Clapeyron-Clausius equation. Clapeyron-Clausius equation Clapeyron-Clausius equation chemistry
Clapeyron-Clausius equation (differential).
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Liquid-vapor equilibrium curve or, which is the same, pressure dependence saturated vapors on temperature, for a given substance can be found by solving the so-called Clapeyron–Clausius equation.
The method of thermodynamic cycles allows us to establish the form of this equation. Let there be two experimental isotherms, which correspond to temperatures and (Fig. 50).
In the initial state 1, one mole of liquid is under pressure at temperature and occupies volume (there is no vapor phase). By adding heat, we evaporate this mole of liquid isothermally. In this case, the liquid will always be under pressure equal to the pressure of its saturated vapor. As a result, the liquid will turn into a gaseous state 2. The amount of heat required to convert one mole of liquid into a gaseous state at constant temperature and pressure is, by definition, the latent heat of vaporization. Thus, at transition 12 the liquid received an amount of heat equal to .
The work done per cycle is numerically equal to the area of the quadrilateral 1234:
(3.3.1)
and cycle efficiency
.
(3.3.2)
On the other hand, the 1234 cycle is a Carnot cycle, so its efficiency is
(3.3.3)
Comparing expressions (3.3.2) and (3.3.3), we obtain
.
(3.3.4)
Equation (3.3.4) can be presented in a slightly different form. To do this, we will find the change in entropy during the transition of a liquid from state 1 to gaseous state 2 (Fig. 50).
,
(3.3.5)
where and are the entropies of liquid in state 1 and vapor in state 2, respectively. Using the last relation, equation (3.3.4) can be written as follows:
.
(3.3.6)
The differential equation (3.3.4) is the Clapeyron–Clausius equation. This equation represents the change in saturated vapor pressure as the temperature changes by . To find a solution to this equation, it is necessary to know the dependence of the latent heat of evaporation on temperature, as well as the dependence of the volumes of liquid and gas on temperature and pressure. In general, this equation is nonlinear differential equation, the solution of which is found by numerical integration.
At temperatures significantly lower than the critical one, the volume occupied by the liquid is much less than the volume occupied by the liquid, i.e. For example, the volume of water vapor at 1600 times the volume of water, boiling liquid oxygen at a temperature of – C is accompanied by an increase in volume by approximately 300 times. Therefore, at temperatures, equation (3.3.4) will take the form
. (3.3.7)
In the same approximation, saturated steam can also be considered an ideal gas. This allows us to write equation (3.3.7) as:
. (3.3.8)
Integrating the left side of equation (3.3.8) in the range from to , and the right side – in the range from to , we obtain
,
(3.3.9)
where is the saturated vapor pressure obtained from experiment at temperature, and is the average value of the heat of evaporation over the temperature range
From formula (3.3.9) it is clear that the pressure of saturated vapors increases exponentially with increasing temperature.
The Clapeyron–Clausius equation rewritten as
, (3.3.10)
as we will show later, it also determines the dependence of the boiling temperature of the liquid on external pressure. In equation (3.3.10)
is the change in boiling point when the external pressure changes by . Therefore, the liquid-vapor equilibrium curve is also called the boiling curve.
52. Van der Waals equation – The equation of state is not ideal gases. Experimental determination of the van der Waals equation constants.
As noted earlier, at low temperatures and high pressures the Mendeleev–Clapeyron equation for one mole of a substance
gives significant deviations from experimentally measured values.
Numerous attempts have been made to find an equation of state for real matter, which could cover, if not all states of matter, then at least gaseous and liquid. Of the many proposed equations, the most famous is the van der Waals equation:
,
(3.4.2)
written for one mole of a substance. For moles this equation is:
.
(3.4.3)
Constants are determined experimentally and have different values for different types of molecules. Equation (3.4.2) is not derived; it is established by introducing into famous equation Mendeleev–Clapeyron two amendments. To justify their introduction, we note that in equation (3.4.2) volume means the volume of a vessel containing one mole of gas. In the case of an ideal gas consisting of material points, this entire volume is available for the movement of molecules. In a real gas, the molecules themselves occupy a certain part of the volume of the container, and this part is inaccessible to all other molecules. This portion of the volume should be subtracted from the volume. Then equation (3.4.2) takes the form
. (3.4.4)
From the last expression it is clear that the correction is equal to the volume that the gas would occupy at infinitely high pressure, i.e., the molecules of a real gas cannot approach each other to a distance equal to zero, even at infinitely high pressure. Therefore, introducing a correction means approximately taking into account the repulsive forces between molecules.
As we know, not only repulsive forces act between molecules, but also attractive forces. Any molecule located near the wall of a vessel experiences a net force of attraction
from the side of molecules located in the sphere of action of attractive forces (Fig. 51).
Let's select a platform on the wall surface. Let there be molecules on it. Then the resulting force acting on the molecules of this area from the gas
,
(3.4.6)
since from the conditions of symmetry all forces have the same magnitude and direction. If we divide the force by the area, we get the so-called molecular pressure, with which the molecules located at the wall act on the rest of the gas mass:
. (3.4.7)
Each of the factors in formula (3.4.7) is obviously proportional to the density of the gas, which, in turn, is inversely proportional to the volume of the gas, so we can write:
, (3.4.8)
Where a– positive constant coefficient.
Thus, as a result of the action of attractive forces, the pressure on the wall from the gas side will be less than the pressure (3.4.4) that the wall would experience if there were no attractive forces between the molecules, i.e.
.
Where do we find the van der Waals equation:
.
(3.4.9)
Let us explain the appearance of additional pressure in formula (3.4.9). Let the gas be in a cylinder under a weightless piston. External pressure tends to compress the gas, that is, to bring its molecules closer together. If gas molecules did not attract each other, the gas would experience only external pressure. But the mutual attraction of molecules, as we have found out, also tends to bring the molecules closer to each other, that is, it acts in the same direction as external pressure. Therefore, the result of the attraction of molecules is reflected in an apparent increase in external pressure on the gas, as if some additional pressure had been added to the pressure on the piston.
Experimental determination of the van der Waals equation constants
For experimental determination of constants a and the gas under study is placed in a closed volume vessel with a built-in pressure gauge and the pressure of this gas is measured at different temperatures. By numerical differentiation of the experimentally obtained curve we determine the partial derivative. From the van der Waals equation we find this derivative
.
(3.5.1)
From here we get the value:
. (3.5.2)
Substituting expression (3.5.2) into the van der Waals equation (3.4.9), we calculate another quantity a:
.
