In constructing the phase diagram from the free energy graphs in Figure 5.30, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas's free energy curve at TA. Draw a qualitatively accurate phase diagram for such a system, showing how you obtained the phase diagram from the free energy graphs. Show that there is a particular composition at which this gas mixture will condense with no change in composition. This special composition is called an azeotrope.

Short Answer

Expert verified

The entropy of a gas increases as the temperature rises; because the gas has more degrees of freedom, it has more entropy.

As a result of the negative sign of entropy, the gas curve falls. Lowering the temperature causes the curve to rise until it coincides with the liquid curve at one point, forming an azeotrope combination.

Step by step solution

01

Given information

In constructing the phase diagram from the free energy graphs in Figure 5.30, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas's free energy curve at TA.

02

Explanation

Consider the following curve, which shows the free energy of the gas and liquid at temperature TA. We can see that the gas curve is more concave than the liquid curve, indicating that the two curves intersect at two points, indicating that the liquid and gas are stable in two different composition ranges.

03

Explanation

Draw a tangent on a graph between x and T (the phase diagram) at the two intersection locations as indicated in the accompanying figure; this tangent intersects with the gas and liquid curves. Then draw perpendicular lines from the four intersection points on a graph between x and T (the phase diagram).

04

Explantion

The Gibbs free energy is given by:

G=U+PV-TS

At constant volume and entropy, the change in Gibbs free energy is as follows:

dG=dU+VdP-SdT

By increasing the temperature, we get

GT=-S

The entropy of a gas increases as the temperature rises; because the gas has more degrees of freedom, it has more entropy.

As a result of the negative sign of entropy, the gas curve falls. Lowering the temperature causes the curve to rise until it coincides with the liquid curve at one point, forming an azeotrope combination.

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Most popular questions from this chapter

Use the result of the previous problem to estimate the equilibrium constant of the reactionN2+3H22NH3at 500° C, using only the room- temperature data at the back of this book. Compare your result to the actual value of K at 500° C quoted in the text.

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 S fixed, each/S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U ) 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, and G ). 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.


Imagine that you drop a brick on the ground and it lands with a thud. Apparently the energy of this system tends to spontaneously decrease. Explain why.

Show that equation 5.40 is in agreement with the explicit formula for the chemical potential of a monatomic ideal gas derived in Section 3.5. Show how to calculate μ°for a monatomic ideal gas.

Repeat the preceding problem with T/TC=0.8

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