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Chapter 5: Q 5.8 (page 158)

Derive the thermodynamic identity for G (equation 5.23), and from it the three partial derivative relations 5.24.

Short Answer

Expert verified

The expression for change in G is (μdN-SdT+VdP) and the relations are

S=-∂G∂TP,N,V=∂G∂PT,Nandμ=∂G∂NT,P.

Step by step solution

01

Explanation

Write the expression for Gibbs free energy.

G=U-TS+PV

Here, G is Gibbs free energy, T is the absolute temperature, S is the entropy, P is the pressure and V is the volume.

Write the expression for the infinitesimal change in G.

dG=dU-TdS-SdT+PdV+VdP……..(1)

Write the expression for the infinitesimal change in U.

dU=TdS-PdV+μdN

Substitute (TdS-PdV+μdN)for dU in expression (1).

dG=μdN-SdT+VdP……..(2)

Rearrange expression (2) for constant P and constant N.

dG=-SdT

Rearrange the above expression.

S=-∂G∂TP,N
02

Calculation

Rearrange expression (2) for constant T and constant N.

dG=VdP

Rearrange the above expression.

V=∂G∂PT,N

Rearrange expression (2) for constant T and constant P.

dG=μdN

Rearrange the above expression.

μ=∂G∂NT,P

Thus, the expression for change in G is (μdN-SdT+VdP) and the relations are

S=-∂G∂TP,N,V=∂G∂PT,Nandμ=∂G∂NT,P.

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Most popular questions from this chapter

Problem 5.58. In this problem you will model the mixing energy of a mixture in a relatively simple way, in order to relate the existence of a solubility gap to molecular behaviour. Consider a mixture of A and B molecules that is ideal in every way but one: The potential energy due to the interaction of neighbouring molecules depends upon whether the molecules are like or unlike. Let n be the average number of nearest neighbours of any given molecule (perhaps 6 or 8 or 10). Let n be the average potential energy associated with the interaction between neighbouring molecules that are the same (4-A or B-B), and let uAB be the potential energy associated with the interaction of a neighbouring unlike pair (4-B). There are no interactions beyond the range of the nearest neighbours; the values of μoandμABare independent of the amounts of A and B; and the entropy of mixing is the same as for an ideal solution.

(a) Show that when the system is unmixed, the total potential energy due to neighbor-neighbor interactions is 12Nnu0. (Hint: Be sure to count each neighbouring pair only once.)

(b) Find a formula for the total potential energy when the system is mixed, in terms of x, the fraction of B.

(c) Subtract the results of parts (a) and (b) to obtain the change in energy upon mixing. Simplify the result as much as possible; you should obtain an expression proportional to x(1-x). Sketch this function vs. x, for both possible signs of uAB-u0.

(d) Show that the slope of the mixing energy function is finite at both end- points, unlike the slope of the mixing entropy function.

(e) For the case uAB>u0, plot a graph of the Gibbs free energy of this system

vs. x at several temperatures. Discuss the implications.

(f) Find an expression for the maximum temperature at which this system has

a solubility gap.

(g) Make a very rough estimate of uAB-u0for a liquid mixture that has a

solubility gap below 100°C.

(h) Use a computer to plot the phase diagram (T vs. x) for this system.

Problem 5.35. The Clausius-Clapeyron relation 5.47 is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ∆V depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ∆V to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:

P= (constant) x e-L/RT

This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.

Use the data at the back of this book to verify the values of ∆Hand ∆G quoted above for the lead-acid reaction 5.13.

In Problems 3.30 and 3.31 you calculated the entropies of diamond and graphite at 500 K. Use these values to predict the slope of the graphite- diamond phase boundary at 500 K, and compare to Figure 5.17. Why is the slope almost constant at still higher temperatures? Why is the slope zero at T = 0?

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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