Classical and statistical thermodynamics carter pdf

 
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  1. Classical and Statistical Thermodynamics, Carter
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Classical And Statistical Thermodynamics Carter Pdf

statement classical and statistical thermodynamics ashley h carter that you are looking for. . Thermodynamics by Ashley H. Carter PDF, ePub eBook D0wnl0ad . Classical and Statistical Thermodynamics (Carter, Ashley H.) Frank Somer Jr. Department of Statistical Mechanics. View: PDF | PDF w/ Links. Related Content. {Review and introduction Kinetic description Classical statistics. Elementary) D dllllte gases duction to the methods of statistical thermodynamics. Depending.

Tt be sure, the inadequacies of c l assic al ther modynamics becorne apparertt upon close scrutiny and invite inquiry about a more fundamen t al descripti on This, of course, exactly reflects the historical. If only the statistical picture is presented, however, it is my observation that the student fails to appreciate fully its more abstract concepts, given no exposure to the related classical ideas first. Not only do clas si c al and st a tistical the rtn odyn amics in this se11se corr1 plemcnt each oth er, they also bea u ti fully illustrate the physicist's perpetu al stri vi ng for descriptions of gr eater power , el egance, uni ver s ali ty, and freedon1 from ambiguity. Chapters 1 th rough r epres e nt a fairly traditional introduction to the classical theory. Early on emphasis is placed on the advantages of expressing the fu!

The kinetic theory of gases, treated in Chapter 1 1 , is con molecuJar basis of such thermody n amic prop erties of gaes as the tempera ture, pressure, and thermal energy. It represe11ts, both logically and historically, the tran sition between classic a l thermodyn am i cs and the s tat istical t heo ry The unde rlying principles of equilibrium statistical thern1odynamics are introduced in Chap ter 12 through consideration of a simple c oin tossing - experiment.

The basic concepts are then defined. The statistical interpretation of a system con taining many molecttles is observed to require a knowledge of the properties of the in div idua l molecules making up the system.

In C h apt er 13, the explication of classical and quantum statistics and the derivation of the particle distribution functions is based on the method of Lagrange multipliers. A di scussio of tl1e connection betwe en cla ssic a l an d statistical the rm o dynamics c om pl etes the development of the mathematica1 formulation of the statistical theory. Chapter 14 is devoted to the statistics of an ideal gas. Chapters 15 througl1 19 present important examples of the application of the statistical method "The.

Preface Throughout the book a serious attempt has been made to keep the level of the chapters as uniform as possible. On the other hand. In preparing the text, my greatest debt is to my students, whose response has provided a practical filter for the refinement of the material presented herein. A thermal electromotive force emf e is generate J when the two j unctions are at different temperatures. In a liquid helium cryostat. Problems 17 Estimate the pressure you exert when standing on the floor.

Repeat the calculation for spiked heels. I 18 b The Nature ofThermodynamics If. Calculate the corresponding value of the temperature on the a Celsius. Every system has its own equation of state. It is an expression of the results of experiments. The situation is analogous to that in mechanics. This very important question will be addressed in Chapter 6.

We saw in Chapter 1 that the concept of temperature is intimately associated with the. In thertnodynamics. The mechanical variables occur in "canonically conjugate" pairs. We note that the equation of state does not involve time. Perhaps the simplest example of a thermodynamic system is a homogeneous fluid say. I n a the isothern1s are eq uilateral hyperbolae: The projections of the surface f P. T plane and the v. In Equation In 1 8 1 1 Avogadro postulated that at a given temperature and pressure equal volumes of all gases con tain e.

Thus the equation of state can be writ ten Pv RT. T 0 on the P-v plane. J Since n 2. The reason for this is that the molecules in a gi ven sample of gas have negligible volumes compared with the volume of the sample itself.. We use whatever diagram is most appropriate for the process we are in terested in. T plane are sh ow n i n Figure For b 0. Equation 2. The correction As T increases. The transition between the two types of curves is a curve having an inflection point CP.

This is an unstable region since the pressure no longer increases as the volume is diminished. The constant b takes into accottnt the finite volume occupied by the molecules. Van der Waals derived his equation from considerations based on kinetic theory. In Figure 2. Among the numerous equations of state that have been introduced to represent the behavior of real gases. For T Tc.. Part of the gas begins to liquefy at b and the pressure remains constant as the volume is further decreased as long as the temperature is held constant.

