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Introduction to the Thermodynamics of Materials Fourth Edition David medical-site.infol School of Materials Engineering Denise medical-site.infok, Vice. View eBook Introduction To The Thermodynamics Of Materials, Sixth Edition By David R. Gaskell, David E. Laughlin [KINDLE PDF. EBOOK. Gaskell Thermodynamics Solutions - [PDF] [EPUB] Gaskell Thermodynamics Solutions. Introduction to the Thermodynamics of Materials.

Skip to main content. Log In Sign Up. Introduction to the Thermodynamics of Materials. Possible spelling error: Si3N4 over range K For compounds, these are enthalpies for formation from elements. The enthalpies of pure elements are taken, by convention to be zero. Some early thermodynamics problems were for very practical problems. For example, in a steam engine heat is supplied to water to create steam. The steam is then used to turn an engine which does work. Finally, the water is exhasted to the environment or in a cyclic engine it can be condensed and recyled to the heating chamber or boiler Work Boiler at T Engine Done 1 Exhaust Condenser at T 2 Steam power plant or steam engine An early goal for thermodynamics was to analyze the steam engine and to figure out the maximum amount of work that could be done for an engine operating between the input temperature T1 and the output temperature T2. Some of the most important work on thermodynamics of heat engines was done by Nicholas Carnot around

More complicated systems require more variables. Thus equations of state follow the rules of mutlivariable mathematics. In thermodynamics, we are often concered with how something changes as we change the independent variables. A general analysis of such a problem can be written down purely in mathematical terms. This Mathematica notation will be used throughout these notes which were prepared in a Mathematica notebook. Many input expressions are followed be semicolons which simple supresses uninteresting Mathematica output.

Any change in volume due to a change in T and P can be calculated by integrating dV: In thermodynamics we are usually dealing with physical quantities. In general, the partial derviatives for the total differentials themselves often have physical significance.

In other words, they often correspond to measurable quanties. Sometimes the physical significance is not clear. In such problems, the partial derivative is defined as having having physical significance or it becomes a new thermodynamic quantity. One good example to be encountered later in this course is chemical potential. For example V is a state variable. It depends only on the independent variables P, T, and perhaps others and not on the path taken to get to the variables.

There are many thermodynamic state variables and they are very important in thermodynamics. There are some thermodynamic quantities that are not state variables. The two most important are heat and work. The heat supplied to a system or the work done by a system depend on the path taken between states and thus by definition, heat and work are not state variables.

The previous sections define equations of state for matter. Equilibrium is the state of the system when the variable reaches the value it should have as defined by the equation of state. For example, a pure gas has an equation of state V[P,T]. Equilibrium is reached when after changing P and T to some new values, the volume becomes equal to the V[P,T] defined by the equation of state. All systems naturally proceed towards equilibrium. They are driven there by natural tendencies to minimize energy and to maximize entropy.

These concepts will be discussed later. Although all systems tend towards equilibrium, thermodynamics says nothing about the rate at which they will reach equilibrium. Some systems, particularly condensed solids as encountered in material science, may not approach equilibrium on a pratical time scale. Boyle found that at constant T that V is inversely proportional to P.

These are the units of work or energy. When R was first measured, P was measured in atm and V in liters; thus P V or work or energy has units liter-atm.

SI units for energy is Joules. The first law of thermodynamics connects the two energy units and allows one to relate heat and work energy or to relate calories and Joules. Extensive variables depend on the size of the system such as volume or mass.

Intensive variables do not depend on the size such as pressure and temperature.

Extensive variables can be changed into intensive variables by dividing them by the mass or number of moles. Such intensive variables are often called specific or molar quantities.

For example, the volume per mole or molar volume is an intensive variable of a system. Similarly, mass is an extensive property, by mass per unit volume or density is an intensive property. Systems are characterized by the number of components and the type of phase diagrams depend on the number of components.

Examples are one-component unary , two-component binary , three-component ternary , four- component quarternary , etc.. In each zone, one state is the most stable state. On lines, two phases can coexist. At triple points, three phases can coexist. Example of unary is water phase diagram.

Unary diagrams usually use two variables like P and T. Binary diagrams add composition as a third variable. Binary diagrams are usually for one variable T, P, or V together with the composition variable. The complete phase space is 3D. Thus, 2D binary plots are sections of the 3D curves. Zones can be single phase solutions or two-phase regions. The relative proportions of phases in two- phase regions are given by the lever rule.

Choice of components is arbitrary. First law connected heat and work and clarified conservation of energy in all systems. The key new energy term that developed from the first law is internal energy. Internal energy often has a nice physical significance; sometimes, it significance is less apparant. The first law says energy is conserved, but it makes no statement about the possible values of heat and work. The second law defines limits on heat and work in processes.

It was used to define the efficiency of heat engines. The second law also lead to the definition of entropy. Entropy was slow to be accepted, because it has less apparant physical significance than internal energy.

Rougly speaking, entropy is the degree of mixed-upedness.

Some thermodynamic problems require an absolute value of entropy, the third law of thermodynamics defines the entropy of a pure substance at absolute zero to be zero. The principles of thermodynamics is are nearly fully defined after defining the laws of thermodynamics, internal energy, and entropy. The rest of the study of thermodynamics is application of those principles to various problems. All systems try to minimize energy and maximize entropy. Most problems we ever encounter can be solved from these basic principles.

