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,

VUS=SUV

where each /Vis taken with Sfixed, each /Sis taken with Vfixed, and Nis always held fixed. From the thermodynamic identity (forU) you can evaluate the partial derivatives in parentheses to obtain

TVS=-PSV

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,andG ). 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.

Short Answer

Expert verified

Maxwell relations are:

TVS=-PSVTPS=VSPTPS=PTVSPT=-VTP

Step by step solution

01

To find

Four Maxwell relations.

02

Keeping N as constant derive the given equation.

We have the thermodynamics identity:

dU=TdS-PdV+μdN

at constant volume and number of molecules (at which dN=0anddV=0)

we have:

T=USV............(1)

and at constant entropy and number of molecules (at which dN=0anddS=0)

we have:

P=-UVS............(2)

In the given we have: role="math" localid="1648414830374" VUS=SUV.......(3)

Now substitute equation (1) and (2) in (3)

TVS=-PSV

03

continuing derivation

We have following the enthalpy identity as:

dH=TdS+VdP+μdN

at constant pressure and number of molecules (at which dN=0anddP=0) we have:

role="math" localid="1648416938771" T=HSP.........(3)

again differentiate equation (3) w.r.t. P

TPS=HPS

Then at constant entropy and number of molecules (at which dN=0anddS=0) we have:

role="math" localid="1648416952292" V=HPS.........(4)

again differentiate equation (4) w.r.t. V

VSP=HPS

combine these two equations together to get the following result:

TPS=VSP

04

continuing derivation 

We have following the Helmholtz free energy is given by:

dF=-SdT-PdV+μdN

at constant pressure and number of molecules (at which dN=0anddP=0) we have:

role="math" localid="1648418905850" S=-FTP......(5)

again differentiate equation (5) w.r.t. V

SVT=-FVT

and at constant entropy and number of molecules (at which dN=0anddS=0) we have:

role="math" localid="1648418958034" P=-FVS......(6)

again differentiate equation (6) w.r.t. T

PTV=-FVT

combine these two equations together to get the following result:

TPS=PTV

05

continuing derivation 

We have following the Gibbs free energy is given by:

dG=-SdT+VdP+μdN

at constant pressure and number of molecules (at which dN=0anddP=0) we have:

role="math" localid="1648419302437" S=-GTP.......(7)

again differentiate the equation (7) w.r.t. P

SPT=-GPT

and at constant temperature and number of molecules (at which dN=0anddT=0) we have:

role="math" localid="1648419374743" V=GPS.......(8)

again differentiate equation (8) w.r.t. T

VTP=GPT

combine these two equations together to get the following result:

SPT=-VTP

06

Final answer

Maxwell relations are:

TVS=-PSVTPS=VSPTPS=PTVSPT=-VTP

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Most popular questions from this chapter

Use the data at the back of this book to verify the values of ΔH and ΔGquoted above for the lead-acid reaction 5.13.

At temp 298K and pressure 1 bar.

Consider the production of ammonia from nitrogen and hydrogen,

N2 + 3H2 2NH3
at 298 K and 1 bar. From the values of Hand S tabulated at the back of this book, compute Gfor this reaction and check that it is consistent with the value given in the table.

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?

Problem 5.35. The Clausius-Clapeyron relation 5.47 is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and V depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take V to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:

P= (constant) x e-L/RT

This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.

Repeat the previous problem for the opposite case where the liquid has a substantial negative mixing energy, so that its free energy curve dips |below the gas's free energy curve at a temperature higher than TB. Construct the phase diagram and show that this system also has an azeotrope.

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