Between b and d. Thereafter the curve rises steeply because it takes a large increase of pressure to compress a liquid.

Above Tc it is impossible to liquefy a gas.. The interface between a liquid and a vapor is not discernible.

The critical values Vc.. The P-v. Notice the regions solid. This is indeed the case see Problem Figure 2. At low temperatures and high pressures transitions occur to the liquid phase and the solid phase.. Elsewhere two phases can exist simultaneously in equilibrium. T surface for a substance that contracts on.

T surface for a pure substance includes these phases as well as the gas phase. The S-L curve is the freezing point curve. No critical point can exist for soJid-Iiquid equil ibrium the S-L curve.

The projections of the surfaces on the P. A gas has neither a fixed volume nor shape. A solid has a definite volume and shape. A normal liquid is isotropic. T diagrams for a a su bs t ance that contracts on freezing: The physical distinction between t he various phases is straightforward.

The L-V curve is the vapor pressure curve. That is because solids and liquids possess diffe rent symmetry properties. By raising the temperature a bove Tc then increasing the pressure. The triple point occurs at. The transition from one sym metry to another is strictly a discontinuous process. For C02 the critical temperature is 3 l. A liquid has a fixed volume but not a fixed shape. T plane are of special in terest Figure 2.

In this way. It is also known as the hsaturated. During a change of state. Since v is the reciprocal of the density. Water expands about 10 percent on freezing. An iceberg floats with one-seventh of its volume above the surface. For H 2 0 the critical point is Changes of state are accompanied by changes in molecular forces. Thus the pressure must be reduced to a low value to reach the triple point.

If water contracted on freezing.. For example. S Sketch of the specific volume of water as a function of temperature. The triple 3 point is 0. But at the steam point the expansion is enormous: Substituting Equation 2. The negative sign is used since the volume always decreases wi t h increasing pressure at consta n t tempera t ure: The expansivity. Simi l a rl y the isothermal compressibility is defined as. For an ideal gas. For a l iqu id or a solid. We wish to know what change in pressure would be necessary to keep the volume constant.

We suppose for a moment that the fluid. T2 is isochoric that is. T1 to another equilibrium state P2. The volume increases linearly with an increase in temperature and decreases linearly with an tncrease tn pressure. For any one of these states we may write Equation 2. This could be achieved by placing in contact with the fluid a series of large bodies reservoirs ranging in temperature from T1 to T2 to effect a quasistatic cooling through a sequence of equilibrium states. For copper in this temperature range.

Note that this result could have been obtai ned immediately from Equation 2. The most con ve nie nt path is a reversible path.. Unfo rtuna tely. The result wou ld be large temperat ure gradients within the fluid itself and between the fl uid and its surroundings. In practice.. This is because a P deterr11i ned only by the end points.. Assume that oxygen behaves like an ideal gas.

Problems ll Ou r fundamental state variables P. The atomic weight of helium is 4. Show that. Show that for large values of T and v. At h ighe r temperatures and l arge specific vol utnes low densities all gases a pproxi ma t e an ideal gas. As noted in the text iJ PI rl l ' Compare the numerical value of in the fo llowi ng table: For water. Assume that the expansion of the glass is negligible.

H the bottle and contents are heated to SOOC. Since P and dV are positive. Let y represent such an intensive variable and X the corresponding extensive variable. We assume that the process is reversible and that the system is in equilibrium at all times. The gas expands against the pressure associated with the external force F possibly due to the weight of the movable part of the piston.

The variables X1. This sign convention will be used throughout the text. The integration is nontrivial since P P V in general.

A is its cross-sectional area. Consider the configuration work done by the system consisting of a gas enclosed in a cylindrical container by a piston Figure 3.

In the most general case. Here dx is the displacement of the piston. P is the pressure the intensive variable. For an isochoric process. For a surface film of liquid. For an isobaric process. Extensive Variab1e.. In an electrolvtic cell such as a lead storage battery.. The First Law ofThermodynamics Chap 3 The configuration work done in the change of volume of an ideal gas can be easily calculated in some special cases.

TABLE The work done is evidently the area under the curve in a P-V diagram. In both the isobaric and the isothermal process.

This important sign convention will be followed throughout the text. Regardless of the sense of rotation of Lhe shaft. Such work is always done on the system.. The total work is the algebraic sum of the two kinds of work.. W depends on the path of integration. The stirrer is attached to a shaft projecting through the wall of the container and an external torque is applied at the outer end of the shaft Figure This is easily seen in Figure a v property of the system: W is not 3.

The stirrer and the fluid together constitute the system. Since f P dV is the area under the curve. The integral around a closed path is not zero: There are two well-known examples of dissipative work.

Thus a. Consider a stirrer immersed in a fluid. If a process is to be reversible In the P. Having defined configuration work and dissipative work. Figure 3. A second example of dissipative work is the work needed to maintain an electric current I in a resistor of resistance R Figure 3. Reversing the direction of current flow does not affect the sign of the work.

A current is passed through a I resistor V diagram of Figure The system is assumed to be thermally insulated from its surroundings. To be sure. Thus Equation 3. In differential form. We can remove the bar from the differentiaL but only if we restrict ourselves to adiabatic work.

We define U as the internal energy of the system. This is not to say. To summarize. For any change between given states. In other words. Heat flow into the system is equal to the total work done by the system minus the adiabatic work done.

One useful way of looking at the first law follows from transposing the terms in Equation 3. In words. Energy is conserved if heat is taken into account. Think of a bank: Heat may be conveyed from one body to another by conduction. The increase in internal energy of a system equals the heat flow into the system minus the work done by the system.

In a calorimetric experiment by the tnethod of mixtures. These properties are summarized as follows: It can be shown that the quantity liQ exhibits the properties that are commonly associated with heat. What's left in the bank is d U. It should also be kept in ind that heat transfer is not the only way to change the temperature of matter: It is defined as follows: When no work is done. One definition of heat is: Heat is energy transferred across the boundary of a system temperature difference only.

The relationship between the joule and tlte kilocalorie is: When the temperature of a body increases. Changes of state e. A more precise definition of heat is: Heat is the change in internal energy of a S 'Stem when no work is done on or by the system. Consider gas in a cylinder. The First Law ofThermodynamics. When energy is added to a system in the form of heat. Energy is conserved.

For a reversible process. Heat is energy transferred to a system causing a change in its internal energy minus any work done in the process. If heat is added through diathe1tt1al walls. Thus work and heat are. The internal energy U is a state variable: Now let the gas be confined in a cylindrical piston with adiabatic walls. Then where Wd is the dissipative work done on the system and is inherently negative. The right-hand sides of the equations in the two examples give rise to the same change in internal energy.

Equation How much work is done? We assume that whereas the pressure change is large. For copper. We need some more basic concepts. The length of the stroke is The First Law ofThermodynamics Chap. Suppose that a. How much work is done by the steam per str oke? If the initial pressure is 1 atm and the final pressure is atm.

Draw a diagram of the process in the P. The gas expands reversibly and isothermally to twice its original volume.

Note that for an ideal gas UrxT. How much work is done against the atmosphere when 10 kg of ice melts into water? Calculate a the work done by the gas. Take 3. V plane and in the P. The gas is then expanded isother1nally until the pressure drops to its original value. Problems 49 An ideal gas originally at a temperature and pressure P1 is compressed 7.

T plane. How much work is done by the gas in the baJJoon? How much heat does it absorb? Find A and B.. The First law ofThermodynamics Chap. If the initial temperature is K what is the final temperature? At constant temperature. What is the heat transfer? Find the work necessary to decrease the volume to 5 m a b c d e At constant pressure. Calculate the work done on the material. What is the temperature at the end of the process in a? What is the pressure at the end of the process in b?

Sketch both processes in the p V plane.. The specific heat capacity depends on the conditions under which the heating takes place that is.

The heat capacity C of a system is defined. We define the specific heat capacity. We interpret Q is not a state variable and 'llQ as an infinitesimally small flow of heat and dT as the resulting change in temperature.

There are two important cases: In general.. Sears and. For high temperatures. As an example. Kinetic Figure Adapted from Thermodynamics. This follows from the Gay-Lussac-Joule experiment see Chapter 5. The result. The substitution of Equations 4.