(3.5.3)
Experience has shown that the values a and are not constants, but depend on temperature, although weakly. In calculations using the van der Waals equation, as constants a and take the average values of the functions and in the temperature range of interest
53. Isotherms of the van der Waals equation and their comparison with experimental isotherms. Determination of critical parameters of a substance from the VdW equation. Metastable states of matter are supersaturated steam and superheated liquid. Wilson chamber and bubble chamber.
In Fig. 52 shows van der Waals gas isotherms.
At very high temperatures they have a shape close to hyperbolas; these isotherms characterize the gaseous state of the substance (almost ideal gas). As the temperature decreases, the shape of the isotherm changes and at a certain temperature (critical) it detects an inflection point of the curve. At even lower temperatures (subcritical), instead of a horizontal section corresponding to the liquid-vapor phase transition, the isotherms have a wave-like section (Fig. 53).
Rice. 52 R and p. 53
Measurements show that the isotherms of real matter practically coincide with the van der Waals isotherm in the sections (gaseous state) and (liquid state). However, in the middle part, instead of the horizontal section 51, corresponding to the liquid-vapor phase transition, the van der Waals isotherm has a wave-like section. This wavy section is characterized as follows. Section 12 corresponds to the metastable state of steam (supersaturated steam), and section 54 corresponds to the metastable state of liquid (superheated liquid). At point 1 there is only saturated steam, and at point 5 there is only boiling liquid. As for section 234 of the wavy curve, it is physically impossible, since there are no substances in nature for which, at a constant temperature, an increase in volume would lead to an increase in pressure. The latter is possible only if the temperature in this area is not constant. Supersaturated steam (section 12) is a gaseous state of a substance in which the pressure p greater than the saturated vapor pressure at a given temperature - can be obtained experimentally by compressing a pure gas to a pressure greater than the saturated vapor pressure, and it will not condense. The state of supersaturated vapor, although it has a certain stability, is less stable than the two-phase state (isobar 135), in which, as we know, part of the substance is in the form of a liquid, and part is in the form of saturated vapor. Therefore, with little external influence the supersaturated vapor partially turns into liquid, and the remaining vapor becomes saturated.
Superheated liquid (section 45) - a condition characterized by the fact that it exists at a pressure lower than the saturated vapor pressure at a given temperature - can be obtained by prolonged boiling of pure liquid, as a result of which gas bubbles (nuclei of vaporization) are removed from the liquid, and the liquid is heated to a temperature above its boiling point at a given pressure. The state of a superheated liquid also turns out to be less stable than the state of equilibrium between the liquid and saturated steam. If particles of a foreign substance are introduced into such a superheated liquid, it quickly transitions to a two-phase state.
The states of supersaturated steam and superheated liquid are used in devices nuclear physics(Cloud chamber and bubble chamber) for recording and measuring the parameters of elementary particles.
If on the van der Waals isotherm the wavy section is replaced by some horizontal straight line 135, then the isotherm obtained in this way will qualitatively correctly describe the two-phase state of the substance. The position of this line can be determined if the second law of thermodynamics in Clausius notation (2.13.7) is applied to the closed reversible cycle 1234531:
. (3.6.1)
Since along the entire path 1234531 the temperature of the substance remains unchanged (since this path is made up of sections of two possible variants of the same isotherm), the last equation can be written in the form
. (3.6.5) with experimental data indicate that the van der Waals equation, from which this value is obtained, is approximate, although the qualitative picture of the change in the state of matter is conveyed by the equation quite correctly. Known big number attempts to obtain a more accurate equation of state of matter. However, these equations contain a large number of correction factors, physical meaning which is unclear, as in the van der Waals equation.
In the most general view Using the methods of statistical physics, Academician N. N. Bogolyubov obtained the equation of state
,
(3.6.12)
where are the so-called virial coefficients, which are functions only of temperature. From the Bogolyubov equation it follows that the larger the value of the molar volume, the smaller the number of terms of the series should be taken into account to obtain a fairly accurate result. When all terms of the power series vanish, and equation (3.6.12) takes the form , i.e., as one would expect, the Bogolyubov equation turns into the Mendeleev–Clapeyron equation. Virial coefficients cannot be calculated purely theoretical methods and therefore must be determined using experimental data. However, this problem turns out to be so complex that it is more appropriate to obtain the equation of state simply in the form of an interpolation formula describing the experimental data.
The heat of vaporization increases as the temperature and vapor pressure increase.
Most often, the Clapeyron-Clausius equation is used to simply calculate or measure the heat of vaporization of various substances. By measuring the vapor pressure at different temperatures and plotting it on a graph with the ln value on one axis p, and on the other hand, the value 1/T, scientists based on the received linear dependence(the angle of inclination of the straight line) determine the heat of evaporation of the substance.
Benoît Paul Émile Clapeyron, 1799-1864
French physicist and engineer. Born in Paris. Graduated from the Polytechnic School and the School of Mining. In 1820-1830 worked at the Institute of Railway Engineers in St. Petersburg. Upon returning to France, he became a professor at the School of Bridges and Roads in Paris. He became famous as a designer railways, designer of railway bridges and steam locomotives. He proved the “three moment theorem”, used to calculate load-bearing structures with three or more points of support. However, Clapeyron made his greatest contribution to science through the study of thermal processes, for which he was elected a full member of the French Academy of Sciences.
Rudolf Julius Emanuel Clausius, 1822-88
German physicist. Born in Kieslin (now Koszalin, Poland) in the family of a pastor. Studied at private school, of which his father was the director. In 1848 he graduated from the University of Berlin. After graduating from university, he preferred physics and mathematics to history, which he initially studied, taught in Berlin and Zurich, and served as a professor of physics at universities in Zurich, Würzburg and Bonn. Since 1884 - rector of the University of Bonn. Clausius's main works are devoted to the fundamentals of thermodynamics and kinetic theory gases Unfortunately, severe injuries received while serving as a volunteer corpsman during Franco-Prussian War, prevented Clausius from realizing his full scientific potential. However, after the war and injuries, it was he who formulated the second law of thermodynamics in its modern form.
THERMAL EFFECTS OF PHASE TRANSITIONS.
CLAPEYRON – CLAUSIUS EQUATION.
The transition of a component from one phase to another is accompanied by the release or absorption of heat, which can be quantified based on the fundamental equation of thermodynamics:
DERIVATION AND ANALYSIS OF THE CLAPEYRON – CLAUSIUS EQUATION.
