Loading in 5 sec....

Chapter 5 Simple Applications of Macroscopic ThermodynamicsPowerPoint Presentation

Chapter 5 Simple Applications of Macroscopic Thermodynamics

Download Presentation

Chapter 5 Simple Applications of Macroscopic Thermodynamics

Loading in 2 Seconds...

- 96 Views
- Uploaded on
- Presentation posted in: General

Chapter 5 Simple Applications of Macroscopic Thermodynamics

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 5Simple Applications of Macroscopic Thermodynamics

Classical, Macroscopic, Thermodynamics

- Drop the statistical mechanics notation for average quantities. We know that All Variables are Averages Only!
- We’ll discuss relationships between macroscopic variables using
The Laws of Thermodynamics

- Some Thermodynamic Variables of Interest:
Internal Energy = E, Entropy = S,Temperature = T

For Gases:

External Parameter = V, Generalized Force = p

(V = volume, p = pressure)

For A General System:

External Parameter = x, Generalized Force = X

- Assume the relevant External Parameter = Volume V in order to have a specific case to discuss. For infinitesimal, quasi-static processes:
The 1st & 2nd Laws of Thermodynamics

1st Law: đQ = dE + pdV

2nd Law: đQ = TdS

The Combined 1st & 2nd Laws

TdS = dE + pdV

- Note that, in this relation, there are 5 Variables:
T, S, E, p, V

- It can be shown that:
Any 3 of these can always be expressed as functions of any 2 others.

- That is, there are always 2 independent variables & 3 dependent variables. Which 2 are chosen as independent is arbitrary.

Now, A Brief, Pure Math Discussion

- Consider 3 variables: x, y, z. Suppose we know that x & y are Independent Variables. Then, It Must Be Possible to express z as a function of x & y. That is, There Must be a Functionz = z(x,y).
- From calculus, the total differential of z(x,y)has the form:
dz (∂z/∂x)ydx + (∂z/∂y)xdy (a)

- Suppose instead that we want to take y & z as independent variables. Then, There Must be a Functionx = x(y,z).
- From calculus, the total differential of x(y,z) has the form:
dx (∂x/∂y)zdy + (∂x/∂z)ydz (b)

- Using (a) & (b) together, the partial derivatives in (a) & those in (b) can be related to each other.
- We always assume that all functions are analytic. So, the 2nd cross derivatives are equal: Such as
(∂2z/∂x∂y) (∂2z/∂y∂x),etc.

Mathematics Summary

- Consider a function of 2 independent variables:f = f(x1,x2).
- It’s exact differential is df y1dx1 + y2dx2, where, by definition:
- Because f(x1,x2) is an analytic function, it is always true that
Most Ch. 5 applications use this with the

Combined 1st & 2nd Laws of Thermodynamics:

TdS = dE + pdV

Some Methods & Useful Math Tools

for Transforming Derivatives

Pure Math: Jacobian Transformations

- A Jacobian Transformation is often used to
- transform from one set of variables to another.
- For functions of 2 variables f(x,y) & g(x,y) it is:

Determinant!

Jacobian Transformations

Have Several Useful Properties

- Suppose that we are only interested in the first partial derivative of a function f(z,g) with respect toz at constant g:
- This expression can be simplified using the chain rule expansion and the inversion property

Properties of the Internal Energy E

dE = TdS – pdV (1)

First, choose S &Vas independent variables:

E E(S,V)

∂E

∂E

(2)

dE

Comparison of (1) & (2) clearly shows that

∂E

∂E

and

Applying the general result with 2nd cross derivatives gives:

Maxwell Relation I!

If S &p are chosen as independent variables, it is convenient to define the following energy:

H H(S,p) E + pVEnthalpy

Use the combined 1st & 2nd Laws. Rewrite them in terms of dH:dE = TdS – pdV = TdS – [d(pV) – Vdp] or

dH = TdS + Vdp

(1)

But, also:

(2)

Comparison of (1) & (2) clearly shows that

and

Applying the general result for the 2nd cross derivatives gives:

Maxwell Relation II!

If T &V are chosen as independent variables, it is convenient to define the following energy:

F F(T,V) E - TSHelmholtz Free Energy

Use the combined 1st & 2nd Laws. Rewrite them in terms of dF:dE = TdS – pdV = [d(TS) – SdT] – pdV or

dF = -SdT – pdV (1)

But, also: dF ≡ (F/T)VdT + (F/V)TdV (2)

Comparison of (1) & (2) clearly shows that

(F/T)V ≡ -S and (F/V)T ≡ -p

Applying the general result for the 2nd cross derivatives gives:

Maxwell Relation III!

If T &p are chosen as independent variables, it is convenient to define the following energy:

G G(T,p) E –TS + pVGibbs Free Energy

Use the combined 1st & 2nd Laws. Rewrite them in terms of dH:dE = TdS – pdV = d(TS) - SdT – [d(pV) – Vdp] or

dG = -SdT + Vdp (1)

But, also: dG ≡ (G/T)pdT + (G/p)Tdp (2)

Comparison of (1) & (2) clearly shows that

(G/T)p ≡ -S and(G/p)T ≡ V

Applying the general result for the 2nd cross derivatives gives:

Maxwell Relation IV!

Summary: Energy Functions

1. Internal Energy:E E(S,V)

2.Enthalpy: H = H(S,p) E + pV

3.Helmholtz Free Energy: F = F (T,V) E – TS

4.Gibbs Free Energy: G = G(T,p) E – TS + pV

1. dE = TdS – pdV

2. dH = TdS + Vdp

3. dF = - SdT – pdV

4. dG = - SdT + Vdp

Combined 1st &

2nd Laws

2.

1.

3.

4.

Another Summary: Maxwell Relations

(a) ΔE = Q + W

(b) ΔS = (Qres/T)

(c) H = E + pV

(d) F = E – TS

(e) G = H - TS

1. dE = TdS – pdV

2. dH = TdS + Vdp

3. dF = -SdT - pdV

4. dG = -SdT + Vdp

Maxwell Relations:“The Magic Square”?

Each side is labeled with an

Energy (E, H, F, G).

The corners are labeled with

Thermodynamic Variables

(p, V, T, S).

Get the

Maxwell Relations

by “walking” around the

square. The partial

derivatives are obtained

from the sides. The

Maxwell Relations

are obtained from the corners.

F

V

T

G

E

S

P

H

16

Summary

The 4 Most Common

Maxwell Relations:

17

18

Maxwell Relations

Maxwell Relations from dE, dF, dH, & dG

∂E

Heat Capacity at Constant Pressure:

Heat Capacity at Constant Volume:

Volume Expansion Coefficient:

Note!!

Reif’s notation

for this is α

Isothermal Compressibility:

The Bulk Modulus is inverse of the

Isothermal Compressibility!!

B (κ)-1

Summary of Results

Derivations are in the text and/or left to the student!

Entropy:

Enthalpy:

Gibbs Free Energy:

- Given the entropy S as a function of temperature T & volume V, S = S(T,V), find a convenient expression for (S/T)P, in terms of some measureable properties.
- Start with the exact differential:
- Use the triple product
rule & definitions:

- Use a
Maxwell Relation:

- Combining these
expressions gives:

- Converting this to a partial derivative gives an identity:
- This can be rewritten as:
- The triple product rule is:
- Substituting gives:

Note again the definitions:

Volume Coefficient of Expansion

βV-1(V/T)p

Isothermal Compressibility

κ -V-1(V/p)T

Note!! Reif’s

notation for

this is α

- Using these in the previous expressionfinally gives the desired result:
- Using this result as a starting point,A GENERAL RELATIONSHIPbetween The Heat Capacity at Constant Volume CV& The Heat Capacity at Constant Pressure Cpcan be found:

- For an Ideal Gas, it’s easily shown (Reif) that the Equation of State(relation between pressure P, volume V, temperature T) is (in per mole units!):Pν = RT.
- With this, it is simple to show that the volume expansion coefficient β & the isothermal compressibility κare:

and

- So, for an Ideal Gas, the volume expansion coefficient & the isothermal compressibility have the simple forms:

and

- We just found in general that the heat capacities at constant volume & at constant pressure are related as

- So, for an Ideal Gas, the specific heats per mole have the very simple relationship:

Consider Two Identical Objects, each of mass m, & specific heat per kilogram cP. See figure next page.

Object 1 is at initial temperatureT1.

Object 2 is at initial temperatureT2.

Assume T2 > T1.

When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf.

Two Identical Objects, of mass m, & specific heat per kilogram cP. Object 1 is at initial temperatureT1. Object 2 is at initial temperatureT2.

T2 > T1. When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf.

After a long enough time, the two objects are at the same temperature Tf. Since the 2 objects are identical, for this case,

Q

Heat Flows

Object 2

Initially at T2

Object 1

Initially at T1

For some time

after the initial

contact:

- Of course, by the 2nd Law,the entropy changeΔSmust be positive!! This requires that the temperatures satisfy:

- NOTE:In the following, various quantities are written in per mole units! Work with the Combined 1st & 2nd Laws:
- Definitions:
- υ Number of moles of a substance. ν (V/υ) Volume per mole.
- u (U/υ) Internal energy per mole. h (H/υ) Enthalpy per mole.
- s (S/υ) Entropy per mole. cv (Cv/υ) const. volume specific heat per mole.
- cP (CP/υ) const. pressure specific heat per mole.

Internal Energyu(T,ν)

Enthalpyh(T,P)

Entropy

1

3

2