This lesson 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 sections of mathematics, from school to higher. Yes, and in physics too. It is for this reason that tasks of this kind are necessarily present in both the Unified State Examination and the OGE. At all levels - both basic and profile.

Actually, the entire theoretical part of such tasks is a single phrase. Universal and simple to disgrace.

We are surprised, but remember:

Any equality with letters, any formula is ALSO an EQUATION!

And where is the equation, there automatically and . So we apply them in the order that is convenient for us and - the case is ready.) Have you read the previous lesson? No? However… Then this link is for you.

Ah, are you aware? Great! Then we apply theoretical knowledge in practice.

Let's start simple.

How to express one variable in terms of another?

This problem comes up all the time when systems of equations. For example, there is an equality:

3 x - 2 y = 5

Here two variables- x and y.

Suppose we are asked expressxthroughy.

What does this task mean? It means that we should get some equality, where pure x is on the left. In splendid isolation, without any neighbors and coefficients. And on the right - what will happen.

And how do we get such equality? Very simple! With the help of all the same good old identical transformations! Here we use them in a convenient way us order, step by step getting to pure X.

Let's analyze the left side of the equation:

3 x – 2 y = 5

Here we are hindered by a triple in front of 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 identity transformation:

3 x = 5 + 2 y

Half done. There was a three in front of the X. How to get rid of it? Divide both parts into this same trio! Those. engage second identical transformation.

Here we share:

That's all. We expressed x through y. On the left - pure X, and on the right - what happened as a result of the "cleansing" of X.

Could it be at first divide both parts by 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.

I remind you that the order of transformations does not play any role. How us convenient, that's what we do. The most important thing is not the order in which identical transformations are applied, but their right!

And it is possible from the same equality

3 x – 2 y = 5

express y in terms ofx?

Why not? Can! Everything is the same, only this time we are interested in a clean Y on the left. So we clean the game from everything superfluous.

First of all, we get rid of the expression 3x. Let's move it to the right side:

–2 y = 5 – 3 x

Left with a minus two. Divide both parts by (-2):

And all things.) We expressedythrough x. Let's move on to more serious tasks.

How to express a variable from a formula?

No problem! Similar! If we understand that any formula - also the equation.

For example, such a task:

From the formula

express variable c.

The formula is also an equation! The task means that through transformations from the proposed formula, we need to get some new formula. In which on the left will stand a clean With, and on the right - what happens, then it happens ...

However ... How can we this very With pull it out?

How-how ... Step by step! It is clear that to select a clean With straightaway impossible: she sits in a fraction. And the fraction is multiplied by r… So, first of all, we clean letter expression With, i.e. the whole fraction. Here you can divide both parts of the formula into r.

We get:

The next step is to take out With from the numerator of a fraction. How? Easily! Let's get rid of the fraction. There is no fraction - there is no numerator either.) We multiply both parts of the formula by 2:

The elementary remains. We will provide the letter on the right 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 an easy task. And now a task based on the real version of the exam:

The locator of a bathyscaphe, evenly plunging vertically downward, emits ultrasonic pulses with a frequency of 749 MHz. The submersion rate of the bathyscaphe is calculated by the formula

where c = 1500 m/s is the speed of sound in water,

f 0 is the frequency of the emitted pulses (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 bathyscaphe sinks at a rate of 2 m/s.

"A lot of bukuff", yes ... But the letters are the 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'll do. Let's look at the formula:

Directly, of course, the letter f you can’t pull it out in any way, it is again hidden in a fraction. And both the numerator and the denominator. Therefore, the most logical step would be to get rid of the fraction. And there you will see. For this we apply second transformation - multiply both parts by the denominator.

We get:

And here is another rake. Please pay attention to the brackets in both parts! Often it is in these very brackets that errors in such tasks lie. More precisely, not in the brackets themselves, but in their absence.)

The brackets on the left mean that the letter v multiplies to the whole denominator. And not in its individual pieces ...

