/*! 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 5.61 Suppose you need a tank of oxyge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Q 5.61 (page 194)

Suppose you need a tank of oxygen that is 95% pure. Describe a process by which you could obtain such a gas, starting with air.

Short Answer

Expert verified

The process is explained.

Step by step solution

01

Given information

A tank of oxygen that is 95% pure.

02

Explanation

The phase diagram for nitrogen and oxygen is given below:

Consider the experimental phase diagram for nitrogen and oxygen at atmospheric pressure in Figure 5.31 of the book. Starting with air (which is a mixture of nitrogen 79 percent N2 and oxygen 21 percent O2), we lower the temperature until we reach a temperature of 81.5 K, at which point the oxygen will start to liquefy, and to find the percentage of oxygen in the liquid, we draw a horizontal line from the upper curve to the lower curve until it intersects with it at this point x= 0.5, which means the oxygen in the liquid is 95 percent pure. Then, while keeping the temperature constant, we pump the gas down the liquid, then let the temperature rise again, increasing the proportion of oxygen in each cycle, and so on.

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 91影视!

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 a hydrogen fuel cell, the steps of the chemical reaction are

at-electrode:H2+2OH-鉄�2H2O+2e-at+electrode:12O2+H2O+2e-鉄�2OH-

Calculate the voltage of the cell. What is the minimum voltage required for electrolysis of water? Explain briefly.

Below 0.3 K the slope of the 掳He solid-liquid phase boundary is negative (see Figure 5.13).

(a) Which phase, solid or liquid, is more dense? Which phase has more entropy (per mole)? Explain your reasoning carefully.

(b) Use the third law of thermodynamics to argue that the slope of the phase boundary must go to zero at T = 0. (Note that the *He solid-liquid phase boundary is essentially horizontal below 1 K.)

(c) Suppose that you compress liquid *He adiabatically until it becomes a solid. If the temperature just before the phase change is 0.1 K, will the temperature after the phase change be higher or lower? Explain your reasoning carefully.

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.

If expression 5.68 is correct, it must be extensive: Increasing both NA and NB by a common factor while holding all intensive variables fixed should increase G by the same factor. Show that expression 5.68 has this property. Show that it would not have this property had we not added the term proportional to In NA!.

The partial-derivative relations derived in Problems 1.46,3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between CPandCV.

(a) With the heat capacity expressions from Problem 3.33 in mind, first considerSto be a function of TandV.Expand dSin terms of the partial derivatives (鈭�S/鈭�T)Vand (鈭�S/鈭�V)T. Note that one of these derivatives is related toCV

(b) To bring in CP, considerlocalid="1648430264419" Vto be a function ofTand P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP=0and note that you have derived a nontrivial expression for (鈭�S/鈭�T)P. This derivative is related to CP, so you now have a formula for the difference CP-CV

(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be

CP=CV+TV尾2魏T

(d) Check that this formula gives the correct value of CP-CVfor an ideal gas.

(e) Use this formula to argue that CPcannot be less than CV.

(f) Use the data in Problem 1.46 to evaluateCP-CVfor water and for mercury at room temperature. By what percentage do the two heat capacities differ?

(g) Figure 1.14 shows measured values of CPfor three elemental solids, compared to predicted values of CV. It turns out that a graph of 尾vs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CPand CVagree at low temperatures but diverge in the way they do at higher temperatures.
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Electric Charge Field and Potential

Read Explanation

Dynamics

Read Explanation

Quantum Physics

Read Explanation

Famous Physicists

Read Explanation

Electromagnetism

Read Explanation
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.