These is equal to 8. From Equation 4. For a monatomic gas at room tempera t u re For a diatomic gas at room temper at ur e. Jiscovercd the equivalence of heat and work and first enunciated the principle of conservation of energy If cv is independent of the temperature. The origin of the word is the Greek verb thalpein meaning ''to heat. Thus Equation 4. Thus is the specific work involved in the process.

The quantity h is called the specific enth. Consider now the three possible phase changes. A change of phase is an isothetultlal and isobaric process and entails a change of volume. Let 1 denote a solid. Equati on 4. I Here l is the latent heat of transformation per kilomole associated with a given phase change.

If two variables have exact differentials. With this definition. The circle denotes a cycJicaJ process around the triple point for which the total change of enthalpy is zero.

Then solid:. This is merely a notational convention. Here the first superscript is associated with the. Applications of the First Law Chap. Consider a cyclical process around the triple point and close enough to it so that only changes in the enthalpy occur during phase transitions. Since enthalpy is a state function. This can be depi cted in a p T. Some values are given in Table 4.

T and Twas arbitrary. This is an important result. Hence we can write T h.: For an ideal gas Setting llq 0 in Equation 4. Analogous relations involving the internal energy and the Reversible process: Dividing Equation 4. Pv1 at bot h limits: Note that for a T2. Work is done by the system at the expense of internal energy. For an adiabatic process. For an ideal monatomic gas. The internal energy decreases. Then T2 is found from the ideal gas law: Take cp for co. Estimate the fractional change in mass when the temperature of a block of copper is rai sed from K to K.

What is the internal energy change of the air in the tire between pressure measurements?

Assume that air is an ideal gas with a constant specific hea t capacity Cv and that the internal volume of the tire remains constant at 0. VT3 is constant. Assume that all the processes are isobaric. The specific heats at constant pressu. Problems 65 where a.. How much heat is transferred in an isobaric process in which a kilomole of gas experiences a temperature increase from T to2T?

Compute the final volume. Assuming that the e xpa n sion is adiabatic and that the fireball remains sphe ri c a L estimate the radius of the ball when the temperature is K. Using the cyclical and reciprocal relations for partial derivatives given in Appendix A.

Before describing the experiments. In the. Consequences of the First Law Chap. Summarizing these resul ts. When t he. The subscript on the d e ri va t i ve is appro priat e since the internal e ne rgy coefficient. The question is: Subsequent experitnents using various gases ind ica t e that the temperature d e cr e as es bv at most a verv sn1all amount.

See Sect ton 7. Since the constant a is zero for an ideal gas. Consider Equation 4. Carrying out the differentiation. This very in1portant experimental result can be proved theoretically. A continuous throttl i ng process can be pe rformed by a pump that mai n tai ns a constant high pressure on one side of a porous wall and a constant low pressure on the other side. It is important.

The experiment involves a throttling process. These matters will be discussed in Chapter 6. Principle of the Joule-Thomson porous plug experiment. Here a corresponds to the initial state and b to the final state, with the piston in the fully advanced position.

Classical and Statistical Thermodynamics, Carter

In the Joule-Thomson experiment, the pressure and temperature of the gas before passing through the plug are kept constant and the pressure on the other side is varied while the corresponding temperature is measured. The results give a locus of points on a T-P diagram representing the isenthalpic state of the gas Figure 5.

L is negative. The point 0 is called the inversion point. The experimental result is that for most gases over a reasonably wide range of temperatures and pressures, the T. P curve is approximately flat and. The heat engine is a highly useful concept in thermodynamics and reminds one of the subject's origins. A heat engine is a system that receives an input of heat at a high temperature..

Before discussing a practical heat engine.. In Figure 5. In this case, the heat out of the system. Examples are stirring heat or the heat generated by the flow of current through a resistor. The concept of a heat engine.

Classical and Statistical Thermodynamics, Carter

In a , work is done on the system and is converted to heat. In b , heat is extracted from a reservoir and is converted to mechanical work. This configuration is not possible. Case b is the reverse of c. Heat is extracted from a reservoir and is converted to mechanical work by the machine. Can the conversion be percent efficient? The answer is no. Thus case b must be modified as shown in Figure 5. Heat is absorbed from the higher temperature reservoir, work is done, and heat is given up by the engine to a lower temperature reservoir.