For any equilibrium transition of a substance from one phase α to another phase β, applying the equation ( * ) to each of the phases, we can write
The indices α and β reflect the parameters belonging to the corresponding phase. Under equilibrium conditions, there is no change in the Gibbs energy between phases α and β, i.e.
,
P 
equating the right-hand sides of equations 1 and 2, we obtain
For an equilibrium reversible process according to the equations 
And 
let's write down
and equation (3) will take the form
,
where ∆H per – heat of phase transition.
The thermal effect accompanying the phase transition is defined as follows:
the equation
Clapeyron–Clausius
where ∆V is the change in volume as a result of the phase transition; dP/dT – change in pressure depending on temperature while maintaining equilibrium between the two phases.
The Clapeyron–Clausius equation relates the thermal effect of the process to changes in saturated vapor pressure, temperature, and volume changes during a phase transition.
For the processes of evaporation l→p and sublimation tv→p, the Clapeyron–Clausius equation can be represented as follows:
where ∆H isp, ∆H sub are the heats of evaporation and sublimation; V p, V liquid, V solid are the molar volumes of vapor, liquid and solid, respectively.
During the process of evaporation and sublimation, a significant change in the specific volume ∆V and a significant change in the value of dP/dT are observed. During melting, on the contrary, the change in ∆V is small and the value of dP/dT is insignificant.
Example 1. Let us calculate the melting temperature of phenol Tm using the Clapeyron–Clausius equation. The density of solid phenol ρ solid at atmospheric pressure is 1.072∙10 3 kg/m 3, and of liquid phenol ρ l = 1.056∙10 3 kg/m 3; heat of fusion ∆H pl = 1.045∙10 5 J/kg; freezing temperature 314.2 K. Let's determine dP/dT and melting temperature at P = 5.065∙10 7 Pa:
The increase in melting temperature with an increase in pressure by 1 atm (1.013∙10 5 Pa) is 4.525∙10 -8 deg/Pa. When the pressure increases to 5.065∙10 7 Pa, the melting temperature increases by ∆T = (dT/dP)∆P = 4.525∙10 -8 ∙ 5.065∙10 7 = 2.29 K, i.e. will be Tpl = 314.2+2.29 = 316.49 K.
It should be borne in mind that during the melting process for most substances V liquid > V solid, then ∆V > 0 and with increasing pressure P the melting temperature T increases.
However, substances such as water (H 2 O), bismuth (Bi), have a volume of the solid phase V tv greater than the volume of the liquid phase V l< V тв. Тогда в процессе плавления этих веществ изменение мольного объема ∆V будет <0 и при повышении давления Р температура плавления будет уменьшаться Т↓
EXAMPLE 2. The sliding of skates on ice is caused by the formation of water in the sliding plane, which acts as a liquid lubricant. Previously, it was believed that the formation of water occurs due to the melting of ice under the pressure of a sharp ridge. However, thermodynamic calculations using the Clapeyron–Clausius equation do not confirm this. Indeed, the specific volume of water (l) and ice (tv) are equal, respectively, to V l beat = 10 -3 m 3 /kg and V hard beat = 1.091·10 -3 m 3 /kg; heat of fusion ∆H pl = 332.4 kJ/kg:
Solution:
This value shows that to lower the melting temperature of ice by one degree Kelvin, it is necessary to increase the pressure by 1.34∙10 7 Pa, i.e. approximately 134 atmospheres, which is unrealistic, since the ice cannot withstand such pressure - it cracks.
Ice melting occurs primarily as a result of friction and the conversion of work into heat as the skate slides across the ice, rather than by increasing pressure on the ice.
Equation for the evaporation process 
can be represented in integral form. The molar volume of vapor significantly exceeds the molar volume of liquid, V p >> V l, i.e. the value of V can be neglected. Then the Clapeyron–Clausius equation will be written as:
Steam obeys the ideal gas laws: PV=RT 
, Then 
, we transform the equation by rearranging the pressure P to the left side of the equation, and dT to the right side. We get:
or
Let us integrate equation (1) in the range from T 1 to T 2 and, accordingly, from P 1 to P 2, provided that in the region of low steam pressures ∆Н isp ≈ const; as a result of integration we obtain:
∆Н isp / R = const, put outside the integral sign
Using equation (2), you can graphically determine the values of the heat of evaporation if the pressures P 1 and P 2 and the corresponding evaporation temperatures T 1 and T 2 are known. To do this, it is necessary to plot the inverse temperature values on the abscissa axis, and lnP on the ordinate axis.
The dependence of lnP on 1/T will be linear, and the slope of this straight line is equal to 
, i.e. 
, A 
The calculated values of ∆Н isp are obtained with an accuracy sufficient for practice, not inferior to the accuracy of direct measurement. It is possible to use equation (2) for reverse calculation, when the change in pressure with a change in temperature during the evaporation process is determined by the value of ∆H evap.
The heat of phase transitions can also be determined by the standard enthalpy of formation, depending on the phase state of the reaction products.
Example. This is best illustrated by the example of the heat of formation of water from gaseous oxygen and hydrogen, which is
H 2 (g) + 1/2O 2 (g) = H 2 O (g), (l), (solv)
for water vapor ∆Н (g) 0 = -241.82 kJ/mol; for water in liquid state∆H (l) 0 = -285.83 kJ/mol; for ice ∆Н (tv) 0 = -291.82 kJ/mol. The heat of condensation of water is equal to:
and the heat of transformation of water into ice:
As can be seen, the thermal effect of phase transitions is significantly less than the heat of formation of substances.
As a result of phase transitions, entropy change. We present such changes depending on temperature in the figure.
As is known, the entropy of an ideal crystal at absolute zero is zero. With increasing temperature, atoms (ions) fluctuate relative to their equilibrium position, the number of possible ways of their placement increases, and entropy increases (ΔS>0). When the melting temperature is reached (point A in the figure), the crystal lattice is destroyed abruptly (segment AB), the thermodynamic probability of the system W increases, and in accordance with the formula S=k∙lnW (where k is the Boltzmann constant) entropy during the transition from solid to liquid state growing. A more significant jump in entropy occurs during the transition from a liquid to a gaseous state (VG segment), when the short-range order of the arrangement of particles relative to each other is violated and the movement of particles becomes chaotic.
Example. Let us estimate the entropy jump using the example of phase transitions of water:
,
when the standard absolute values of entropy are known S TV 0 =39.4; S f 0 =69.9; S g 0 =188.7 J/(mol K).
