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Q 5.72

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An Introduction to Thermal Physics
Found in: Page 200
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

Repeat the previous problem for the diagram in Figure 5.35 (right), which has an important qualitative difference. In this phase diagram, you should find that β and liquid are in equilibrium only at temperatures below the point where the liquid is in equilibrium with infinitesimal amounts of α and β . This point is called a peritectic point. Examples of systems with this behaviour include water + NaCl and leucite + quartz.

Starting at x=0 on the left, theα phase and the liquid phase are stable, then the liquid phase is stable, then the β phase and the liquid phase are stable, then a narrow range of x where only the β phase is stable, and ultimately just the liquid phase is stable.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given information

β and liquid are in equilibrium only at temperatures below the point where the liquid is in equilibrium with infinitesimal amounts of α and β . This point is called a peritectic point.

Step 2: Explanation

Consider the Gibbs free energy graph below, which shows a system with three solid phases: α,β and γ One is a pure A substance, one is a pure B substance, and one is a mixture of the two.

Step 3: Explanation

First, as shown in the diagram, we draw three tangent lines from left to right: first, the a phase plus the liquid phase are stable, then the liquid phase is stable, then the β phase plus the liquid phase are stable, then a narrow range of x where only the β phase is stable, and finally only the liquid phase is stable.

Step 4: Calculations

The Gibbs free energy is given as:

G=U-TS+PV

At constant entropy and constant pressure, differentiate the Gibbs free energy to get:

dG=dU-SdT+PdV

By increasing the temperature

∂G∂T=-S

We can see from this equation that as the temperature rises, the stability ranges of α and β vanish. When we lower the temperature, the stability of γ plus the liquid appears, as shown in the preceding figure for large x. As the temperature drops, the liquid's stable range narrows until it evaporates at two locations, one of which is known as the eutectic point (at which all the liquid freezes). Only at temperatures below the red point, where the liquid is in equilibrium with tiny amounts of α and β and the liquid in equilibrium. This point is known as the peritectic point.

Most popular questions for Physics Textbooks

For a magnetic system held at constant TT and HH (see Problem 5.17 ), the quantity that is minimized is the magnetic analogue of the Gibbs free energy, which obeys the thermodynamic identity

Phase diagrams for two magnetic systems are shown in Figure 5.14 ; the vertical axis on each of these figures is μ0Hμ0H (a) Derive an analogue of the Clausius-Clapeyron relation for the slope of a phase boundary in the HH - TT plane. Write your equation in terms of the difference in entropy between the two phases. (b) Discuss the application of your equation to the ferromagnet phase diagram in Figure 5.14. (c) In a type-I superconductor, surface currents flow in such a way as to completely cancel the magnetic field (B, not H)(B, not H) inside. Assuming that MM is negligible when the material is in its normal (non-superconducting) state, discuss the application of your equation to the superconductor phase diagram in Figure 5.14.5.14. Which phase has the greater entropy? What happens to the difference in entropy between the phases at each end of the phase boundary?

Graphite is more compressible than diamond.

(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?

(b) The isothermal compressibility of graphite is about 3 x 10-6 bar-1, while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.46.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

Let the system be one mole of argon gas at room temperature and atmospheric pressure. Compute the total energy (kinetic only, neglecting atomic rest energies), entropy, enthalpy, Helmholtz free energy, and Gibbs free energy. Express all answers in SI units.

When solid quartz "dissolves" in water, it combines with water molecules in the reaction

SiO2( s)+2H2O(l)⟷H4SiO4(aq)

(a) Use this data in the back of this book to compute the amount of silica dissolved in water in equilibrium with solid quartz, at 25° C

(b) Use the van't Hoff equation (Problem 5.85) to compute the amount of silica dissolved in water in equilibrium with solid quartz at 100°C.

Consider a fuel cell that uses methane ("natural gas") as fuel. The reaction is

CH4+2O2⟶2H2O+CO2

(a) Use the data at the back of this book to determine the values of ΔH and ΔG for this reaction, for one mole of methane. Assume that the reaction takes place at room temperature and atmospheric pressure.

(b) Assuming ideal performance, how much electrical work can you get out of the cell, for each mole of methane fuel?

(c) How much waste heat is produced, for each mole of methane fuel?

(d) The steps of this reaction are

at - electrode: CH4+2H2O→CO2+8H++8e-at - electrode: 2O2+8H++8e-→4H2O

What is the voltage of the cell?
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