/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q.5.43 Repeat the preceding problem wit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Repeat the preceding problem with T/TC=0.8

Short Answer

Expert verified

The pressure of phase transition is 0.38 Pa

Step by step solution

01

Step 1

From problem 5.51 we know that

p=8t3v-1-3v2
02

substituting the value of t=0.8 and iterating the value and plotting the graph

p=8(0.8)3v-1-3v2p=6.43v-1-3v2

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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.

When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables, Rewrite the van der Waals equation in terms of these variables, and notice that the constants a and b disappear.

In the previous section I derived the formula (∂F/∂V)T=-P. Explain why this formula makes intuitive sense, by discussing graphs of F vs. V with different slopes.

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,
∂∂V∂U∂S=∂∂S∂U∂V

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

∂T∂VS=-∂P∂SV

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.


A mixture of one part nitrogen and three parts hydrogen is heated, in the presence of a suitable catalyst, to a temperature of 500° C. What fraction of the nitrogen (atom for atom) is converted to ammonia, if the final total pressure is 400 atm? Pretend for simplicity that the gases behave ideally despite the very high pressure. The equilibrium constant at 500° C is 6.9 x 10-5. (Hint: You'l have to solve a quadratic equation.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.