of a molecule is equal to the sum of the weights of the atoms that form it, adding the weights of the atoms that make up the molecule, we determine its weight in oxygen units, that is, the molecular weight of the substance. The accuracy of the molecular weight found will be the same as the accuracy of the atomic balance.
Finding the formula of a chemical compound in many cases can be greatly simplified if we use the concept of the ovalence of elements.
Recall that the valence of an element is the property of its atoms to attach to or replace a certain number of atoms of another element.
What is valence
an element is determined by a number indicating how many hydrogen atoms(oranother monovalent element) adds or replaces an atom of this element.
The concept of valence applies not only to individual atoms, but also to entire groups of atoms that make up chemical compounds and participating as a whole in chemical reactions. Such groups of atoms are called radicals. V inorganic chemistry the most important radicals are: 1) an aqueous residue, or hydroxyl OH; 2) acidic residues; 3) basic residues.
The water residue, or hydroxyl, is obtained if one hydrogen atom is removed from the water molecule. In a water molecule, hydroxyl is bonded to one hydrogen atom, therefore, the OH group is monovalent.
Acid residues are called groups of atoms (and sometimes one atom), "remaining" from acid molecules, if mentally subtract from them one or more hydrogen atoms, which are replaced by a metal. of these groups is determined by the number of removed hydrogen atoms. For example, it gives two acidic residues - one is bivalent SO 4 and the other is monovalent HSO 4, which is part of various acid salts. Phosphoric acid H 3 PO 4 can give three acid residues: trivalent PO 4, bivalent HPO 4 and monovalent
H 2 PO 4 etc.
We will call the main residues; atoms or groups of atoms "remaining" from base molecules, if mentally subtract one or more hydroxyls from them. For example, sequentially subtracting hydroxyls from the Fe (OH) 3 molecule, we obtain the following basic residues: Fe (OH) 2, FeOH, and Fe. they are determined by the number of taken away hydroxyl groups: Fe (OH) 2 - monovalent; Fe (OH) is divalent; Fe is trivalent.
Basic residues containing hydroxyl groups are part of the so-called basic salts. The latter can be considered as bases in which some of the hydroxyls are replaced by acidic residues. So, when two hydroxyls in Fe (OH) 3 are replaced with an acid residue SO 4, the basic salt of FeOHSO 4 is obtained, when one hydroxyl is replaced in Bi (OH) 3
acid residue NO 3 gives the basic salt Bi (OH) 2 NO 3, etc.
Knowledge of the valences of individual elements and radicals allows, in simple cases, to quickly compose formulas of very many chemical compounds, which frees the chemist from the need to memorize them mechanically.
Chemical formulas
Example 1. Make up the formula of calcium bicarbonate - an acidic salt of carbonic acid.
The composition of this salt should include calcium atoms and monovalent acid residues HCO 3. Since it is divalent, two acid residues must be taken for one calcium atom. Therefore, the formula for the salt will be Ca (HCO 3) g.
Physics is the science of nature. It describes the processes and phenomena of the surrounding world on the macroscopic tier - the tier of small bodies comparable to the size of a person himself. Physicists use a mathematical aggregate to describe the processes.
Instructions
1. Where do physical formulas? In a simplified way, the scheme for acquiring formulas can be presented as follows: a question is posed, guesses are put forward, a series of experiments is carried out. The totals are processed, certain formulas and this gives a preface to the new physical theory either continues and develops the existing one more closely.
2. A person who comprehends physics does not need to go through every given difficult path again. It is enough to master the central concepts and definitions, get acquainted with the scheme of the experiment, learn to deduce the fundamental formulas... Of course, one cannot do without strong mathematical knowledge.
3. It turns out, learn the definitions of physical quantities related to the topic under consideration. Every size has its own physical meaning, one that you must understand. Let's say 1 coulomb is a charge that passes through the cross section of a conductor in 1 second at a current of 1 ampere.
4. Understand the physics of the process in question. By what parameters is it described, and how do these parameters change over time? Knowing the basic definitions and understanding the physics of the process, it is easy to get the simplest formulas... As usual, between the values or the squares of the values, directly proportional or inversely proportional dependencies are established, the proportionality indicator is introduced.
5. By means of mathematical reforms it is allowed to derive secondary formulas from primary formulas. If you learn to do this easily and quickly, it will be permissible not to memorize the latter. The core method of reforms is a substitution method: some value is expressed from one formulas and substituted into another. The main thing is that these formulas corresponded to the same process or phenomenon.
6. Equations are also allowed to be added to each other, divided, multiplied. Time functions are often integrated or differentiated, getting new dependencies. Taking logarithms is good for exponential functions. In the end formulas rely on the result, the one that you want to get as a result.
Every human life is surrounded by most of the different phenomena. Physicists are engaged in the comprehension of these phenomena; their tools are mathematical formulas and achievements of their predecessors.
Natural phenomena
The study of nature helps to be smarter about the available sources, to discover new sources of energy. So, geothermal springs heat approximately all of Greenland. The very word "physics" goes back to the Greek root "physics", which means "nature." Thus, physics itself is the science of nature and natural phenomena.
Forward to the future!
Often physicists in literally“Ahead of time”, discovering laws that find use only tens of years (and even centuries) later. Nikola Tesla discovered the laws of electromagnetism that are being used today. Pierre and Marie Curie discovered radium with virtually no support, under conditions incredible for a modern scientist. Their discoveries have helped save tens of thousands of lives. Now physicists of every world are focused on the issues of the universe (macrocosm) and the smallest particles of matter (nanotechnology, microcosm).
Understanding the world
The most important engine of society is curiosity. This is why the experiments at the Large Hadron Collider are of such high importance and are sponsored by an alliance of 60 states. There is a real possibility of revealing the secrets of society. Physics is a fundamental science. This means that any discoveries of physics are allowed to be applied in other spheres of science and technology. Small openings in one branch can have a dramatic effect on the entire “adjacent” branch. In physics, the practice of research by groups of scientists from different countries is famous, a policy of help and cooperation has been adopted. The mystery of the universe, matter worried the great physicist Albert Einstein. He proposed the theory of relativity, explaining that gravitational fields distort space and time. The apogee of the theory was the well-known formula E = m * C * C, which combines energy with mass.
Union with mathematics
Physics relies on the latest mathematical tools. Often, mathematicians discover abstract formulas, deriving new equations from existing ones, applying more high levels of abstraction and the laws of logic, making brave guesses. Physicists follow the development of mathematics, and occasionally scientific discoveries abstract science helps explain hitherto unfamiliar natural phenomena. Sometimes, on the contrary, physical discoveries push mathematicians to create guesses and a new logical aggregate. The connection between physics and mathematics, one of the most important scientific disciplines, reinforces the authority of physics.