ME581: Homework Two
Fall 2008

Prof. W. H. Warnes
Office: Dearborn Hall 303E
Phone and Voice Mail: (541) 737-7016
FAX (541) 737-2600
matsci.oregonstate.edu

DUE Thursday, October 23, 2008

1) An Einstein solid has 100 distinguishable particles, each of which can sit in one of five energy levels, denoted e₁ = e*, e₂ = 2e*, e₃ = 3e*, e₄ = 4e*, e₅ = 5e*.

a) What is the multiplicity of the macrostate for which there are 91 particles in e₁, 4 particles in e₂, 3 particles in e₃, 1 particle in e₄, and 1 particle in e₅?

b) What is the total energy of the system in part a?

c) We can keep the total energy the same by moving one particle from e₂ to e₁, and one particle from e₂, to e₃. This is a different macrostate, and has a different multiplicity. What is the multiplicity of this new macrostate?

d) What is the ratio of the multiplicity of the macrostate in part (c) to that in part (a)? What does this tell you about the change in ENTROPY of the system in moving from macrostate (a) to (c)?

EXTRA CREDIT: The Boltzmann distribution tells us how the number of particles in each energy level change. This can be simplified (for a large enough system) to

Together with a couple of equations fixing the total number of particles, and holding the total energy of the system constant it is possible for one to solve for the equilibrium distribution of particles (the values of the N_i which maximizes the multiplicity.) Calculate the equilibrium distribution of particles for this system.

2) Chapter 4 in Gaskell derives the change in configurational entropy () on mixing for a (non-interacting) random mixture of two elements in a binary solution (see equation 4.18 for example.)

a) Calculate the for a binary mixture of a total of 1 [mole] of atoms that is 50% A atoms and 50% B atoms.

b) Would you expect the to increase or decrease for 1 [mole] of a quaternary alloy that is made of equal parts of elements A, B, C, and D? Explain in a few short sentences.

3) I am interested in learning about the reaction to make zirconia: Zr(solid) + O₂(gas) = ZrO₂(solid). To try to get a handle on it, follow the procedure from the book and lecture to determine the heat of reaction as a function of temperature.

a) Determine an equation for the enthalpy as a function of temperature for O₂(gas) over the temperature range of T = 298 - 1400 [K].

b) Determine an equation for the enthalpy as a function of temperature for Zr(solid) over the same temperature range (NOTE that solid Zr goes through a phase transformation at about T = 1136 [K].)

c) Determine an equation for the enthalpy as a function of temperature for ZrO₂(solid) over the same temperature range.

d) Using your results from parts (a), (b), and (c), construct a plot (similar to Figure 6.8 from the book) of the enthalpy of reaction over the temperature range T = 298 - 1400 [K].

e) At T = 1000 [K], is the reaction endo- or exo-thermic?

f) How much heat is required or given off for the reaction at T = 1000 [K]?

EXTRA CREDIT: Imagine we have a sealed, adiabatic container with 1 mole of Zr(s) and 1 mole of O₂(gas) initially held at 1000 K (for instance, in a reaction furnace in Dr. Cann's ceramics lab.) Assume the reaction goes to completion (100% ZrO₂(solid)). Estimate the final temperature of the ZrO₂(solid).

4) Lithium has a very large coefficient of thermal expansion of about 56 x 10^-6 [(m/m)/K] at room temperature. Calculate the pressure dependence of the enthalpy of solid Li at room temperature.

5) Following a procedure similar to that in class, use the Clausius-Clapeyron equation to find the saturated vapor pressure of liquid Al as a function of temperature between 1000-2000 [K]. BE SURE TO LIST ANY APPROXIMATIONS OR ASSUMPTIONS YOU HAVE USED.

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SOLUTIONS

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1) An Einstein solid has 100 distinguishable particles, each of which can sit in one of five energy levels, denoted e₁ = e*, e₂ = 2e*, e₃ = 3e*, e₄ = 4e*, e₅ = 5e*.

1a) What is the multiplicity of the macrostate...

Given the multiplicity equation from lecture and the book (eq. 4.1);

1b) What is the total energy of the system in part a?

Adding up the energy of each particle gives:

1c) ...What is the multiplicity of this new macrostate?

1d) What is the ratio of the...

The important feature here is that, for such a small change in the configuration, there is a large change in the multiplicity. This also implies a large change in the thermal entropy, through the relation . This gives an increase in the entropy of

EXTRA CREDIT: Calculate the equilibrium distribution...

Five equations and five unknowns will give the answer. Here are the five equations I used:

I used Mathematica to get the answer:

N₁ = 86
N₂ = 12
N₃ = 2
N₄ = 0
N₅ = 0.

2a) Calculate the for a binary mixture...

The configurational entropy is given (in one form) by equation 4.18, and in another form (from lecture) as

2b) Would you expect the to increase or decrease...

