/*! 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 5.74 Check that equations 5.69 and 5... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

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

Let the system be one mole of argon gas at room temperature and atmospheric pressure. Compute the total energy (kinetic only, neglecting atomic rest energies), entropy, enthalpy, Helmholtz free energy, and Gibbs free energy. Express all answers in SI units.

Write down the equilibrium condition for each of the following reactions:

(a)2H↔H2(b)2CO+O2↔2CO2(c)CH4+2O2↔2H2O+CO2(d)H2SO4↔2H++SO42-(e)2p+2n↔He4

Derive the van't Hoff equation,

dlnKdT=ΔH°RT2

which gives the dependence of the equilibrium constant on temperature." Here ∆H°is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if ∆H°is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of∆H°; solve the equation in this case to obtain

lnKT2-lnKT1=ΔH°R1T1-1T2

As you can see from Figure5.20,5.20,the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives ofPPwith respect toVV(at fixedTT) are zero. Use this fact to show that

Vc=3Nb, Pc =127ab2 and kTc=827ab

When solid quartz "dissolves" in water, it combines with water molecules in the reaction

SiO2(s)+2H2O(l)⟷H4SiO4(aq)

(a) Use this data in the back of this book to compute the amount of silica dissolved in water in equilibrium with solid quartz, at 25° C

(b) Use the van't Hoff equation (Problem 5.85) to compute the amount of silica dissolved in water in equilibrium with solid quartz at 100°C.

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.