Then we have
According to the picture for water
Based on the known enthalpy of the phase transition, the change in entropy can be calculated in accordance with the formula 
Example. Let's calculate the change in entropy during the vaporization of 1 mole of ethyl chloride at 12.3 0 C, when the heat of evaporation ∆H exp = 24.16 kJ/mol.
Molecular mass 
= 64.5 g/mol.
In conclusion, I note that we considered only first-order phase transitions. During phase transitions of the first kind, the properties of substances, expressed, for example, through the chemical potential, by the first derivatives of one of the characteristic functions, change abruptly with a continuous change in the corresponding parameters: temperature, pressure, volume and entropy. In this case, the transition heat ∆Н per is released or absorbed in accordance with the Clapeyron–Clausius equation.
In addition to them, however, there are phase transitions of the second order. They are not accompanied by the release or absorption of heat; for them the Clapeyron–Clausius equation loses its meaning. These transitions characterize changes in the system that are not determined by the volume and energy supply. In this case, the first derivatives of one of the characteristic functions are continuous, and the second derivatives (for example, heat capacity) change abruptly. Phase transitions of the second kind include transitions from a paramagnet to a ferromagnet, a dielectric to a ferroelectric, as well as the processes of superfluidity, superconductivity, etc.
Currently, there are about 400 solid minerals for which phase transitions of the second order are observed: rutile, anatase, diamond and especially quartz, which has seven modifications, and along with phase transitions of the first order, phase transitions of the second order are observed. Thus, at 573 0 C and the transition of quartz modification β 
α heat capacity and the coefficient of linear expansion change abruptly (1st order), but at the same time heat of 10.9 kJ/mol is absorbed (2nd order).
Determination of the heat of vaporization of a liquid
For equilibrium between liquid and vapor, the Clapeyron-Clausius equation can be derived as follows.
According to the second law of thermodynamics, if a system is in equilibrium, then under given conditions ( R, T) no work is done
Ap = 0; G and -G P = 0; G and = G P (2)
that is, the Gibbs free energies of a pure liquid and its vapor are equal (these are equilibrium conditions).
Let's change one of the conditions that determine equilibrium, for example, temperature (that is, we heat the system by T), as a result of this a new pressure will be established, a new liquid-vapor equilibrium:
G " and = G " P (3)
or G and = G P (4)
If a change in the external parameter occurs by an infinitesimal amount - dT, then the isobaric-isothermal potentials will change by an infinitesimal value:
dG and =dG P (5)
The isobaric-isothermal potential represents the free energy of the system and is a function of pressure and temperature:
dG=VdP–SDT(6)
therefore, it can be used to establish a quantitative relationship between saturated vapor pressure and temperature.
After substituting (6) into (5) we have in the equilibrium state
V and dP-S and dT = V P dP-S P dT(7)
from where (8)
Where ( S P -S and) And ( V P - V and) - change, respectively, in the entropy and volume of the system during the transition of a substance from a liquid to a vapor state. According to the second law of thermodynamics, the change in entropy of a system during evaporation is equal to the reduced heat of evaporation
(9)
Substituting expression (9) into (8) we get
(10)
where ∆ H isp and ∆ V isp- respectively, the change in enthalpy and volume during evaporation, and T bale- boiling temperature.
Equation (10) called the Clapeyron-Clausius equation . According to (10), the slope of the lines in the phase diagram of water (Fig. 1) is determined by the sign of the derivative dP/dT or its reciprocal dT/dP- characterizing the change in temperature with increasing pressure.
Of the phase transitions, we consider evaporation and melting. The heat of evaporation - the transition of the liquid phase into the gaseous phase - is positive. Molar heat of vaporization is the amount of heat expended to evaporate one mole of a liquid substance. The volume of gas during evaporation is always greater than the corresponding volume of liquid, that is, in equation (10) V P >V and. That's why dP/dT, and therefore dT/dP are also always positive ( dT/dP> 0). Consequently, the evaporation temperature always increases with increasing pressure (curve OK in Fig. 1 or see table. 1 Appendix). With increasing temperature, the saturated vapor pressure above the liquid increases, reaching its maximum value at the critical temperature. The latter is the limiting temperature (for example, for water it is 374.12 o C) at which equilibrium between the liquid and vapor phases of a substance is possible. At higher temperatures, a substance can only be in a gaseous state, and the concept of saturated vapor loses its meaning.
The heat of fusion - the transition of the solid phase into the liquid phase is also always positive.
Rice. 1. Diagram of the state of water at low pressures
Regions: 1 - solid phase (ice); II- liquid; III - par.
Curves: AO - sublimation; OK - evaporation; OV - melting.
O is a triple point corresponding to the equilibrium of three phases.
The volume of the liquid phase in the general case can be greater or less than the volume of the same amount of solid phase. From here, in accordance with equation (10), it follows that the value dP/dT or its inverse value dT/dP, can be positive or negative. This means that the melting point can rise or fall with increasing pressure. Magnitude dT/dP positive for most substances. It has a negative value only for water, bismuth and a few other substances for which the density of the liquid at the melting point is greater than the density of the solid phase ( V and -V T) < 0. В связи с этим при увеличении давления температура плавления льда понижается (криваяOB).
It should be noted that the considered patterns are valid for low pressures.
The Clapeyron-Clausius equation (10) can be transformed by taking the following approximations:
1) Since ∆ V isp =(V P -V and) >> 0 (for example, for water, the molar volume in the vapor state at no. V P≈ 22400 cm 2, and in the liquid state V and≈ 18 cm 3), then without a large error we can neglect the value V and and accept that ∆ V isp ≈ V P .
2) At not too high pressures and temperatures (far from critical), the equation of state for ideal gases can be applied to real systems. The error obtained in this case turns out to be insignificant.
(11)
Substituting (11) into (10) we get:
(12)
which after transformation
(13)
takes the form
(14)
Heat of vaporization depends on temperature : As the temperature increases, the heat of vaporization decreases. At the critical temperature, the heat of evaporation is zero. However, at temperatures far from critical, changes in ∆ N isp with temperature are not very great. In a not too wide temperature range∆ N isp can be considered constant.
Integration of the Clapeyron-Clausius equation (14) within the temperature range T 1 and T 2, which correspond to pressures R 1 and R 2 at constant ∆ N isp, gives
(15)
or when moving to decimal logarithms
(16)
(R- universal gas constant equal to 8.314 J/mol K).
Equations (15), (16) allow us to calculate the heat of evaporation. To do this, based on experimental data, a dependence is built lnP=f(1/T) or lgP=f(1/T) and two points are selected on the resulting straight line (Fig. 2). Substitute the values of the logarithm of pressure and inverse temperature corresponding to these points into equation (17):
(17)
∆ value N isp according to equation (17) depends on the temperature interval taken and the closer to the true one, the smaller this interval. However, such a calculation requires very precise measurements of boiling point and vapor pressure.
To calculate ∆ N isp in a relatively wide temperature range (50...100 o C), you should select points that most accurately fit the straight line lgP=f(1/T).
The boiling point of a liquid, the vapor pressure at a given temperature and the heat of vaporization are specific constants of matter , the values of which are necessary for many theoretical and practical calculations. Based on these data, it is possible to determine the purity of chemicals, calculate the separation of mixtures by distillation, and calculate the energy costs for the evaporation of liquid required to carry out the reaction in the gas phase.
THERMODYNAMICS
PART II
PHASE EQUILIBRIA.
THERMAL ANALYSIS
Educational and methodological manual
Berezniki 2011
Reviewer:
Candidate of Technical Sciences, Associate Professor of the Department of Chemical Technology and Energy Dyblin B.S.
(Berezniki branch of Perm State Technical University)
Kolbasina, V.D.
K60 Thermodynamics. Part II. Phase equilibria. Thermal analysis: educational method. allowance / V.D. Sausage. – Berezniki branch of Perm State Technical University. – Berezniki, 2011. – 53 p.
The benefit is fully consistent with the programs for physical chemistry for engineering students and is intended to acquire skills independent work when solving problems, as well as to prepare for laboratory practical work.
The manual gives an idea of the phase components of thermodynamic systems and the essence of thermal analysis, explains the terms used in their definition and the basic principles of the method of thermal analysis of thermodynamic systems. It provides examples of solutions to thermodynamic equilibrium and thermal analysis examples, along with the construction of fusibility diagrams. Examples of registration and execution of calculation work are given.
Intended for students studying the course “Physical Chemistry”.
ISBN © State Educational Institution of Higher Professional Education
"Perm State
Technical University", 2011
1. Clapeyron–Clausius equation. 4
1.1. Melting. 6
1.2. Evaporation (sublimation) 9
2. Thermodynamic equilibrium. Gibbs phase rule. 12
2.1. Diagram of the state of water in the medium pressure region. 16
3. Two-component systems... 18
3.1. Systems with complete insolubility of components in the solid state 20
3.1.1. Non-isomorphic two-component systems with simple eutectics 20
3.1.2. Non-isomorphic two-component systems forming a stable chemical compound. 29
3.1.3. Non-isomorphic two-component systems that form two new chemical compounds. 31
3.1.4. Non-isomorphic two-component systems forming an unstable chemical compound. 32
3.2. Isomorphic systems (systems with solid solutions) 36
4. Three-component systems... 43
5. Thermal analysis. 47
5.1. Experimental part. 51
Bibliographic list. 52
Clapeyron–Clausius equation
Processes that involve the transformation of one phase of a substance into another of the same substance, occurring without chemical reactions, are called phase transformations (melting, sublimation, evaporation, polymorphic transformations).
When a system consisting of several phases reaches equilibrium, the transition of molecules from one phase to another does not stop. For example, in an equilibrium water-vapor system, molecules constantly move from liquid to vapor and back. Equilibrium is characterized by equality of the rates of evaporation and condensation. Equilibrium is thus maintained by two opposing processes proceeding at the same speed.
Phase equilibria, of course, can be established in other systems, for example, in liquid-liquid systems. solid, or solid-gas, etc.
Equilibrium state at constant R And T thermodynamically characterized by the equality of the Gibbs energies of one and the other phase: , i.e. isobaric-isothermal potentials of a pure substance in two phases that are in equilibrium, with equal molecular weights, are equal to each other.
When the Gibbs energy of one phase changes to maintain equilibrium in the system, the Gibbs energy of another phase changes by the same amount, i.e.
Change in isobaric-isothermal potential DG can only happen through change R And T, because G = ƒ ( P, T).
This dependence is expressed in general form by the equation
Therefore, for two adjacent phases that are in equilibrium, we write
since (equilibrium condition), then
let's separate the variables
Where S I And S II– entropy of 1 mole of a substance in the first and second phases;
V I And V II– volume of 1 mole of substance in the first and second phases;
– temperature coefficient of change in saturated pressure
– entropy of phase transition,
Where DH f.p. – enthalpy of phase transition;
T f.p. – phase transition temperature.
Then the equation will take the form
. (1)
This relationship was found by Clapeyron even before the discovery of the first law of thermodynamics, and then derived by Clausius. Equation (1) is called the Clapeyron–Clausius equation in differential form. It is a general thermodynamic equation applicable to all phase transitions of pure substances (single-component systems), i.e. to the processes of melting (solid-liquid equilibrium), evaporation (liquid-vapor equilibrium), sublimation (solid-vapor equilibrium), polymorphic transformation (equilibrium of their forms), as well as their reverse processes.
The Clapeyron–Clausius equation can be applied to any amount of matter by assigning extensive quantities ( DH And D.V.) to the same amount. Usually these quantities are referred to either moles or grams.
In order to use it to find one of the dependencies, you need to know the other three. For example, to find the dependence of saturated vapor pressure on temperature, you need to know the dependence of the heat of phase transition ( DH f.p) on temperature and the dependence of the molar volumes of equilibrium phases ( V I–V II) on temperature.
Let us consider the applicability of the Clapeyron–Clausius equation to phase transitions (melting, evaporation, sublimation), which are of the most general interest.
Melting
Solution.
Let us determine the change in melting temperature with an increase in pressure by 1 atm, i.e. .
From the Clapeyron–Clausius equation
.
According to the condition here:
T f.p – melting temperature under pressure of 1 atm;
D.V.– difference in volumes (specific) of liquid and solid tin;
DH f.p.ud – specific heat of fusion of tin.
For our case
T f.p = 231.9 + 273 = 504.9K,
Then , 
.
The problem statement gives the molar heat of fusion. It must be converted into specific heat of fusion, since M r (Sn) = 118.7 g/mol, then
.
Considering that , then
.
After substitution we get:
This means that with an increase in pressure by 1 atm, the melting point of tin will increase by 3.35∙10 -3 degrees.
