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The methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:

NaAlSi3O8NaAlSi2O6+SiO2

Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both S and V.

Short Answer

Expert verified

The temperature is 100 K above room temperature, an extra 1.7 kbar of pressure is required to keep Jadeite+Quartz stable.

Step by step solution

01

Given information

The Gibbs free energy, Entropy and Molar Volumes for Albite, Jadeite, quartz is given as:

AlbiteGa=-3711.5KJSa=204.4J/KVa=100.07cm3JadeiteGj=-28521KJSj=1335J/KVj=6040cm3QuartzGq=-856.64KJSq=4184J/KVq=2269cm3

Albitejadeite+quartz

For the above reaction

role="math" localid="1646935903662" G=Gfinal-Ginitial=Gj+Gq-Ga=(-28521-85664+37115)kJ=276kJ>0

02

Explanation

As a result, under normal settings, the Albite is more stable.

At high pressures, the jadeite + quartz combination becomes more stable.

The change in volume is calculated as

V=Va-Vj-Vq=(10007-6040-2269)cm3=1698cm3=1.69810-5m31kJkbar=10-5m3V=1698kJkbar

03

Explanation

At standard temperatures, the pressure at which jadeite + quartz becomes stable is:

P=GVP=2761698KJKJKbarP=1.6254Kbar

The coexistence line with the P-axis in the P-T diagram is thus of relevance.

The Clausius-Clapeyron equation's slope is

dPdT=SVS=Sa-Sj-Sq=(2044-1335-4184)J/K螖厂=2906J/KSlope=螖厂螖痴=2906J/K1698J/bar=1711barK

04

Conclusion

The phase diagram for Jadeite+Quartz and Albite is

Jadeite+Quartz is stable at pressures more than 1.6254kbar at room temperature. The Jadeite& Quartz-Albite phase border has a slope of 17.11 bar/K.

If the temperature is 100 K above room temperature, an extra 1.7 kbar of pressure is required to keep Jadeite+Quartz stable.

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Most popular questions from this chapter

In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout.

(a) Expand the van der Waals equation in a Taylor series in , keeping terms through order . Argue that, for T sufficiently close to Tc, the term quadratic in (V-VC)becomes negligible compared to the others and may be dropped.

(b) The resulting expression for P(V) is antisymmetric about the point V = Ve. Use this fact to find an approximate formula for the vapor pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary,dP/dT

( c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find Vg-VlTc-T.8, where (3 is known as a critical exponent. Experiments show that (3 has a universal value of about 1/3, but the van der Waals model predicts a larger value.

(d) Use the previous result to calculate the predicted latent heat of the transformation as a function of temperature, and sketch this function.

The shape of the T = Tc isotherm defines another critical exponent, called P-PcV-VcCalculate 5 in the van der Waals model. (Experimental values of 5 are typically around 4 or 5.)

A third critical exponent describes the temperature dependence of the isothermal compressibility, K=-t This quantity diverges at the critical point, in proportion to a power of (T-Tc) that in principle could differ depending on whether one approaches the critical point from above or below. Therefore the critical exponents 'Y and -y' are defined by the relations

T-Tc-Tc-T-'

Calculate K on both sides of the critical point in the van der Waals model, and show that 'Y = -y' in this model.

Use the data at the back of this book to calculate the slope of the calcite-aragonite phase boundary (at 298 K). You located one point on this phase boundary in Problem 5.28; use this information to sketch the phase diagram of calcium carbonate.

Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,

VUS=SUV

where each /Vis taken with Sfixed, each /Sis taken with Vfixed, and Nis always held fixed. From the thermodynamic identity (forU) you can evaluate the partial derivatives in parentheses to obtain

TVS=-PSV

a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H,F,andG ). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you've done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems.

Use a Maxwell relation from the previous problem and the third law of thermodynamics to prove that the thermal expansion coefficient (defined in Problem 1.7) must be zero at T=0.

Is heat capacity (C) extensive or intensive? What about specific heat (c) ? Explain briefly.

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