AdvThermo_PS01_2013_Soln

MECH 5403 FAdvanced Thermodynamics

1. (20 marks). See Callen (1960), Question 1.9-1 or Callen (1985), Question 1.10-1

The following equations are purported to be fundamental equations of various thermodynamicsystems. Use the information provided in Callen to determine if any of the equations below arephysically permissible and sketch the shape of the curve S = S (U) assuming that both the volumeV and the number of moles N are constants.

(A)

(B)

(C)

Quantities and R are positive constants and symbols N, V and U denote the total molenumber, volume and internal energy. In all cases where fractional exponents appear only the real,positive root is to be taken as physically acceptable.

Demonstrate if any of these expressions is also monotonically increasing?

Solutions

The requirements for a suitable fundamental equation (in the entropic form) as summarized byCallen can be stated as follows:

Postulate I: There exist particular states (called equilibrium states) of simple systems that,macroscopically, are characterized completely by the internal energy U, the total system volume V,and the mole numbers N1, N2, ... , Nr of the chemical components. This postulate is satisfied alreadyin the sense that we have been given functional relationships.

Postulate II: There exists a function (called the entropy and denoted by the symbol S) of the extensiveparameters of any composite system, defined for all equilibrium states and having the followingproperty - The values assumed by the extensive parameters in the absence of an internal constraintare those that maximize the entropy over the manifold of constrained equilibrium states.

Postulate III: The entropy of a composite system is additive over the constituent subsystems. Also,the entropy is continuous and differentiable and it is a monotonically increasing function of theenergy. This postulate contains the following mathematical tests that any suitable fundamental eqmust satisfy; namely,

(1) The entropy must be a continuous, differentiable and monotonically increasing function

of the internal energy so that, for a single component system, we require

and ,

(2) The entropy must be a homogeneous, first-order function of its extensive parameters sothat where is a constant.

Postulate IV (Nernst postulate): The entropy of any system vanishes in the limit of the system goingto a state of absolute zero temperature. Mathematically, we require

(3) The entropy must be bounded in the limit of low temperature so that

as .

(A) Given the proposed function we have

(1)

provided , and

which does not satisfy the monotonically increasing condition.

(2)

(3) can only be achieved if either or and this is not

possible because would require that . Similarly, requiring that also leads to unacceptable conclusion that this can only be achieved if

. Thus, this equation is not an acceptable fundamental equation becauseit does not satisfy requirement (1b) and (3).

Finally, a sketch of S ( U ) versus U would show a simple ln ( U ) dependence.

(B) Given we have

(1)

if only if or

which represents a limited range of validity and thus this proposedfunction is not suitable as a fundamental eq.

(2)

(3) will occur if either U or faster

than or if faster than . At

constant V and N we see that where are

constants which means that as we see that in a suitable fashion. Also, as .

as since . Thus, we achieve

the result that the entropy as the temperature . Finally, the equationin this part is not a suitable fundamental eq since it does not satisfy the first twoconditions (1) and (2).

Finally, a sketch of S ( U ) would show the entropy S varying with dependence on the internalenergy U as .

(C) Given we have

(1) for and this obviously

does not satisfy the first requirement.

(2)

(3) as from the negative T

region. Thus, this proposed relation is not suitable as it violates all conditions.

2. (10 marks). The free electrons responsible for conduction in a metal can be regarded as an exotickind of ideal gas. In this case the equation of state is determined by Tf, a constant, called the Fermitemperature, given by

,

where Np is the number of free electrons in a volume V, m is the electron mass, h is Plancksconstant and k is Boltzmanns constant. When the actual temperature T

where we have used sodiums density and the molar mass in kg mol-1 and the result has units ofparticles (in this case electrons) per unit volume (units of m3). Inserting this value in the expressionfor the Fermi temperature yields

K

indicating that the electrons within a metal are in an incredible, agitated state of motion. The reasonfor this behaviour is complex and related to the Pauli exclusion principle.

U. Mizutami, Introduction to the Electron Theory of Metals, Cambridge Univ. Pr., Cambridge(2001), pg. 26 and hyperphysics.phy-astr.gsu.edu/hbase/tables/fermi.html

(b) The pressure of the free electron gas within the sodium metal is estimated from

.which indicates that the electron gas in under enormous pressure.