The engine works in a cycle, so that the state of the system is the same at the end of the cycle as at the beginning. Fig11re S. S Modification of Figure 5. In writing this, we arc mindful of our sign convention in which the heat flow into the system and the work done b. Since the state. Referring to Tahle 5. The slopes of the isotherrns and adiabats are exaggerated.

However, we note that the use of Equations 5. It can be easily shown that this is just the area enveloped by the curves in the P- V diagram delimiting the cyclical process. It is important to emphasize that a Carnot engine operates between. Also, if a working substance other than a11 ideal gas is used, the shape of the cunyes in the P- V diagram will, of course, be different.

Carnot theorized that the efficiency given by Equation 5. We note that the efficiency would be percent if we were able to obtain a low temperature reservoir at absolute zero. However, this is forbidden by the third law. A Carnot refrigerator is a Carnot engine in reverse Figure. In this case we define a coefficient of performance. The last step follows from Equation 5. A schematic dia gram of a typical refrigerator is shown in Figu re 5.

The liquid undergoes a throttling process in. The vaporization is completed in the evaporator: Refrigerators are designed to extract as much heat as possible from a cold reservoir with small expenditure of work. The coefficient of performance of a household refrigerator is in the range 5 to Show that the Joule coefficient 71 is zero.

Show that the Jouie-Thomson coefficient JL is zero. Carefully sketch a Carnot cycle for an ideal gas on a a u-v diagram; b a u-T diagram; c a u-h diagram; d a P T diagram. S-6 A Carnot engine is operated between two heat reservoirs at temperatures of K to K. If the engine receives kilocalories from the reservoir at Kin each cycle, how much heat does it reject to the reservoir at K?

If the engine is operated as a refrigerator i. How much work is done by the engine in each case? The latent heat of fusion of water is 3. S-8 W hich gives the greater increase in the efficiency of a Carnot engine: The ratio of the final volume to the initial volume in the adiabatic expansion is 5.

Compute the temperature of the reservoirs between which the engine ope ra t e s.

A ssutne. A Carnot engine. The processes are as follows: Assume that the working substance is an ideal gas and show that the effi.. For greater values, the rise of temperature upon compression is large enough to cause an explosion. This is called preignition. For diesel engines preignition is not a problem and higher compression ratios are possible. This is partly the reason that diesel engines are inherently more efficient than gasoline engines.

Assume that the working substance is an ideal gas with constant specific heat capacities Cp and c, Show that the efficiency is.. The first law of thermodynamics is remarkable in its generality, simplicity, and utility. Yet our statement of it is in some respects less than satisfactory. Furthermore, in our discussion of heat engines in the previous chapter, we hinted that the first law does not constitute a complete theory because certain processes that it perntits do not occur in nature.

The problems are as follows. First, classical thermodynamics is concerned with states of equilibrium and various processes connecting them. We have seen that there is a very substantial advantage in dealing with state variables, changes that are expressible as exact differentials.

However, two of the three quantities appearing in our statement of the first law the work perfottned and the heat exchanged are inexact differentials. Is there any way in which we can write the first law in terms of state variables only? But reversible processes are idealizations whereas real processes are irrev. This leads us to ask: Is there some state variable by which we can distinguish between a reversible and an irreversible process?

Finally, in Chapter 5 we noted that certain processes are impossible that are by no means prohibited by the first law. It is not possible to construct a machine that convet1s all the heat it extracts from a reservoir into useful mechanical work.

It is not possible to transfer heat from a cold body to a hot body without supplying energy in the forrn of work. The first law says nothing about these prohibitions, these "principles of impotency," as they have been called. Is there, then, something missing from our theory? The answer is yes to all of these questions. We will begin by introducing the concept of entropy as a useful mathematical variable.

Phenomenological observations will give the concept its physical meaning. Neither aQ nor ztW is an exact differential: We note that Pis an intensive state. The quantity aW,jP is therefore extensive. Also, two of the three fundamental state variables. Both of these problems will be addressed later in this chapter. The reverse process does not take place Heat flows from a high temperature body to a low temperature reservoir in the absence of other effects Figure 6.