I would expect the configurational entropy to increase, because adding any additional elements will increase the number of ways the atoms can be arranged. In fact, the entropy increases by a factor of two:

3) I am interested in learning about the reaction to make zirconia...

3a) Determine an equation for the enthalpy as a function of temperature for O₂(gas)...

This is identical to what we did in class, and to the example in the book:

3b) Determine an equation for the enthalpy as a function of temperature for Zr(solid)...

For the temperature range of 298-1136 [K] (in the ALPHA-phase):

For the temperature range of 1136-1400 [K] (in the BETA-phase): We use the same procedure (with the c_p of BETA-Zr), but we need to add the enthalpy of the ALPHA-phase at 1136 [K] and the latent heat of transformation:

3c) Determine an equation for the enthalpy as a function of temperature for ZrO₂(solid)...

And, again, this time using the data for ZrO₂(solid):

3d) ...construct a plot of the enthalpy of reaction...

First, the FULL equation: Start with Hess' Law (products minus reactants) to find the enthalpy of reaction

For the temperature range of 298-1136, we have:

For the temperature range of 1136-1400[K], we have:

I used EXCEL to give me a plot:

3e) At T = 1000 [K], is the reaction endo- or exo-thermic?

Because the enthalpy of reaction is negative (about -1095 [kJ/mole] at T = 1000 [K]), the reaction is EXOTHERMIC.

3f) How much heat is required or given off for the reaction at T = 1000 [K]?

Plugging into the equation gives the exact value of 1095.11 [kJ/mole] given off by the reaction.

EXTRA CREDIT: Estimate the final temperature of the ZrO₂(solid).

When the reaction occurs, because we are at constant temperature, the heat generated is equal to the reaction enthalpy calculated in part (f). Since the container is adiabatic, none of the heat can be released to the outside, and so the heat will raise the temperature of the ZrO₂(solid) ceramic. We can use the specific heat equation to estimate the temperature rise due to the heat evolution:

Solve this for T, either by iteration or using a solver program (I used Mathematica again...)

T = 10658 [K].

WOW! That's pretty hot! But is it reasonable? There are two ideas to get from this problem. First, the metal oxidation process (and many other materials reactions) are powerful, so it is important to think about this kind of thing in doing laboratory experiments and materials fabrication that involve reaction processes. Making ZrO₂ from elemental Zr and O is potentially very dangerous, and could lead to large temperatures and large pressures if not properly controlled. Second, it turns out that nature only provides us with Zr chemically bound as oxide, so we are always starting from the ZrO₂ and working hard to make the elemental metal, Zr. Making bulk ZrO₂ objects involves sintering of ZrO₂ powder, rather than oxidizing a metallic part.

4) Calculate the pressure dependence of the enthalpy...

From thermodynamic relations, we can write:

We need to know the molar volume for solid Li. The density at room temperature is 534 [kg/m³], with an atomic weight of 6.941 [g/mole].

The final equation for the pressure dependence of the enthalpy of solid Li is

(Remember that, by convention, the enthalpy of the stable phase is zero at 298 [K].)

5) ...use the Clausius-Clapeyron equation to find the saturated vapor pressure of liquid Al...

The Clausius-Clapeyron equation gives the relationship between P and T at equilibrium for a two-phase mixture of a gas phase and a condensed phase. The equation is:

where we have made the following assumptions:

1) that the molar volume of the condensed phase is negligible compared to the molar volume of the gas phase, and

2) that the gas phase is an IDEAL gas.

The specific heat at constant pressure for the liquid Al is found in the back of the book, and the specific heat for gaseous Al is approximated to be that of an ideal gas:

Using these values for the specific heat gives us the following expression for the difference in enthalpy between the phases as a function of temperature:

We need to find the value of the constant A. We get this in the same way as problem 3 part (b). From the CRC Handbook data, we have the heat of transition (latent heat) of vaporization as:

With these values, and the specific heats of the liquid and vapor phases, we can calculate the enthalpy difference between the vapor and liquid Al at 298 as follows:

The quantity we want is the distance from point (a) to point (d). We'll get there by finding the enthalpy change along the path (a)-(b)-(c)-(d). The (a)-(b) path is the enthalpy change in the liquid from 298 to the boiling temperature. The (b)-(c) path is just the latent heat of vaporization. The (c)-(d) path is the change in the enthalpy of the vapor from the boiling temperature to 298. Writing this all out explicitly gives the enthalpy difference at 298 as:

Now we plug this into the Clausius-Clapeyron equation and do another integral to get:

We can find the integration constant by knowing one value of P(T) for Aluminum. The easiest point to find is the standard boiling point, which has a vapor pressure of 1 [atm] at a temperature of 2740 [K]. Plugging these values in for P and T, and keeping the units consistent between DELTA H and R, we get the value of the integration constant, B, and the final equation: