- #1

zenterix

- 619

- 81

- Homework Statement
- In following the book "Physical Chemistry" by Silbey, Alberty, and Bawendi, I did not understand how they derived conditions on differentials of ##U, H, A##, and ##G## that lead to extremum principles on certain very specific processes.

- Relevant Equations
- In what follows, I will show the desired results and also what I came up with to justify the results. My question is if my calculations are correct.

For the internal energy function ##U(S,V,\{n_i\})## we have

$$dU=TdS-pdV+\sum\limits_{i=1}^{N_s}\mu_id n_i\tag{1}$$

where ##N_s## is the number of species in the system.

We also have

$$dU=\delta q+\delta w\tag{2}$$

by the 1st law of thermodynamics. I am using ##\delta## to denote an inexact differential.

The 2nd law of thermodynamics tells us that

$$dS\geq \frac{\delta q}{T}\tag{3}$$

Suppose we have a system at constant ##S, V##, and ##\{n_i\}##.

Then, somehow, using (1), (2), and (3) we can conclude that

$$(dU)_{S,V,\{n_i\}}\leq 0\tag{4}$$

Why can we conclude (4)?

What I came up with is

$$dU=\delta q+\delta w \leq TdS-pdV+\sum\limits_{i=1}^{N_s}\mu_idn_i$$

and so for ##dS=dV=dn_i=0## we reach ##dU\leq 0##.

However, I am not sure this is correct.

For example, consider Helmholtz energy

$$dA=dU-TdS-SdT$$

$$=-SdT-PdV+\sum\limits_{i=1}^{N_s}\mu_i dn_i$$

Somehow, we should be able to conclude that

$$(dA)_{T,V,\{n_i\}}\leq 0$$

for a spontaneous change at fixed ##T,V##, and ##\{n_i\}##.

What I came up with is

$$dA=\delta q+\delta w-TdS-SdT$$

$$=(\delta q-TdS)-PdV-SdT+\sum\mu_i dn_i$$

and for fixed ##T,V##, ##n_i## we have

$$(dA)_{T,V,\{n_i\}}=\delta q-TdS\leq 0$$

The analogous calculations for Gibbs energy are

$$dG=\delta q+\delta w-TdS-SdT+PdV+VdP$$

$$=\delta q +(-PdV+\sum \mu_i dn_i)-TdS-SdT+PdV+VdP$$

$$=(\delta q-TdS)-SdT+VdP+\sum\mu_i dn_i$$

and since ##\delta q\leq TdS## we have that for constant ##T,P##, and ##\{n_i\}## we have

$$(dG)_{T,P,\{n_i\}}=\delta q-TdS\leq 0$$

The analogous calculations for enthalpy are

$$dH=dU+PdV+VdP$$

$$=\delta q+\delta w+PdV+VdP$$

$$=\delta q+VdP+\sum\mu_i dn_i$$

At fixed ##S,P##, and ##\{n_i\}## we have

$$(dH)_{S,P,\{n_i\}}=\delta q\leq TdS=0$$

My question boils down to if these calculations are correct?

I kinda think it is although for me at this point this is just an exercise in algebra, I don't see the big picture of these calculations very well.

I mean, each result means that for the spontaneous process in question we have a condition on the relevant differential of a thermodynamic potential. In all cases, the result is that at equilibrium we have the minimum of a thermodynamic potential.

These are extremum principles associated with equilibrium after each specific type of process.

$$dU=TdS-pdV+\sum\limits_{i=1}^{N_s}\mu_id n_i\tag{1}$$

where ##N_s## is the number of species in the system.

We also have

$$dU=\delta q+\delta w\tag{2}$$

by the 1st law of thermodynamics. I am using ##\delta## to denote an inexact differential.

The 2nd law of thermodynamics tells us that

$$dS\geq \frac{\delta q}{T}\tag{3}$$

Suppose we have a system at constant ##S, V##, and ##\{n_i\}##.

Then, somehow, using (1), (2), and (3) we can conclude that

$$(dU)_{S,V,\{n_i\}}\leq 0\tag{4}$$

Why can we conclude (4)?

What I came up with is

$$dU=\delta q+\delta w \leq TdS-pdV+\sum\limits_{i=1}^{N_s}\mu_idn_i$$

and so for ##dS=dV=dn_i=0## we reach ##dU\leq 0##.

However, I am not sure this is correct.

For example, consider Helmholtz energy

$$dA=dU-TdS-SdT$$

$$=-SdT-PdV+\sum\limits_{i=1}^{N_s}\mu_i dn_i$$

Somehow, we should be able to conclude that

$$(dA)_{T,V,\{n_i\}}\leq 0$$

for a spontaneous change at fixed ##T,V##, and ##\{n_i\}##.

What I came up with is

$$dA=\delta q+\delta w-TdS-SdT$$

$$=(\delta q-TdS)-PdV-SdT+\sum\mu_i dn_i$$

and for fixed ##T,V##, ##n_i## we have

$$(dA)_{T,V,\{n_i\}}=\delta q-TdS\leq 0$$

The analogous calculations for Gibbs energy are

$$dG=\delta q+\delta w-TdS-SdT+PdV+VdP$$

$$=\delta q +(-PdV+\sum \mu_i dn_i)-TdS-SdT+PdV+VdP$$

$$=(\delta q-TdS)-SdT+VdP+\sum\mu_i dn_i$$

and since ##\delta q\leq TdS## we have that for constant ##T,P##, and ##\{n_i\}## we have

$$(dG)_{T,P,\{n_i\}}=\delta q-TdS\leq 0$$

The analogous calculations for enthalpy are

$$dH=dU+PdV+VdP$$

$$=\delta q+\delta w+PdV+VdP$$

$$=\delta q+VdP+\sum\mu_i dn_i$$

At fixed ##S,P##, and ##\{n_i\}## we have

$$(dH)_{S,P,\{n_i\}}=\delta q\leq TdS=0$$

My question boils down to if these calculations are correct?

I kinda think it is although for me at this point this is just an exercise in algebra, I don't see the big picture of these calculations very well.

I mean, each result means that for the spontaneous process in question we have a condition on the relevant differential of a thermodynamic potential. In all cases, the result is that at equilibrium we have the minimum of a thermodynamic potential.

These are extremum principles associated with equilibrium after each specific type of process.

Last edited: