Q. 5.12 Functions encountered in physics... [FREE SOLUTION]

91影视

An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 5: Q. 5.12 (page 158)

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,
$\frac{鈭�}{鈭� V} (\frac{鈭� U}{鈭� S}) = \frac{鈭�}{鈭� S} (\frac{鈭� U}{鈭� V})$
where each $鈭� / 鈭� V$ is taken with S fixed, each $鈭� / 鈭� S$ 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
${(\frac{鈭� T}{鈭� V})}_{S} = - {(\frac{鈭� P}{鈭� S})}_{V}$
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.

Short Answer

Expert verified

Maxwell's relations are

${(\frac{鈭� T}{鈭� V})}_{S} = - {(\frac{鈭� P}{鈭� S})}_{V} {(\frac{鈭� T}{鈭� P})}_{S} = {(\frac{鈭� V}{鈭� S})}_{P} {(\frac{鈭� T}{鈭� P})}_{S} = {(\frac{鈭� P}{鈭� T})}_{V} {(\frac{鈭� S}{鈭� P})}_{T} = - {(\frac{鈭� V}{鈭� T})}_{P}$

Step by step solution

Given information

Maxwell relation is given.

Derive equation for constant volume

We have the thermodynamics identity:

$d U = T d S - P d V + 渭 d N$

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

we have

$T = {(\frac{鈭� U}{鈭� S})}_{V} . . . . . . . . . . . . (1)$

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

$P = - {(\frac{鈭� U}{鈭� V})}_{S} . . . . . . . . . . . . (2)$

In the given we have:

$\frac{鈭�}{鈭� V} (\frac{鈭� U}{鈭� S}) = \frac{鈭�}{鈭� S} (\frac{鈭� U}{鈭� V}) . . . . . . . (3)$

Now substitute (1) and (2) in (3) We get

${(\frac{鈭� T}{鈭� V})}_{S} = - {(\frac{鈭� P}{鈭� S})}_{V}$

Derive equation for constant pressure

We have following the enthalpy identity as:

$d H = T d S + V d P + 渭 d N$

t constant pressure and number of molecules (at which dN=0 and dP=0),

we have

$T = {(\frac{鈭� H}{鈭� S})}_{P} . . . . . . . . . (4)$

again differentiate equation (4) w.r.t. P, we get

${(\frac{鈭� T}{鈭� P})}_{S} = \frac{鈭� H}{鈭� P 鈭� S}$

Then at constant entropy and number of molecules (at which dN=0, dS=0),

we have

$V = {(\frac{鈭� H}{鈭� P})}_{S} . . . . . . . . . (5)$

again differentiate equation (5) w.r.t. V, we get

${(\frac{鈭� V}{鈭� S})}_{P} = \frac{鈭� H}{鈭� P 鈭� S}$

Combine these two we get

${(\frac{鈭� T}{鈭� P})}_{S} = {(\frac{鈭� V}{鈭� S})}_{P}$

Derivation continued

We have following the Helmholtz free energy is given by:

$d F = - S d T - P d V + 渭 d N$

at constant pressure and number of molecules (at which dN=0 and dP=0)

we have

$S = - {(\frac{鈭� F}{鈭� T})}_{P} . . . . . . (6)$

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

${(\frac{鈭� S}{鈭� V})}_{T} = - \frac{鈭� F}{鈭� V 鈭� T}$

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

we have

$P = - {(\frac{鈭� F}{鈭� V})}_{S} . . . . . . (7)$

again differentiate equation (7) w.r.t. T

${(\frac{鈭� P}{鈭� T})}_{V} = - \frac{鈭� F}{鈭� V 鈭� T}$

combine these two equations together to get the following result we get

${(\frac{鈭� T}{鈭� P})}_{S} = {(\frac{鈭� P}{鈭� T})}_{V}$

continuing derivation

We have following the Gibbs free energy is given by:

$d G = - S d T + V d P + 渭 d N$

at constant pressure and number of molecules (at which dN=0 and dP=0)

we have

$S = - {(\frac{鈭� G}{鈭� T})}_{P} . . . . . . . (8)$

again differentiate the equation (8) w.r.t. P

${(\frac{鈭� S}{鈭� P})}_{T} = - \frac{鈭� G}{鈭� P 鈭� T}$

and at constant temperature and number of molecules (at which dN=0 and dT=0)

we have:

$V = {(\frac{鈭� G}{鈭� P})}_{S} . . . . . . . (9)$

again differentiate equation (9) w.r.t. T

${(\frac{鈭� V}{鈭� T})}_{P} = \frac{鈭� G}{鈭� P 鈭� T}$

combine these two equations together to get the following result:

${(\frac{鈭� S}{鈭� P})}_{T} = - {(\frac{鈭� V}{鈭� T})}_{P}$

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

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Given information

Derive equation for constant volume

Derive equation for constant pressure

Derivation continued

continuing derivation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Wave Optics

Energy Physics

Kinematics Physics

Electricity and Magnetism

Medical Physics

Magnetism and Electromagnetic Induction

Study anywhere. Anytime. Across all devices.