Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Q 5.74 (page 202)

Check that equations 5.69 and 5.70 satisfy the identityG=NAμA+NBμB (equation 5.37)

Short Answer

Expert verified

Hence, the identity is satisfied.

G=NAμA+NBμB

Step by step solution

01

Given information

The identityG=NAμA+NBμB

02

Explanation

The chemical potentials of the solvent and solute are determined by the following equations:

μA=μ0-NBkTNAμB=f+klnNBNA

Where, μ0andfare functions of T and P.

We need to check that these equation satisfy equation 5.37, which is given by:

G=NAμA+NBμB

substitute with the above equations to get:

NAμA+NBμB=NAμ0-NBkT+NBf+NBklnNBNA

By using ln(A/B)=ln(A)-ln(B), we can write G as

NAμA+NBμB=NAμ0-NBkT+NBf+NBklnNB-NBkln(N)(1)

The Gibbs free energy for pure solvent is given by (from equation 5.68):

G=NAμ0+NBf-NBkTlnNA+NBkTlnNB-NBkT(2)

Comparing (1) and (2)

G=NAμA+NBμB

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.

Start your free trial

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

Plot the Van der Waals isotherm for T/Tc = 0.95, working in terms of reduced variables. Perform the Maxwell construction (either graphically or numerically) to obtain the vapor pressure. Then plot the Gibbs free energy (in units of NkTc) as a function of pressure for this same temperature and check that this graph predicts the same value for the vapor pressure.

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 (∂S/∂T)Vand (∂S/∂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 (∂S/∂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+TVβ2κT

(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 βvs.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.

Sketch a qualitatively accurate graph of G vs. T for a pure substance as it changes from solid to liquid to gas at fixed pressure. Think carefully about the slope of the graph. Mark the points of the phase transformations and discuss the features of the graph briefly.

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.)

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.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Physics of Motion

Read Explanation

Scientific Method Physics

Read Explanation

Kinematics Physics

Read Explanation

Turning Points in Physics

Read Explanation

Electricity and Magnetism

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.

Sign-up for free