The melting point of tin under a pressure of 100 atm will be equal to
Example 2. Specific volume ( V) of ice at 0 0 C is 1.091 cm 3 /g, and water 1 cm 3 /g. The heat of fusion of ice is 34.292 J/g. How will it change T pl of ice when the pressure changes by 1 atm? At what temperature does ice melt under its own saturated vapor pressure of 4.6 mmHg?
Solution.
We need to determine
since it has the dimension deg/atm, and the value ( V V – V ice) cm 3 /g, then the value DH pl should be expressed in atm cm 3 /g. Considering that , then we get
Therefore, if the pressure increases by 1 atm, the melting point decreases by 0.073 0 C.
If the pressure decreases from 1 atm to 4.6 mmHg (), then
– the melting temperature will increase by 0.0726 0 C.
Example 3. Under a pressure of 0.1013 MPa, ice melts at a temperature of 273 K. The specific volume of ice at 273K is 991.1∙10 -3 cm 3 /g, and that of water is 916.6∙10 -3 cm 3 /g. The molar heat of fusion of ice is 6010 J/mol. Calculate the pressure at which ice will melt at 271K.
Solution.
Let's use the Clapeyron–Clausius equation:
Where DV = V and – V t = 916.6 ∙ 10 -3 – 991.1 ∙ 10 -3 = –74.5 ∙ 10 -3 cm 3 /g – the minus sign shows that when ice melts, the volume of the system decreases;
DH pl – heat of fusion. The problem gives the molar heat of fusion. Must be converted to specific heat of fusion.
Mr(H 2 O) = 18.01 g/mol, then 
,
but for the dependence of the unit of measurement – , and 1 J = 9.867 cm 3 ∙atm. Or 1 J = 9.867 ∙ 0.1013 cm 3 MPa.
Then DN pl = 333.70 ∙ 9.867 ∙ 0.1013.
Let's calculate
A negative value of the dependence shows that with increasing pressure (), the melting temperature of ice decreases ().
The pressure at which ice will melt at 271K will be found from equation (3).
From here, but DT= (271–273) K = – 2K, and calculated earlier ( 
), Thus
Hence R = P 0 + DP= 0.1013 + 33.7 = 33.8 (MPa) – at a pressure of 33.8 MPa, ice will melt at 271K.
Evaporation (sublimation)
At moderate temperatures and pressures, not too close to critical, the volume of boiling liquid is small compared to the volume of dry saturated steam, so the change in volume D.V. = V P - V f in the Clapeyron–Clausius equation can be replaced by volume V P– dry saturated steam. The Clapeyron–Clausius equation in this case will take the form
If at moderate pressures to dry saturated steam the ideal gas equation of state applies PV = RT, replace and then
,
let's separate the variables
.
Based on reasoning, the Clapeyron–Clausius equation for the sublimation process can be obtained, which is approximate. Its advantage lies in the ease with which it can be integrated
.
On the diagram in ln coordinates P – 1/T this equation is expressed by a straight line with a tangent of the angle of inclination to the axis 1/ T, equal to – .
This circumstance can be used to find the approximate value of the average molar heat of vaporization (sublimation) in a certain temperature range.
Integration of the approximate Clapeyron–Clausius equation under the assumption that DH does not depend on T within P 1 – R 2 gives
This equation is suitable for use in a small temperature range.
Example 1. The normal boiling point of iodine is 185 0 C. Heat of vaporization DH used.ud = 164.013 J/g. Approximately to what temperature should iodine be heated in an apparatus in which pressure is maintained? 
to ensure distillation?
Solution.
Let us use the Clapeyron–Clausius equation
. (4)
The equation gives the molar heat of evaporation ( DH used), and in the condition of the problem the specific heat of vaporization, but
Let's translate t 0 C in T TO. T= 185 0 C + 273 = 458K.
Let us substitute the available data into equation (4) and solve for T 2.
,
; 
t 0 C = 386.4 – 273 = 113.4 0 C.
Conclusion. At a pressure of 100 mm Hg, iodine will boil at a temperature of 113.4 0 C.
Example 2. At atmospheric pressure diethylamine boils at 58 0 C. At what pressure will diethylamine boil at 20 0 C if the normal heat of vaporization is 27844.52 J/mol?
Solution.
Let us use the Clapeyron–Clausius equation
. (4)
Let's translate t 0 C in T TO.
T 1= 273 + 58 = 331K.
T 2= 273 + 20 = 293K.
Let's substitute the data into equation (4) and solve for R 2.
,
Conclusion. At a pressure of 208.5 mmHg. diethylamine will boil at 20 0 C.
Example 3. The vapor pressure of crystalline acetylene at 132K is 1.7 mmHg, and at 153K it is 27.6 mmHg. Calculate the molar heat of fusion of acetylene if its specific heat of vaporization is 828.014 J/g.
Solution.
According to the condition, acetylene goes from a solid state to a vapor state, i.e. DH f.p. = DН pl + DH Spanish Let us use Clapeyron–Clausius equation (4):
and decide regarding DH f.p,
.
Let's substitute the data,
.
Then DH pl = DН f.p. – DН isp
The problem gives the specific heat of evaporation. It must be converted into molar heat of evaporation, since Mr(C 2 H 2) = 26 g/mol
DH pl = 22281.44 – 21528.364 = 753.056.
Conclusion. The molar heat of fusion of acetylene is 753.056.
Two-component systems
The interaction of substances in multicomponent systems without isolating the resulting products is studied by physical and chemical analysis method, the essence of which is to study the relationship between numerical values physical properties equilibrium chemical system and concentrations of components that determine the state of equilibrium.
Based on the study of the physical properties of an equilibrium system, diagrams are constructed in composition-property coordinates. By the geometric features of the diagrams, by the totality of lines, surfaces, etc. one can clearly judge not only the chemical nature of the substances formed, but also the number, limits of stability, conditions of coexistence different phases in system.
The foundations of this method were laid by D.I. Mendeleev, Le Chatelier, G. Tamman, comprehensively developed by N.S. Kurnakov, and have found wide application in the production of steel, other alloys, galurgy and the production of silicate materials.
Let us consider two-component condensed systems where liquid and solid phases are present.
The Gibbs phase rule will be expressed in this case by the formula:
but in such systems the pressure usually remains constant ( R= const) therefore the number of free variables becomes equal to 1 and then
those. Such a phase diagram can be constructed on a plane, expressing the composition-temperature relationship.
