Derivation of the formula. Preparation for the Unified State Exam. Express a variable from a formula How to express numbers from formulas
To derive the formula of a compound, you must first of all establish, through analysis, 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 compound is expressed as a percentage, but it can be expressed in any other numbers indicating the ratio difference between the weight quantities of the elements forming a given substance. For example, the composition of aluminum oxide, containing 52.94% aluminum and 47.06% oxygen, will be completely defined if we say that and are combined in a weight ratio of 9:8, i.e., that at 9 wt. parts of aluminum account for 8 weight. including oxygen. It is clear that the ratio of 9:8 should equal the ratio of 52.94:47.06.
Knowing the weight composition of a complex substance and the atomic weights of its constituent elements, it is not difficult to find the relative number of atoms of each element in the molecule of a given substance 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 a calcium chloride molecule by X, and the number of chlorine atoms through u. Since a calcium atom weighs 40, and a chlorine atom weighs 35.5 oxygen units, the total weight of the calcium atoms that make up the calcium chloride molecule will be equal to 40 X, and the weight of chlorine atoms is 35.5 u. The ratio of these numbers, obviously, must be equal to the ratio of the weight quantities of calcium and chlorine in any amount of calcium chloride. But last attitude equals 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 the calcium chloride molecule, but they are fractional, whereas the molecule can only contain an integer number of atoms. To express attitude X:at two integers, divide both terms of the second ratio by the smallest of them. We get
X: at = 1:2
Consequently, in a calcium chloride molecule 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 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 these, CaCl 2, indicating the smallest possible number of atoms in a molecule of calcium chloride.
However, arbitrariness in choosing a formula disappears if, along with the weight composition of the substance, its molecular composition is also known weight. In this case, it is not difficult to derive a formula expressing the true composition of the molecule. Let's give an example.
By analysis it was found that glucose contains 4.5 wt. parts of carbon 0.75 wt. parts of hydrogen and 6 wt. including 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 the glucose molecule. Denoting the number of carbon atoms by 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, simplest formula glucose would be CH 2 O. But calculated from it would be 30, whereas in reality glucose is 180, i.e. six times more. Obviously, for glucose you need to take the formula C 6 H 12 O 6.
Formulas based, in addition to analytical data, also on the determination of molecular weight and indicating real number atoms in a molecule are called true or molecular formulas; formulas derived only from analysis data are called simplest or empirical.
Having become familiar with the conclusion chemical formulas,” it is easy to understand how precise molecular weights are determined. As we have already mentioned, 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 the substance, it is possible to establish its formula, expressing atomic composition molecules. Since the weight of a molecule is equal to the sum of the weights of the atoms that form it, by adding 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 molecular weight found will be the same as the accuracy of atomic weights.
Finding the formula of a chemical compound in many cases can be greatly simplified if we use the concept of ovality of elements.
Let us 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) adds or replaces an atom of that element.
The concept of valence extends 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) aqueous residue, or hydroxyl OH; 2) acid residues; 3) main balances.
An aqueous residue, or hydroxyl, is formed when one hydrogen atom is removed from a water molecule. In a water molecule, the hydroxyl is bonded to one hydrogen atom, therefore the OH group is monovalent.
Acidic residues are groups of atoms (and sometimes even one atom) that “remain” from acid molecules if you mentally subtract from them one or more hydrogen atoms replaced by a metal. of these groups is determined by the number of hydrogen atoms removed. For example, it gives two acidic residues - one divalent SO 4 and the other monovalent HSO 4, which is part of various acid salts. Phosphoric acidH 3 PO 4 can give three acidic residues: trivalent PO 4, divalent HPO 4 and monovalent
N 2 PO 4 etc.
We will call the main residues; atoms or groups of atoms that “remain” from base molecules if one or more hydroxyls are mentally subtracted 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 hydroxyl groups removed: Fe(OH) 2 - monovalent; Fe(OH) is divalent; Fe is trivalent.
Major 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 acid residues. Thus, 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 valencies of individual elements and radicals allows simple cases quickly compose formulas for many chemical compounds, which frees the chemist from the need to memorize them mechanically.
Chemical formulas
Example 1. Write the formula for calcium bicarbonate - an acid salt of carbonic acid.
The composition of this salt should include calcium atoms and monovalent acid residues HCO 3. Since it is divalent, then for one calcium atom you need to take two acidic residues. Therefore, the formula of 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 level - the level of small bodies comparable to the size of a person himself. To describe processes, physics uses a mathematical unit.
Instructions
1. Where do physical formulas? A simplified scheme for acquiring formulas can be presented as follows: a question is posed, guesses are made, a series of experiments is carried out. The results are processed and certain formulas, and this gives a preface to the new physical theory or continues and develops an existing one.
2. A person who comprehends physics does not need to go through each given difficult path again. It is enough to master the central concepts and definitions, become familiar with the experimental design, learn to derive fundamental formulas. Of course, you can’t do without strong mathematical knowledge.
3. It turns out, learn the definitions physical quantities related to the topic under consideration. Every quantity has its own physical meaning, one that you must understand. Let's say 1 coulomb is a charge passing 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. What parameters does it describe, 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 relationships are established between quantities or squares of quantities, and a proportionality index is introduced.