On the right, after multiplication, the fraction disappeared and left a single numerator. Which, again, the whole entirely multiplies by letter With. Which is expressed in parentheses on the right side.)

And now you can open the brackets:

Great. The process is underway.) Now the letter f left became common multiplier. Let's take it out of brackets:

There is nothing left. Divide both parts by 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 outside the bracket in the numerator and reduce the whole fraction by (-1), thereby getting rid of unnecessary minuses:

Here is an expression. And now you can substitute numerical data. We get:

Answer: 751 MHz

That's all. I hope the general idea is clear.

We make elementary identical transformations in order to isolate the variable of interest to us. The main thing here is not the sequence of actions (it can be any), but their correctness.

In these two lessons, only two basic identical transformations of equations are considered. 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, taking the root of both sides) will be an identical transformation if both sides of the equation are known to be non-negative.

Or, say, taking the logarithm of both sides of the equation will be the identical transformation if both sides obviously positive. And so on…

Such transformations will be considered 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 laps is calculated by the formula:

WhereV 1 AndV 2 are the average speeds (in km/h) for the first and second laps, respectively. What was average speed a 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 turned out to be 12 km/h?

Task based on real OGE option:

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. Use this formula to find the radiusR(in meters) if the angular velocity is 8.5 s -1 and the centripetal acceleration is 289 m / s 2.

Task based on a real variant profile exam:

To a source with EMF ε=155 V and internal resistancer\u003d 0.5 ohm they want to connect a load with resistanceROhm. The voltage across this load, expressed in volts, is given by:

At what load resistance will the voltage across it be 150 V? Express your answer in ohms.

Answers (in disarray): 4; 15; 2; 10.

And where are the numbers, kilometers per hour, meters, ohms - it's somehow themselves ...)

There are many ways to derive the unknown from the formula, but as experience shows, they are all ineffective. Reason: 1. Up to 90% of graduate students do not know how to correctly express the unknown. Those who know how to do this perform cumbersome transformations. 2. Physicists, mathematicians, chemists - people who speak different languages, explaining the methods of transferring parameters through the equal sign (they offer the rules of a triangle, a cross, etc.) The article discusses a simple algorithm that allows you to one reception, without repeated rewriting of the expression, draw the conclusion of the desired formula. It can be mentally compared with undressing a person (to the right of equality) in a closet (on the left): you cannot take off your shirt without taking off your coat, or: what is put on first is taken off last.

Algorithm:

1. Write down the formula and analyze the direct order of the actions performed, the sequence of calculations: 1) exponentiation, 2) multiplication - division, 3) subtraction - addition.

2. Write down: (unknown) = (rewrite inverse of equality)(the clothes in the closet (to the left of equality) remained in place).

3. The formula conversion rule: the sequence of transferring parameters through the equal sign is determined reverse sequence of calculations. Find in expression last action And postpone it through the equal sign first. Step by step, finding the last action in the expression, transfer here from the other part of the equality (clothing from a person) all known quantities. In the reverse part of the equality, the reverse actions are performed (if the trousers are removed - “minus”, then they are placed in the closet - “plus”).

Example: hv = hc / λm + mυ 2 /2

express frequencyv :

Procedure: 1.v = rewriting the right sidehc / λm + mυ 2 /2

2. Divide by h

Outcome: v = ( hc / λm + mυ 2 /2) / h

express υ m :

Procedure: 1. υ m = rewrite left side (hv ); 2. Sequentially transfer here with the opposite sign: ( - hc /λ m ); (*2 ); (1/ m ); (√ or degree 1/2 ).

Why is it transferred first - hc /λ m ) ? This is the last action on the right side of the expression. Since the entire right side is multiplied by (m /2 ), then the entire left-hand side is divisible by this factor: therefore, brackets are placed. The first action on the right side - squaring - is transferred to the left side last.

Each student knows this elementary mathematics with the order of operations in calculations. That's why All students quite easily without repeated rewriting of the expression, immediately derive a formula for calculating the unknown.