Such diagrams are obtained by thermal analysis. The essence of this method is that a molten mixture of two substances is cooled, measuring the temperature at equal intervals of time and constructing a cooling curve in time-temperature coordinates, using the fact that while no changes (transformations) occur in the cooled system, the temperature drops practically With constant speed. Processes accompanied by the release of heat (crystallization, chemical reactions, polymorphic transformations, etc.), are reflected in the cooling curve either by a break (a section with a slow cooling rate) or horizontal sections with a constant temperature, as shown in Fig. 2.
Rice. 2. Types of cooling curves:
a – pure substance;
b – mixture of isomorphic substances;
c – mixture of non-isomorphic substances
Characteristic points on the cooling curves:
§ Curve a: t kr – crystallization temperature of a pure substance. The duration of the temperature stop and thus the size of the horizontal section on the cooling curve depends on the amount of substance and on the rate of heat removal. As the last drop of liquid disappears, the temperature begins to drop.
§ Curve b: t 1 – temperature of the onset of crystallization of the isomorphic system, t 2 – temperature of the end of crystallization of the isomorphic system.
§ Curve in: t 1 – temperature of the beginning of crystallization of one component of a non-isomorphic system, t 2 –t 3 – temperature of the beginning and end of crystallization of the eutectic mixture. When the melt of a two-component system is cooled, solidification begins with the crystallization of the component relative to which the liquid melt becomes saturated. Cooling curves b and c show that at point t 1, the beginning of crystallization of one of the components leads to a break in the curve and a decrease in the cooling rate due to the release of the heat of crystallization. The absence of a temperature stop is explained by the fact that the composition of the liquid phase changes during crystallization. When the temperature is reached at which the liquid solution becomes saturated relative to the second component, simultaneous crystallization of both components occurs, and another kink appears in the cooling curve ( t 2). In this case, the composition of the liquid phase remains constant. Therefore, a temperature stop is observed on the cooling curve ( t 2–t 3). After the entire mixture has hardened ( t 3) the temperature drops again.
Consequently, any break in the cooling curve indicates the beginning of some transformation.
To obtain a phase diagram, cooling curves are first experimentally obtained for a number of mixtures with different known concentrations of components A And IN and on their basis they are already building a system state diagram A–IN. To do this, all temperature stops and break points on the cooling curves are plotted on the composition-temperature coordinate grid, and then the resulting points are connected.
Let us consider the basic diagrams of equilibrium two-component condensed systems.
Solution.
The total mass of the system is given (10 kg), therefore, 
. According to the lever rule, we measure the segments N II O And T 3 N II and we get solve equation 33 m t = 130 Þ m t = 3.94 kg.
Conclusion: when cooling 10 kg of the composition mixture n up to temperature T 3 3.94 kg of crystals will be released A.
At the point N III The first crystals of the substance begin to fall out IN Therefore, there are 3 phases in the system: one liquid (melt composition n E) and two solid (crystals A and crystals IN), i.e. , Then 
, the system has no options. The value shows that these three phases can only be in equilibrium when completely certain conditions, when the temperature is equal to the eutectic temperature ( T E), and the solution has a eutectic composition ( n E). Neither the temperature nor the composition can be changed in this case without changing the number and type of phases. According to the Gibbs law of phase equilibrium, crystallization of the eutectic from solution should occur at a constant temperature, in addition, the ratio of the masses of crystals A And IN in the precipitated eutectic must be the same as the mass content of substances A And IN in a melt of eutectic composition.
Due to the fact that crystals A And IN during crystallization, eutectics precipitate simultaneously and they do not have the conditions for crystal growth, solid eutectic has a fine-crystalline structure. Frozen melt composition N III(below temperature T E) consists of relatively large crystals A, which fell in the temperature range between the points N I And N III and fine crystalline mixture of crystals A And IN solid eutectic mixture.
The crystallization process ends at the figurative point N III at a temperature T E complete solidification of the liquid solution (melt).
After the disappearance of the liquid phase, only two phases remain in the system: crystals A and crystals IN. This means that only the temperature can change arbitrarily.
At the point N IV cooling of the two solid phases continues.
The cooling curve for the considered case will look like this (see Fig. 3 - III).
Plot N–N I: , , cooling proceeds uniformly according to Newton's law.
Plot N I -N II: incl. N I a break is observed, which indicates the appearance of a new phase - the substance begins to crystallize A. The cooling curve drops more flatly than in the section N–N I. This is explained by the fact that during crystallization of a substance A heat is released from the melt, which slows down the temperature rise, therefore, , .
With further cooling the temperature drops to the eutectic temperature T E, the system reaches the figurative point N III, while the composition of the liquid phase at the point E becomes eutectic and the precipitation of solid eutectic begins, i.e. mixtures of crystals A And IN. (cr A+ cr IN+ F), therefore . N III– the beginning of crystallization of the eutectic, N IV– end of eutectic crystallization.
Tammann triangle
If we consider the cooling curves of mixtures of a two-component non-isomorphic system 2, 3, 4, 6 (see Fig. 3 - I), then we can note that each of them has two kinks.
The first break indicates the appearance of a new phase - this is the beginning of crystallization of the pure component (for cooling curves 2, 3, 4 - this is the beginning of the precipitation of crystals A, for cooling curve 6 – this is the beginning of crystal precipitation IN).
The second break, which turns into a horizontal segment, characterizes the beginning of crystallization of the eutectic.
On the cooling curves, the sizes of these segments (a, b, c, d, e) are different. They are directly proportional to the amount of eutectic deposited. In our example, the largest segment r ( EAT) is on cooling curve 5 (composition: 30% A and 70% IN), which has only one kink, turning into a horizontal platform. This suggests that we are initially dealing with a eutectic mixture, which explains the large size of g.
If on the diagram we plot the segments (a, b, c, d, e) from the solidus line and connect the resulting points to each other, then a triangle is formed T E FM– Tammann triangle, it allows you to calculate the mass of the eutectic mixture that falls out when cooling a system of any composition AB.
Example. Let us initially take 5 kg of a mixture with a composition of 70% A and 30% IN. Determine how much eutectic is released when this mixture is cooled.
Solution.
Consider two triangles. Triangle T E N III N IV similar to a triangle T E EM, from which it follows
,
Where EAT– 5kg, and segments T E E And T E N III we measure.
.
Conclusion: when cooling 5 kg of a mixture of 70% composition A and 30% IN 2.22 kg of eutectic mixture is released.