It turns out, however, that direct use of internal energy and entropy can be difficult. Instead, we define new functions called free energy - Gibbs free energy or Helmholz free energy.

These new energies perform the same function as other thermodynamics functions, but that are physcially much more relevant to typical problems of chemistry and material science. In particular, Gibbs free energy is the most common term needed for chemical and material science problems that are typically encounted in various states of applied temperature and pressure.

Chapter 2: The total change in internal energy is thuys always given by: The total change in enthalpy can be written two ways as: The values for heat and work during a change of state, however, will depend on path. This section gives some results for heat and work during some common processes: Heat and work are thus: Any Processes For any other process, w can be calculated for the P dV integral and q from the first law of thermodynamics.

This result applies for both reversible and irreversible processes; P, however, will be given by an equation of state only for reversible processes. Here are some constants defined in a table used to get numerical results: All results are given in Joules: DU for isothermal step is zero because of the ideal gas.

The constant volume step is same as above and thus obviously gives the same result.

This extra work is not needed because in all cases, DU can be calculated directly from the same information used to first get DH. Finally, adiabatic compression to 1 atm returns to initial volume.

Isobaric compression to 25C returns the gas to its initial state state 1 above. After returning to the intial state, the total work comes from the isobaric steps only; the constant volume steps do no work. Simpler expressions hold in some special cases. From the general equation above, DS is also obviously zero because PVg is constant during a reversible adiabatic processes.

There are two results for each term; either can be used, depending on which one is easier: For free expansion of ideal gas, temperature remains constant. Here the volume triples. For any isothermal expansion to triple the volume, the state functions results are the same as part a. But here the process is reversible.

At constant pressure q is equal to DH and work follows from that results: Heat flow is an integral of the constant-pressure heat capacity. If the heat capacity is independent of temperature, the final temperature will be the average of the two initial temperature.

The engine operates in a cycle and thus must have no entropy change. Chapter 4: Chamber 1 has 1 mole of A and chamber 2 has 1 mole of B. These ideal gases do not interact and thus the total energy change is the sum of entropy changes for each type of gas: When there are 2 moles of A in chamber 1, the entropy change for that gas doubles giving: Notes on Gaskell Text 33 c.

When each chamber has gas A, we can not use the methods in parts a and b because they no longer act independently. When each chamber has 1 mole of A, removing the partition does not change anything.

When one chamber has 2 moles of A and the other has 1 mole of A, the two chambers will be at different pressures and removing the partition will causes changes and a non-zero change in entropy. This problem is best solved by first moving the partition to equalize pressures. Thus the total change in entopy can be calculated from the initial change in volumes done to equalize pressures: Chapter 6: This answer can be obtained by using 0.

The Handbook of Chemistry and Physics gives the thermal expansion of Copper as as 0. Thus the text gave the wrong value in the problem, but used the correct value to derive the solution. Using the correct thermal expansion changes the above results to: This method works because total enthalpy is conserved for adiabatic, constant pressure conditions.

For the reaction in air starting with one total mole of reactants, the fractions are. The DGsimp agrees with the book, but the DG in the book is different.

Some books have 9 problems that correctly correspond to the 9 solutions. Other books probably early printings of the thrid edition are missing the problem that goes with the first solution and have an extra problem that has no solution.

These notes give the solutions to the 8 problems in common to all books. Some books have them as 7. Once the vapor-liquid equilibrium is reached at constant volume, the P and T will remain on the transition curve but the vapor pressure will change with temperature.

Also, Table A-4 gives the second equation as the vapor pressure curve for liquid Zn. The leading constant of Thus under atmospheric conditions, solid CO2 vaporizes into gaseous CO2.

You are also given the slopes of the lines emanating from the triple point by using the Clapeyron equation: This problem is solved in the text see page Mixing of ideal gases is puring due to entropy effects.

The maximum increase in entropy occurs when there are equal parts of each gas. From partial molar results eq.

If you pay by the mole, you would prefer the ideal gas because it would be cheaper. If you pay by the container, you would prefer the van der Waals gas because you would get more moles per dollar. Pressure is given by the virial expansion, so all we need are the initial and final volumes.

To find work done by the gas, we integrate P from V1 to V2 or to find the work done on the gas we reverse the integration and go from V2 to V1. The result after convertion to joules is. Positive work must be done on a system to compress it. Finally, the ideal gas result can come for either above result by setting extra constants to zero, or by directly integrating the ideal gas result: Solve the equation and take the non-zero root: It will be seen later to be assumed to be independent of temperature, but it may depend on pressure.

We can plot the activity coefficients of A and B for various values of W: Here is blow up for some positive W: Inverse functions are being used by Solve, so some solutions may not be found. The results are: Some plots of excess enthalpy which are actually total enthalpy of mixing are: If we take the results in Fig 9. The proportionality constants, however, are different which means they are not equal and furthermore the entropy change must differ from ideal results.

The simplest model is to let W be linear in XB but we introduce this linear dependence in the excess free energy for simplicity instead of in the activity coefficient of A this other method could be used if desired. Here is a sample: David r. Upcoming SlideShare. Like this document? Why not share!

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