Outcome: υ = (( hv - hc /λ m ) *2/ m ) 0.5 ` (or write Square root instead of degree 0,5 )

express λ m :

Procedure: 1. λ m = rewrite left side (hv ); 2. Subtract ( mυ 2 /2 ); 3. Divide by (hc ); 4. Raise to a power ( -1 ) (Mathematicians usually change the numerator and denominator of the desired expression.)

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 the person himself. To describe the processes, physics uses a mathematical aggregate.

Instruction

1. Where do physical formulas? In a simplified way, the scheme for acquiring formulas can be presented as follows: a question is posed, conjectures are put forward, a series of experiments is carried out. The results are processed, certain formulas, and this prefaces the new physical theory or continues and develops more closely the existing one.

2. A person who comprehends physics does not need to go through each given difficult path again. Pretty master central views and definitions, familiarize yourself with the scheme of the experiment, learn to derive fundamental formulas. Of course, one cannot do without strong mathematical knowledge.

3. Comes out, learn the definitions physical quantities related to the topic under consideration. Every quantity has its own physical sense, one that you must understand. Let's say 1 pendant is the charge passing through the cross section of the conductor in 1 second at a current strength of 1 ampere.

4. Understand the physics of the process under consideration. What parameters describe it, and how do these parameters change over time? Knowing the basic definitions and understanding the physics of the process, it is easy to obtain the simplest formulas. As usual, directly proportional or inversely proportional dependences are established between values ​​or squares of values, and an indicator of proportionality is introduced.

5. Through mathematical reforms, it is allowed from primary formulas bring out the secondary. If you learn to do it easily and quickly, the latter will not be allowed to be remembered. The core method of reforms is the substitution method: some value is expressed from one formulas and is substituted into another. The main thing is that these formulas correspond to the same process or phenomenon.

6. Equations can also be added together, divided, multiplied. Time functions are often integrated or differentiated, getting new dependencies. Logarithm is suitable for power functions. At the end formulas rely on the result, the one that you want to get as a result.

Each human life surrounded by many different phenomena. Physicists are engaged in the comprehension of these phenomena; their tools are mathematical formulas and the 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 sources heat almost all of Greenland. The very word "physics" goes back to the Greek root "physis", which means "nature". Thus, physics itself is the science of nature and natural phenomena.

Forward to the future!

Often, physicists are literally "ahead of the times" by discovering laws that are put to use only decades (and even centuries) later. Nikola Tesla discovered the laws of electromagnetism, which are used today. Pierre and Marie Curie discovered radium with virtually no support, under conditions that are incredible for a modern scientist. Their discoveries helped save tens of thousands of lives. Now the physicists of every world are focused on the issues of the Universe (macrocosmos) and the smallest particles of matter (nanotechnology, microcosmos).

Understanding the world

The most important engine of society is curiosity. That is why the experiments at the Large Andron Collider are of such high importance and are sponsored by an alliance of 60 states. There is a real chance to reveal the secrets of society. Physics is a fundamental science. This means that any discoveries of physics can be applied in other areas of science and technology. Small discoveries in one branch can have a striking effect on the entire “neighboring” branch. In physics, the practice of research by groups of scientists from various countries, a policy of assistance and cooperation was adopted. The secret of the universe, matter worried the great physicist Albert Einstein. He proposed the theory of relativity, explaining that gravitational fields bend 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 laws of logic, making bold guesses. Physicists follow the development of mathematics, and occasionally the scientific discoveries of abstract science help explain hitherto unfamiliar natural phenomena. It also happens the other way around - physical discoveries push mathematicians to create guesses and a new logical unit. The connection between physics and mathematics, one of the most important scientific disciplines, reinforces the authority of physics.

Using the record of the first law of thermodynamics in differential form (9.2), we obtain an expression for the heat capacity arbitrary process:

Let us represent the total differential of internal energy in terms of partial derivatives with respect to the 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 obtain the general relationship between and .

Based on the obtained formula, 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 Equation

We express the partial derivatives on the right 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 the gas, therefore

From the thermal equation it is easy to obtain

We substitute (9.9) and (9.10) into (9.8), then

Let's finally write down

You, I hope, have learned (9.11). Yes, of course, this is Mayer's equation. We recall once again that Mayer's equation is valid only for an ideal gas.

9.3. Polytropic processes in an ideal gas

As noted above, the first law of thermodynamics can be used to derive equations for processes occurring in a gas. big practical use finds a class of processes called polytropic. polytropic is a process that takes place at a constant heat capacity .

The process equation is given by the functional relationship of two macroscopic parameters that describe the system. On the respective coordinate plane the process equation is visually represented in the form of a graph - a process curve. A curve representing a polytropic process is called a polytrope. The equation for a polytropic process for any substance can be derived from the first law of thermodynamics using its thermal and caloric equations of state. Let us demonstrate how this is done using the derivation of the process equation for an ideal gas as an example.

Derivation of the equation for a polytropic process in an ideal gas

The requirement of constant heat capacity in the process allows us to write the first law of thermodynamics in the form

Using Mayer's equation (9.11) and the ideal gas equation of state, 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 get

This is the polytropic equation in variables

Eliminating () from the equation, using equality, we obtain the polytropic equation in variables

The parameter is called the polytropic index, which can take, according to (), a variety of values, positive and negative, integer and fractional. The formula () hides many processes. 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 . An adiabatic process is a process that takes place without heat transfer (). There are two ways to implement this process. 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 a gas can be considered adiabatic due to its high speed.

It follows from the definition of heat capacity that in an adiabatic process . According to

where is the adiabatic exponent.

In this case, the polytropic equation takes the form

The adiabatic process equation (9.20) is also called the Poisson equation, so the parameter is often called the Poisson constant. The constant is an important characteristic of gases. From experience it follows that its values ​​for different gases lie in the range of 1.30 ÷ 1.67, therefore, on the diagram of processes, the adiabat "falls" more steeply than the isotherm.

Graphs of polytropic processes for different meanings are presented in fig. 9.1.

On fig. 9.1, the process schedules are numbered in accordance with Table. 9.1.

In order to derive a formula for a complex one, 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 the complex is expressed as a percentage, but it can also be expressed in any other numbers indicating the relationship the difference between the weight amounts of the elements that form a given substance. For example, the composition of alumina, containing 52.94% aluminum and 47.06% oxygen, will be completely determined if we say that and are connected in a weight ratio of 9:8, i.e., that by 9 wt. hours of aluminum accounts for 8 wt. hours of oxygen. It is clear that the ratio of 9:8 should equal the ratio of 52.94:47.06.

Knowing the weight composition of the complex and the atomic weights of the elements that form it, it is not difficult to find the relative number of atoms of each element in the molecule of the taken substance and thus establish its simplest formula.

Suppose, for example, that you want to derive the formula of 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 a calcium chloride molecule through X, and the number of chlorine atoms through y. Since a calcium atom weighs 40, and a chlorine atom 35.5 oxygen units, the total weight of the calcium atoms that make up the calcium chloride molecule will be 40 X, and the weight of chlorine atoms is 35.5 y. 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 ratios, we get:

40x: 35.5y = 36:64

Then we get rid of the coefficients for the unknowns X 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 a calcium chloride molecule, but they are fractional, while only an integer number of atoms can be contained in a molecule. To express attitude X:at two integers, we divide both terms of the ^ second relation by the smallest of them. We get

X: at = 1:2

Therefore, in a calcium chloride molecule, there are two chlorine atoms per calcium atom. This condition is satisfied whole line formulas: CaCl 2, Ca 2 Cl 4, Ca 3 Cl 6, etc. Since we do not have 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 them CaCl 2, indicating the smallest possible number of atoms in a calcium chloride molecule.

However, the arbitrariness in the choice of the formula disappears if, along with the weight composition of the substance, its molecular weight is also known. weight. In this case, it is not difficult to derive a formula expressing the true composition of the molecule. Let's take an example.

By analysis, it was found that glucose contains 4.5 wt. hours of carbon 0.75 wt. hours of hydrogen and 6 wt. hours of oxygen. Its molecular weight was found to be 180. It is required to derive the formula for glucose.

As in the previous case, we first find the ratio between the number of carbon atoms (atomic weight 12), hydrogen and oxygen in a glucose molecule. Denoting the number of carbon atoms through X, hydrogen through at and oxygen through z, make up the proportion:

2x :y: 16z=4.5:0.75:6

where

Dividing all three terms of the second half of the equation by 0.375, we get:

X :y:z= 1: 2: 1

Hence, the simplest formula glucose will be CH 2 O. But calculated from it would be 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 actual number of atoms in a molecule are called true or molecular formulas; formulas derived only from the data of analysis are called simple or empirical.

Acquainted with the conclusion chemical formulas," it is easy to understand how accurate molecular weights are established. As we have already mentioned, the existing methods for determining molecular weights in most cases do not give quite accurate results. But, knowing at least the approximate and percentage composition of a substance, it is possible to establish its formula, expressing the atomic composition of the molecule. Since the weight of a molecule is equal to the sum of the weights of the atoms that form it, adding up the weights of the atoms that make up the molecule, we determine its weight in oxygen units, i.e., the molecular weight of the substance. The accuracy of the found molecular weight will be the same as the accuracy of the atomic weights.

Finding the formula of a chemical compound in many cases can be greatly simplified by using the concept of ovality of elements.

Recall that the valency of an element is the property of its atoms to attach to themselves or replace a certain number of atoms of another element.

What is valence

element is determined by a number indicating how many hydrogen atoms(oranother monovalent element) attaches or replaces an atom of that element.

The concept of valency 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. IN inorganic chemistry the most important radicals are: 1) an aqueous residue, or hydroxyl OH; 2) acid residues; 3) basic balances.

An aqueous residue, or hydroxyl, is obtained if one hydrogen atom is taken away from a water molecule. In a water molecule, the hydroxyl is bonded to one hydrogen atom, therefore, the OH group is monovalent.

Acid residues are called groups of atoms (sometimes even one atom), "remaining" from acid molecules, if one or more hydrogen atoms, which are replaced by a metal, are mentally taken away from them. of these groups is determined by the number of taken away hydrogen atoms. For example, it gives two acid residues - one divalent SO 4 and the other monovalent HSO 4, which is part of various acid salts. Phosphoric acid H 3 RO 4 can give three acid residues: trivalent RO 4, divalent HPO 4 and monovalent

H 2 RO 4 etc.

We will call the main residues; atoms or groups of atoms "remaining" from base molecules, if one or more hydroxyls are mentally taken away from them. For example, successively subtracting hydroxyls from the Fe (OH) 3 molecule, we obtain the following main residues: Fe (OH) 2, FeOH and Fe. they are determined by the number of hydroxyl groups taken away: Fe (OH) 2 - monovalent; Fe(OH)-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 replacing two hydroxyls in Fe (OH) 3 with an acidic residue SO 4, the basic salt FeOHSO 4 is obtained, when replacing one hydroxyl in Bi (OH) 3

the acidic residue NO 3 produces the basic salt Bi(OH) 2 NO 3, etc.

Knowledge of the valences of individual elements and radicals allows in simple cases to quickly draw up formulas for very many chemical compounds, which frees the chemist from the need to memorize them mechanically.

Chemical formulas

Example 1 Write the formula for calcium bicarbonate, the acid salt of carbonic acid.

The composition of this salt should include calcium atoms and monovalent acid residues of HCO 3 . Since it is divalent, two acidic residues must be taken per calcium atom. Therefore, the salt formula will be Ca (HCO 3) g.