A battery will discha: The gas will not compress itself back into its original volume. The reverse will not happen: The gases will not separate spontaneously once mixed.

Two gases.. Consider the following processes and the results observed in nature: We consider two engines. M Carnot and M' hypothetical..

We can arrange the position of the adiabats T.. Using the Clausius statement of the second law. Since a Carnot engine is reversible. We conclude that the compos' I Q 1 I from the cold reservoir ite engi1. In this sense the two statements are equivalent. This conclusion vio. The Hcomposite" engine is shown in Figure 6.

It fol- lows 77'. Referring to Figure 6. Since both engines are reversible.

Classical and Statistical Thermodynamics, Carter | Second Law Of Thermodynamics | Thermodynamics

The Carnot engine M absorbs heat Q2 from work W. This violates the Planck-Kelvin statement. Continuing the proof of Carnot's theorem. Applying the Carnot ratio to an infinitely narrow Carnot diagram finite temperature difference but infinitesimally small quantities of heat extracted and rejected.

By an extension of Equation 6. V diagram of a reversible cycle. The process is replaced by infinitesimal Camot cycles. We replace the process by infinitely narrow Carnot cycles Figure 6. This is Carnot's theorem. Consider next an irreversible cycle. Then J 2 T. Equation 6. An irreversible Camot cycle that operates between the same temperatures and produces the same work has a smaller efficiency 11' than the efficiency 71 of a reversible Carnot cycle.

Following the same reasoning as before. The subscript r emphasizes the reversible nature of the cycle. It does not say that the entropy of part of the system cannot decrease. Suppose now that the sys t e m is is ]ated.

This is the principle o. I I I Figure 6. As far as these equations are concerned. This is not to be confused with the real universe. Clearly this is not so. Suppose that the temp er a ture of a body is T2 and that a reve rsible engine works between T2 a nd T1. If the engine is irreversible. The laws of mechanics are second-order equations in time t and are unaltered by the replacement oft with. The property of increasing entropy is eqttivalent to saying that energy is always being degraded into forms that are more and more difficult to use for the production of work.

The special fact about a reversible Carnot cycle is that the efficiency is independent of the nature of the working substance.

The significance of entropy becomes clearer when we consider heat to be the energy of molecular motion. The lack of complete availability of that energy for the production of work is due to the randomness of the m olecular motion.

Carnot's theorem shows that the ratio Q1jQ2 has the same value for all reversible engines that operate between the same temperatures. V diagram represent an isothern1al expansion during which heat Q2 flows into the system. V plane of a Camot cycle divided into two sub-cycles. For the cycle abcda. To do th is consider cycle ahefa. For this subcycle. Thus or Because the right-hand is a function of fJ1 and 82 only. This is only possible if and Then independent of the properties of any given substance.

S im ilarly for the subcy cle fecdf. The Second Law of Thermodynamics Chap. TI ! Since the scale of temperature is arbitrary. Then 6. The proof is long. Now a av as -- - iJT v r - we can therefore differentiate Equation important relation a as aT av ' T v 6. IS worth the effort. Using the definition of the entropy. The smallest value of Q is zero and the corresponding value of T is zero.

This is a fundamental definition of absolute zero. Comparing Equation 6. An appropriate choice of A' makes 8 equal to T. In a Carnot cycle. Combined First and Second Laws Sec. The definition of the Kelvin scale is completed by assigning to T1 in Equation In its most general fortn.

For a Carnot engine operating between temperatures Tand We can write llQ. Take y rn u m volun1e c anJ on I kg of rne: The latter appears as heat in Equation 6. Finu the entropy increase of the methane during the isothertnal expansion. The l aten t heat of fusion is 3. The Second Law ofThermodynamics Chap. S imil arly. In fact. For the free expansion of a gas. For adiabatic stirring. In the actual process. The mass of the resistor is 5 g. Calculate as if the process is isothermal.

Which is larger? By how much? The atomic weight of carbon is Would you advise investing money to put this engine on the market? I magine a reversible process taking it between the same equilibrium states. Assume that the specific heats of the gas are constants. Of the universe? The resistor is submerged in a large volume of water.

Show that cv is a function ofT on ly. What is the change in the entropy of the resistor? Of the water?