3. (15 marks). The figure below shows five states of a fixed amount of an ideal, monatomic gas,for which the specific heat . For the states a and b: Va = 2 m3, Pa = 103 Pa, Ta = 300 K,Vb = 1 m3, and Pb = 104 Pa. These states are linked by three different quasi-static processes. Thepurpose of this exercise is to calculuate Sb - Sa by evaluating the ratio dQ / T along these three pathsto confirm that the numerical value obtained in each case agrees with

.

The process segments are defined as:a 6 c 6 e 6 d is an adiabatic, quasi-static path;d 6 b is an isothermal process;e 6 b is a constant pressure heating process; andc 6 b is a constant volume heating process.

Throughout your calculations set Sa = 0.

4. (10 marks). See Callen (1960), Question 2.2-6 or Callen (1985), Question 2.2-6

Find the three equations of state for a system with the fundamental equation

where the symbols vo, and R denote positive, real valued constants.

Solution

The three equations of state can be obtained from the fundamental eq provided. Thus,

where the symbol denotes a variation of the internal energy while holding both the volumeand the mole number fixed. The temperature is then obtained as a limiting ratio

or

Of course, this expression may also be obtained by more conventional calculus techniques but thevariational technique of differentiation is helpful when the expressions are complex. The pressureand chemical potential are obtained directly as

so that or , and

where this final expression for the chemical potential can be written as either

or .

5. (10 marks). See Callen (1985), Question 2.2-7

A particular system is characterized by the fundamental equation

where A is an arbitrary constant. If N moles of substance, initially at temperature To and pressure Po,is expanded isentropically (at fixed entropy; s = a constant) until the pressure is halved, then whatwill be the final temperature?

Solution

We begin with the given fundamental eq and determine the temperature and pressure equations ofstate as follows:

or .

or .

As stated in the question, PI = Po and TI = To (the initial conditions) while 2PF = PI and we areasked to determine TF when the volume V is expanded isentropically (at fixed entropy S ). Thus,

and

where .

which means that .

6. (10 marks). Calculate the work and energy input via heat when 2 kg of water undergoes anisothermal process in which the volume changes quasi-statically from 200 m3 to 0.2 m3 at 100oC.

Solution From any table or graph for the properties of water and steam is should be apparent thatthe initial state os om tje vapour region while the final state is in the mixed two-phase region. Weconsider the process in two steps; denoted by below by symbols (1) and (2). The specific volumeof saturated vapour is 1.673 m3 kg-1 at 100oC.

(1) For isothermal compression in the vapour phase we have: VI = 200 m3; VF = 2 1.7 m3 = 3.4 m3. Also, for the vapour phase compression in this case the specific internal energy u is approximatelyconstant, and for the purposes of this calculation we shall treat the vapour as an ideal gas. Now,from the tabulated or graphical representation of water and steam, we have that P = 1.0 105 Pawhen the specific volume v = 1.7 m3 kg-1 at 100oC. Hence for 2 kg of water vapour at 100oC,

P V = 1.01 105 1.673 2 = 3.38 105 Jand thus the work required is

and since we have that .

(2) For isothermal, isobaric (constant pressure) compression within the two-phase region we have:VI = 3.4 m3; VF = 0.2 m3 while P = 1.01 105 Pa. The work performed is

Now, UI = 2 2.51 MJ and UF = 2 0.55 MJ so the difference is . Hence, the

heat outflow for this process is . Finally, for the complete process, one obtains

and .

7. (10 marks). A vessel with a volume of 15 m3 contains liquid water and steam in equilibrium at50oC. The mass of water in the liquid phase is 1.0 kg. Determine the mass of water in the vapourphase.

Solution Let m denote the mass of vapour phase steam. We know the volume is an extensivevariable and we shall assume that the total volume is the sum of the volumes of the individualphases, whether they are mixed or not. From either tabulated or graphical water-steam data we findthat the specific volume of saturated water at 50oC is 0.001 m3 kg-1 while, for saturated steam thecorresponding value is 12.03 m3 kg-1. Hence, the total volume is given by

V = 0.00101 1.0 + 12.03 m = 15 m3.Thus, m ~ 1.25 kg. Please notice that the mass of steam is not very sensitive to the mass of liquidwater in the vessel because the volume occupied by the liquid phase is only a small fraction of thevessel. Not until the mass of water is of the order of several hundred kilograms or more will thevolume and hence the mass of steam in the vessel become significantly affected.

8. (15 marks). The figure below shows isobaric (constant pressure) equilibrium states for water,using the specific volume and the Celsius temperature as thermodynamic coordinates. These statesare near the well-known anomalous point for water where the density is a maximum with respect tochanges in the temperature. It is evident that if we try to use the pressure P and the volume V as thestate coordinates in this region, then two values of the temperature will be obtained for a given pair( P, V ). Does this situation then imply that the temperature is not a function of state ... contrary tothe zeroth law? Explain.

Solution The zeroth law allows us to show formally that the temperature is a property of thethermodynamic state of a system. This principle was recognized by Maxwell (1872) who referredto it as the law of equal temperatures. The modern name was adopted much later in response toa suggestion by Fowler and Guggenheim (1939) who offered the view that This postulate of theexistence of temperature could with advantage be known as the zeroth law of thermodynamics. The statement of the law is If the bodies B1 and B2 are in mutual thermal equilibrium, and the bodies B2and B3 are in thermal equilibrium as well, then bodies B1 and B2 will also be in thermal equilibrium.

The figure above appears to cast doubt about the zeroth law and the statement that the temperatureis a function of state as this system can have two temperatures while in the same specific volumestate. However, the problem is not with the zeroth law but with our interpretation of it.

While it may seem to be improper to enlarge upon the interpretation of laws when inconsistenciesarise, physical laws should be discarded only when there is no avenue to resolve them. In effect thefocus of this problem moves from the law itself to an understanding of the objects to which itapplies. Here, the difficulty occurs because the coordinates ( P, V ) does not constitute a proper pairof state variables. Usually the coordinates ( P, V ) are satisfactory but they cannot be used in allsituations, although it is commonly assumed otherwise.

The selection of a consistent set of thermodynamic variables is rather like the choice of coordinatesfor a point in Euclidean space. For example, we can use Cartesian coordinates ( x, y, z ) or we canuse spherical coordinates ( , , ) to represent a given point. The choice depends on the uniqueapplication, such as the geometric symmetry of the problem of interest. However, points inEuclidean space cannot be represented, in general, by an arbitrary selection of three coordinateschosen from these six. For example, using the coordinates ( , x, y ) would normally identify justtwo points. Similarly, a consistent set of thermodynamic coordinates for a system cannot beestablished by selecting variables from a menu in an arbitrary manner. Thus, the state of a fluidcannot always be represented by ( P, V ) and that other coordinates, such as ( V, T ), ( P, T ) or ( U,V ) should be used instead.

9. (10 marks). A body is cooled from an initial temperature, Ta, to a final temperature, Tb. Themass of the body is m and the specific heat of the body for this process is a constant, denoted by c. The temperature of the heat sink for the refrigerator (involved in the cooling process) is also Ta. Show that the minimum work input to the refrigerator for this process is given by

.

Solution This problem is discussed in Callen (1985), Section 4-5. As stated there the maximumwork theorem states that for all processes leading from the specific initial state to the specified finalstate of the system, the delivery of work is maximum and the delivery of heat is a minimum for areversible process. The applicable equation is (4.13) which reads

where, for our particular case, we have

,

,

since ,

10. (10 marks). Two identical bodies, denoted as Bx and By, both with mass m are specific heat care at temperatures Tx and Ty, respectively. The bodies are located within an adiabatic enclosure. At a particular moment in time, the bodies are placed in thermal contact.

(a) Determine the final, common temperature, Tw, of the composite system Bx + By, and(b) Show that the change in entropy of the composite system is given by

.

Solution We have two identical systems whose heat capacity CV = mc is independent of temperature. If the systems are brought into thermal contact through a rigid, diathermal wall and isolated from thesurroundings, then heat flows from the high T system to the low T system until thermodynamicequilibrium is reached at a temperature Tf. We know from experience that such a process isirreversible and we can calculate the total energy change of the two systems and show that it ispositive to confirm this conclusion. The final temperature is determined from the first law ofthermodynamics as

whereupon .

