Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Everything in this section assumes that the total pressure of the system is fixed. How would you expect the nitrogen-oxygen phase diagram to change if you increase or decrease the pressure? Justify your answer.

Short Answer

Expert verified

The phase area will get smaller.

Step by step solution

01

Given information

The total pressure of the system is fixed.

02

Explanation

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

As a result, as the pressure rises, the Gibbs free energy rises, resulting in:

GP=V>0

Because the volume of the liquid is less than that of the gas,:

role="math" localid="1647026301082" Vliq>VgasGPgas>GPliq

As a result, if the slope of the curve increases or decreases, it will not increase or decrease in the same trend, implying that the phase area will shrink and phase area will get smaller

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 Vaia!

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

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?

The compression factor of a fluid is defined as the ratio PV/NkT; the deviation of this quantity from 1 is a measure of how much the fluid differs from an ideal gas. Calculate the compression factor of a Van der Waals fluid at the critical point, and note that the value is independent of a and b. (Experimental values of compression factors at the critical point are generally lower than the Van der Waals prediction, for instance, 0.227 for H22O, 0.274 for CO22, and 0.305 for He.)

The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S =kln(4) , since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F=0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very

The formula for CP-CV derived in the previous problem can also be derived starting with the definitions of these quantities in terms of U and H. Do so. Most of the derivation is very similar, but at one point you need to use the relation P=-(F/V)T.

Suppose you have a box of atomic hydrogen, initially at room temperature and atmospheric pressure. You then raise the temperature, keeping the volume fixed.

(a) Find an expression for the fraction of the hydrogen that is ionised as a function of temperature. (You'll have to solve a quadratic equation.) Check that your expression has the expected behaviour at very low and very high temperatures.

(b) At what temperature is exactly half of the hydrogen ionised?

(c) Would raising the initial pressure cause the temperature you found in part (b) to increase or decrease? Explain.

(d) Plot the expression you found in part (a) as a function of the dimension- less variable t = kT/I. Choose the range of t values to clearly show the interesting part of the graph.

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.

Sign-up for free