Task. Based on the cooling curves of the aluminum – silicon system (Fig. 4), construct a composition – melting point diagram. Determine from the diagram:
1. At what temperature will crystallization of a system containing 60% silicon begin?
2. Which element will go into solid state?
3. What amount of solid phase will be formed when 2 kg of a system containing 60% silicon is cooled to 1000K?
4. At what temperature will crystallization end?
5. Determine the composition of the last drop of liquid.
6. Find the mass of the eutectic when cooling 2 kg of a mixture containing 60% silicon.
Solution.
Based on the cooling curves, we construct a composition-temperature diagram. During the crystallization of pure silicon (cooling curve 1), a temperature stop is observed at 1693 K (the melting point of silicon). We plot this temperature on the temperature axis corresponding to pure silicon ( L).
Curve 2, containing 80% silicon, at 1593K reveals a decrease in the cooling rate. In this case, pure silicon begins to precipitate in the form of crystals, and the liquid phase is enriched in aluminum. As the aluminum content increases, the melting point of the system decreases. At 845K, a temperature stop is observed on curve 2 (horizontal area - a), after which the entire system passes into the solid state. In this case, both aluminum and silicon precipitate simultaneously in the form of crystals, i.e. The eutectic crystallizes. Both types of crystals are clearly visible under a microscope.
When cooling a system containing 40% silicon (curve 4), a change in the cooling rate is observed already at 1219 K, and the horizontal area (c) is observed at the same temperature as in curve 2 (845 K), which indicates crystallization of the eutectic, and since the composition of the eutectic is constant, then the length of the horizontal area is proportional to the amount of crystallizing eutectic.
When cooling a system containing 10% silicon (cooling curve 5), a temperature stop is observed at 845K. The length of the horizontal area (g) is maximum on curve 5, which means that a system containing 10% silicon corresponds to a eutectic composition.
When cooling pure aluminum (curve 7), a temperature stop is observed at 932K, which corresponds to the melting point of pure aluminum.
Having completed the construction of the diagram for all cooling curves, we obtain two liquidus curves ( NE, EL) and the horizontal line of solidus SEM, which intersect at one so-called eutectic point E.
Above the curves NEL(zone I) the system is in a liquid state.
Fig.4. Aluminum-silicon phase diagram
In zone II, aluminum crystals and melt coexist, the composition of which at each temperature is determined by the curve NE.
In zone III, silicon crystals and melt coexist, the composition of which is determined by the curve EL.
In zone IV, the system is in a solid state. Since the length of the horizontal area corresponding to the crystallization of the eutectic is proportional to the amount of eutectic, this can be used to determine the mass of eutectic that can be separated from a mixture of any composition. To do this, you need to construct a Tammann triangle.
The lengths of the horizontal platforms (a, b, c, d, e) are laid vertically down from SEM at points corresponding to the compositions of the mixtures. Connecting the bottom ends and points WITH And M, we get the Tammann triangle.
1. Crystallization of a mixture containing 60% silicon will begin at a temperature of 1421K.
2. Silicon will pass into the solid phase. The melt will be enriched in aluminum.
3. When a system containing 60% silicon is cooled to 1000K, some silicon will precipitate in the form of crystals. To determine the amount of solid and liquid phases, the lever rule is applied.
The weight of silicon crystals is related to the weight of the liquid phase as a segment OF refers to the segment FP. If the system weight is 2 kg, then
m T +m w = 2 kg,
m f = 2 – m T.
solve the equation for m T:
34 m t + 28 m t = 56,
m t = 0.903 kg.
At T= 1000K crystalline silicon will be released from a system containing 60% silicon.
4. Crystallization of this mixture will end at a temperature of 845 K - the eutectic temperature.
5. The composition of the last drop of liquid corresponds to the composition of the eutectic (10% silicon and 90% aluminum).
6. The mass of the eutectic is determined from the Tammann triangle. Triangle MKD similar to a triangle MES, hence:
by condition m systems = 2 kg => ES= 2, we measure all segments,
When cooling 2 kg of a mixture containing 60% silicon, 0.923 kg of eutectic will be released.
Solution.
Based on the cooling curves, we construct a fusibility diagram (Fig. 9).
Curve 1 corresponds to the cooling of pure gold. At T= 1336K a temperature stop is observed on the curve. It corresponds to the melting point of gold. Pure substances crystallize at a constant temperature until the liquid phase turns into a solid. On the ordinate axis we plot the point (1336), corresponding to the melting point of gold.
Curve 2 corresponds to the cooling of a system consisting of 20% Pt and 80% Au. At T= 1567K there is some kink in the cooling curve (the cooling rate decreases). This is explained by the release of heat during crystallization of the mixture. At T= 1405K crystallization ends. Heat is no longer released, so a certain break in curve 2 at this temperature indicates a slight increase in the cooling rate (the resulting solid system is simply being cooled).
Rice. 9. State diagram of the isomorphic system gold - platinum
On the ordinate axis corresponding to the composition of 20% Pt and 80% Au, we plot T= 1567K (temperature of the onset of crystallization) and T= 1405K (temperature of the end of crystallization). Similarly, we find points corresponding to other compositions. By connecting these points, we get two curves AkSV– liquidus line and AmdB– solidus line, converging at the melting points of pure substances, which represent the diagram of an isomorphic two-component system.
Above the liquidus line, the entire system is in a liquid state (,), below the solidus - in a solid state (,). Between the curves AkSV And AmdB part of the system is in the liquid state, and part has passed into the solid state ( , ) - the region of equilibrium coexistence of liquid and solid solutions. The amount of substances in both liquid and solid states is determined by the lever rule.
1. Crystallization of a system containing 75% Pt and 25% Au ( WITH"), will start at T= 1925K.
2. Crystallization will end when the composition of the solid phase is equal to the composition of the initial liquid phase, i.e. at 1688K ( WITH"").
3. The composition of the first crystal is determined by the point of intersection of the isotherm of the onset of crystallization with the solidus line ( d), which corresponds to the composition d".
4. A system containing 40% Pt and 60% Au is heterogeneous when cooled to 1650K - it consists of a liquid phase of the composition k" and crystals of composition m". The mass of crystals and liquid phase is determined by the lever rule: the weight of the solid phase is related to the weight of the liquid phase as a shoulder kl refers to the shoulder lm, i.e.
The total weight of the mixture is 1.5 kg, and if we assume that
1) the number of degrees of freedom of the system at points A, b, With, d;
2) indicate within what limits the composition of the liquid and solid solutions upon hardening of a 60% alloy;