Since we can calculate changes in entropy in reversible processes we have to construct a reversibleprocesses which takes both systems from their initial state to their final state reversibly. When theheat capacity CV doe s not depend on T, the change in the entropy of the system in a reversible,isochoric process is given by

where the second integral vanishes because the systems volumes do not change and the final integralhas been evaluated assuming the specific heat is constant. Thus, using this result for our two bodies

and, since always, the inequality holds in accord with our

expectation of the second law of thermodynamics. Thus, the process of heat flow from a system athigher T to a system at lower T is irreversible.

11. (20 marks). The fundamental relation for a photon gas is given by the expression

(1)

where c is the speed of light in a vacuum and is the Stefan-Boltzmann constant.

(a) Obtain the Helmholtz free energy F ( T, V ) where F is defined as F = U - TS.(b) Show that U = U ( S, V ) can be obtained from the Helmholtz free energy by taking the inverseLegendre transform.(c) Obtain an expression for the pressure, P, as a function of the entropy S and the volume V and asa function of the temperature T and the volume V. That is, use the two definitions of energy U and

F to obtain the pressure.

Solution (a) We seek to eliminate the entropy S from the Helmholtz potential definition F = U -TS and make the temperature T the independent variable. First, we obtain the temperature as

(2)

so that we can express the entropy in terms of the temperature as

. (3)

Hence, and . (4a.b)

Thus, the Helmholtz potential can be written as

(5) .

Solution (b) Notice that the expressions for S and U as functions of T and V are not fundamentalrelations because, taken alone, they cannot be used in general to reconstruct the relation U ( S, V ). However, the Helmholtz relation F ( T, V ) is a fundamental relation, as we can illustrate by takingthe inverse Legendre transform. Begin by calculating S using

. (6)

Hence, we can express T as a function of S, yielding Eq (2) again. Next, by eliminating T from thedefinition U = F + TS, we finally recover the original fundamental equation, Eq. (1) as required.

Solution (c) We can determine the pressure P in two ways. First, as a function of S and V

(7) .

Please notice that ; instead, from Eq (4b), we get

.

This illustrates the importance of clearly indicating the fixed variables when taking partialderivatives, unless they are really obvious. However, we can obtain P as a function of T and V byusing Eq (3) to eliminate S. Alternatively, we can simply take the derivative of F with respect to Vusing Eq (5) so that

(8)

and in this form we see that P is a function of T alone.

12. (10 marks). Show that the two energies U and F are related by

.Solution We use two facts to demonstrate this relation. First, the definition of the Helmholtzpotenial F = U - TS permits us to write U = F + TS. Next, we recall from the form for dF that

which permits us to identify the definition of S from the Helmholtz potential as

.

Using this definition for S in the relation U = F + TS yields the desired relation above. This is oneof several relations that are collectively known as Gibbs-Helmholtz equations.

13. (10 marks). When the volume of a simple fluid changes in a reversible, adiabatic process theassociated temperature change is determined by the derivative , assuming that N is aconstant. Obtain an expression for this derivative in terms of the quantities cV, and T.

Solution There are two approaches available. In both cases one must bring the entropy S insidethe partial derivative. Thus, one possibility is to recognize the Maxwell relation

which arises from the internal energy relation dU = T dS - P dV + dN. The next step is todivide both numerator and denominator by dT to obtain

which simplifies after using the definitions , and the

triplet relation

.

Alternatively, one may proceed directly using the triplet relation

and then re-arrange as

using the Helmholtz potential dF = - S dT - P dV + dN to alter the numerator. This relation is asderived perviously. The relation also appears reasonable because we expect that an increase in

volume under isentropic conditions (i.e. adiabatic boundaries) would produce a temperature drop.

14. (10 marks). When a simple fluid undergoes an adiabatic throttling process the enthalpy H isconstant. The temperature change associated with this type of process is represented by the Joule-Thomson coefficient which is defined as

where H = U + PV and the mole number N is assumed fixed. Obtain an expression for .

Solution The procedure for simplifying the Joule-Thomson coefficient is to first bring the enthalpywithin the partial derivative using the triplet relation

so that

after using the expression dH = T dS + V dP + dN for the enthalpy. The denominator simplifiesto an expression involving cP since the second partial vanishes while the numerator simplifies afterusing the Maxwell relation . As a consequence, we obtain

for the Joule-Thomson coefficient. This relation shows that the sign can change depending on thevalue of the bracketed term. For an ideal gas this term is zero since always.