Because osmotic pressures can be quite large, you may wonder whether the approximation made in equation5.74is valid in practice: Is μ0really a linear function of Pto the required accuracy? Answer this question by discussing whether the derivative of this function changes significantly, over the relevant pressure range, in realistic examples.

μ0T,P2μ0T,P1+P2-P1μ0P......equation(5.74)

Short Answer

Expert verified

If a solution has high density then it will produces more pressure than the solution of low density.

Step by step solution

01

Given information

We have been given the difference between pressures

02

Explanation

It does not matter how large the osmotic pressure is μ0T,P2μ0T,P1+P2-P1μ0Pequation(5.74),

If the pressure in two parts of the system does not change significantly,

So we can say that the derivative of the function does not change significantly.

Therefore, we can say that an approximation is good enough.

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

Repeat the preceding problem with T/TC=0.8

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 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

In this problem you will derive approximate formulas for the shapes of the phase boundary curves in diagrams such as Figures 5.31 and 5.32, assuming that both phases behave as ideal mixtures. For definiteness, suppose that the phases are liquid and gas.

(a) Show that in an ideal mixture of A and B, the chemical potential of species A can be written μA=μA°+kTln(1-x)where A is the chemical potential of pure A (at the same temperature and pressure) and x=NB/NA+NB. Derive a similar formula for the chemical potential of species B. Note that both formulas can be written for either the liquid phase or the gas phase.

(b) At any given temperature T, let x1 and xgbe the compositions of the liquid and gas phases that are in equilibrium with each other. By setting the appropriate chemical potentials equal to each other, show that x1and xg obey the equations =1-xl1-xg=eΔGA°/RTandxlxg=eΔGB°/RT and where ΔG°represents the change in G for the pure substance undergoing the phase change at temperature T.

(c) Over a limited range of temperatures, we can often assume that the main temperature dependence of ΔG°=ΔH°-TΔS°comes from the explicit T; both ΔH°andΔS°are approximately constant. With this simplification, rewrite the results of part (b) entirely in terms of ΔHA°,ΔHB° TA, and TB (eliminating ΔGandΔS). Solve for x1and xgas functions of T.

(d) Plot your results for the nitrogen-oxygen system. The latent heats of the pure substances areΔHN2°=5570J/molandΔHO2°=6820J/mol. Compare to the experimental diagram, Figure 5.31.

(e) Show that you can account for the shape of Figure 5.32 with suitably chosenΔH° values. What are those values?

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 ofH°; solve the equation in this case to obtain

lnKT2-lnKT1=ΔH°R1T1-1T2

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