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Q 5.39

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An Introduction to Thermal Physics
Found in: Page 176
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

Consider again the aluminosilicate system treated in Problem 5.29. Calculate the slopes of all three phase boundaries for this system: kyanite andalusite, kyanite-sillimanite, and andalusite-sillimanite. Sketch the phase diagram, and calculate the temperature and pressure of the triple point.

The temperature at the triple point is 690 K, and the pressure is 3.3 kbar.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given information

By using the relation we will calculate the slope of all three phase boundaries

dPdT=ΔSΔV

Where,

∆S is the change in entropy

∆V is the change in volume

P is the pressure

T is the temperature

The slope of the phase for the kyanite and andalusite boundary is as follows:

dPdT=Sa-SkVa-Vk

Where,

Sa and Va are the entropy and volume of the andalusite

Sk and Vk are the entropy and volume of the kyanite.

Substituting the values

dPdT=93.22 J/K-83.81 J/K5.153 J/bar-4.409 J/bar =12.65bar/K

Step 2: Explanation

The slope of the phase for the andalusite and sillimanite boundary is as follows:

dPdT=Sa-SsVa-Vs

Where,

Ss and Vs are the entropy and volume of the sillimanite

Substituting the values

dPdT=93.22 J/K-96.11 J/K5.153 J/bar-4.990 J/bar =-17.73bar/K

The slope of the phase for the kyanite and sillimanite boundary is as follows:

dPdT=Sk-SsVk-Vs

Substituting the values

dPdT=96.11 J/K-83.81 J/K4.99 J/bar-4.409 J/bar =21.17bar/K

Step 3: Explanation

The transition from kyanite to andalusite happens at 1 bar at T= 427K, giving up one point on the phase boundary, according to problem 5.29. Now you must come up with a term for this couple.

The transition pressure for kyanite and andalusite is as follows:

P=dPdT·T+CC=P-dPdT·TC=1bar-(12.6bar/K)(427 K)C=-5.38kbar

The transition pressure for andalusite and sillimanite is as follows:

C=P-dPdT·TC=1 bar -(-17.73bar/K)(875 K)C=15.51 kbar

The transition pressure for kyanite and sillimanite is as follows:

C=P-dPdT·TC=1bar-(21.17bar/K)(532 K)C=-11.26 kbar

Step 4: Explanation

Calculate the temperature T and pressure p at the places where the kyanite andalusite (k↔a) line intersects the andalusite ↔ sillimanite (a↔ s) line.

role="math" localid="1646933788841" P=(12.6bar/K)T-5.38kbar for k↔aP=(-17.73bar/K)T+15.51kbar for a↔s

Equating both the values

(12.6bar/K)T-5.38kbar=(-17.73bar/K)T+15.51barT=15.51kbar+5.38kbar12.6bar/K+17.73bar/KT=688.7KT=690K

Step 5: Explanation

As a result, the pressure at the place where the kyanite ↔andalusite (k ↔a) line intersects with the andalusite ↔sillimanite (a ↔s) line is:

P=(12.6bar/K)(688.7 K)-5.38kbarP=3.298 kbarP=3.3 kbar

Calculate the temperature T and pressure P at the places where the andalusite↔sillimanite (a↔s) line interacts with the kyanite↔ sillimanite (k ↔s) line.

P=(-17.73bar/K)T+15.51kbar for a↔sP=(21.17bar/K)T-11.26k bar for k↔s

Equating both the values

(-17.73bar/K)T+15.51kbar=(21.17bar/K)T-11.26kbarT=15.51kbar+11.26kbar21.17bar/K+17.73bar/KT=688.2KT=690K

As a result, the pressure at the place where andalusite↔ sillimanite (a ↔s) and kyanite ↔sillimanite (k↔s) lines connect is as follows:

role="math" localid="1646934600818" P=(-17.73bar/K)(688.2 K)+15.51kbarP=3.30kbar

Step 6: Calculation

Calculating the temperature T and pressure P at k↔a line with k↔s line.

P=(12.6bar/K)T-5.38kbar for k↔aP=(21.17bar/K)T-11.26k bar for k↔s

Equating both the values and solving for T

(21.2bar/K)T-11.26kbar=(12.6bar/K)T-5.38kbarT=11.26kbar-5.38kbar21.17bar/K-12.6bar/KT=686.1KT=690K

Therefore, the pressure of the intersection point of k ↔ a line with k↔s line will be:

P=(12.6bar/K)(686.1 K)-5.380kbarP=3.26kbarP=3.3 kbar

Step 7: Conclusion

The phase diagram for the three boundaries of kyanite, andalusite, and sillimanite is shown below.

The temperature at the triple point is 690 K, and the pressure is 3.3 kbar, according to the above diagram and calculations.

Most popular questions for Physics Textbooks

Calcium carbonate, CaCO3, has two common crystalline forms, calcite and aragonite. Thermodynamic data for these phases can be found at the back of this book.

(a) Which is stable at earth's surface, calcite or aragonite?

(b) Calculate the pressure (still at room temperature) at which the other phase

should become stable.

The partial-derivative relations derived in Problems 1.46,3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between CPandCV.

(a) With the heat capacity expressions from Problem 3.33 in mind, first considerSto be a function of T andV. Expand dS in terms of the partial derivatives (∂S/∂T)Vand (∂S/∂V)T. Note that one of these derivatives is related toCV

(b) To bring in CP, considerlocalid="1648430264419" Vto be a function ofTand P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP=0 and note that you have derived a nontrivial expression for (∂S/∂T)P. This derivative is related to CP, so you now have a formula for the difference CP-CV

(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be

CP=CV+TVβ2κT

(d) Check that this formula gives the correct value of CP-CV for an ideal gas.

(e) Use this formula to argue that CP cannot be less than CV.

(f) Use the data in Problem 1.46 to evaluateCP-CV for water and for mercury at room temperature. By what percentage do the two heat capacities differ?

(g) Figure 1.14 shows measured values of CP for three elemental solids, compared to predicted values of CV. It turns out that a graph of β vs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CP and CV agree at low temperatures but diverge in the way they do at higher temperatures.

Sketch a qualitatively accurate graph of G vs. T for a pure substance as it changes from solid to liquid to gas at fixed pressure. Think carefully about the slope of the graph. Mark the points of the phase transformations and discuss the features of the graph briefly.

In a hydrogen fuel cell, the steps of the chemical reaction are

at - electrode: H2+2OH-⟶2H2O+2e-at + electrode: 12O2+H2O+2e-⟶2OH-

Calculate the voltage of the cell. What is the minimum voltage required for electrolysis of water? Explain briefly.

In a hydrogen fuel cell, the steps of the chemical reaction are at - electrode: H2+2OH-⟶2H2O+2e-at + electrode: 12O2+H2O+2e-⟶2OH-

Calculate the voltage of the cell. What is the minimum voltage required for electrolysis of water? Explain briefly.
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