5. Through mathematical reforms it is possible to derive secondary ones from primary formulas. If you learn to do this easily and quickly, you won’t have to remember the latter. The core method of reform is the method of substitution: some value is expressed from one formulas and is substituted into another. The main thing is that these formulas corresponded to the same process or phenomenon.
6. Equations can also be added, divided, and multiplied. Time functions are often integrated or differentiated, obtaining new dependencies. Logarithm is suitable for power functions. In the end formulas rely on the result, the one you want to get as a result.
Every human life is surrounded by the most varied phenomena. Physicists are dedicated to understanding these phenomena; their tools are mathematical formulas and the achievements of their predecessors.
Natural phenomena
Studying nature helps us to be smarter about existing sources and discover new sources of energy. So, geothermal sources heat approximately the entire Greenland. The word “physics” itself comes from 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 their time,” discovering laws that are used only tens of years (and even centuries) later. Nikola Tesla discovered the laws of electromagnetism, which are used today. Pierre and Marie Curie discovered radium virtually without support, under conditions incredible for a modern scientist. Their discoveries helped save tens of thousands of lives. Now physicists of every world are focused on questions 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 experiments at the Large Hadron Collider are so important and are sponsored by an alliance of 60 countries. There is a real chance of revealing 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 dramatic effect on the entire “neighboring” branch. In physics, the practice of research by groups of scientists from different countries is famous; a policy of assistance and cooperation has been adopted. The mystery of the universe and matter worried the great physicist Albert Einstein. He proposed the theory of relativity, which explains that gravitational fields bend space and time. The apogee of the theory was the well-known formula E = m * C * C, combining energy with mass.
Union with mathematics
Physics relies on the latest mathematical tools. Often, mathematicians discover abstract formulas by deriving new equations from existing ones, using higher levels of abstraction and the laws of logic, making bold guesses. Physicists follow the development of mathematics, and occasionally scientific discoveries Abstract science helps explain hitherto unknown natural phenomena. It also happens, on the contrary, that 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.
There are many ways to derive an unknown from a formula, but as experience shows, all of them are 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 methods for transferring parameters through the equal sign (they offer the rules of triangle, cross, etc.) The article discusses a simple algorithm that allows one reception, without repeated rewriting of the expression, deduce the desired formula. It can be mentally compared to a person undressing (to the right of equality) in a closet (to 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 the inverse of the equality)(the clothes in the closet (to the left of the equality) remained in place).
3. 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 equals sign first. Step by step, finding the last action in the expression, transfer here all known quantities from the other part of the equation (clothing per person). In the reverse part of the equation, the opposite actions are performed (if the trousers are taken off - “minus”, then they are put in the closet - “plus”).
Example: hv = hc / λ m + mυ 2 /2
Express frequencyv :
Procedure: 1.v = rewrite the right sidehc / λ m + mυ 2 /2
2. Divide by h
Result: v
= (
hc
/
λ m
+
mυ
2
/2) /
h
Express υ m :
Procedure: 1. υ m = rewrite left side (hv ); 2. Consistently move 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 hand side is multiplied by (m /2 ), then the entire left side is divided by this factor: therefore, parentheses are placed. The first action on the right side, squaring, is transferred to the left side last.
Every student knows this elementary mathematics with the order of operations in calculations very well. That's why All students quite easily without rewriting the expression multiple times, immediately derive a formula for calculating the unknown.
Result: υ = (( hv - hc /λ m ) *2/ m ) 0.5 ` (or write square root instead of a 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.)
Using the notation of the first law of thermodynamics in differential form (9.2), we obtain an expression for the heat capacity of an arbitrary process:
Let's imagine full differential internal energy through partial derivatives with respect to parameters and:
After which we rewrite formula (9.6) in the form
Relationship (9.7) has independent significance, 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.
Let us consider the process at constant pressure and obtain a general relationship between and .
Based on the formula obtained, one can easily find the relationship between the heat capacities 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 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
From the thermal equation it is easy to obtain
Let's substitute (9.9) and (9.10) into (9.8), then
We'll finally write it down
I hope you found out (9.11). Yes, of course, this is the Mayer equation. Let us 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 application 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 connection between two macroscopic parameters describing the system. On the corresponding coordinate plane the process equation is clearly presented in the form of a graph - a process curve. A curve depicting a polytropic process is called a polytrope. The equation of a polytropic process for any substance can be obtained based on 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 constant heat capacity during the process allows us to write the first law of thermodynamics in the form
Using the Mayer 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
Integrating (9.15), we obtain
This is a 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. Behind the formula () there are many processes hidden. The isobaric, isochoric and isothermal processes known to you are special cases of 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 that can change its volume. The second is to carry out such a fast process that 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 an 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 Poisson’s equation, therefore the parameter is often called Poisson’s constant. Constant is an important characteristic of gases. From experience it follows that its values for different gases lie in the range 1.30 ÷ 1.67, therefore, on the process diagram, the adiabatic “falls” more steeply than the isotherm.
Graphs of polytropic processes for different meanings are presented in Fig. 9.1.
In Fig. 9.1 process graphs are numbered in accordance with table. 9.1.
