Thermodynamics- Property Relations

Question 1

What is dz if z=z(x,y)?

Answer 1

All δ symbols are actually curly ds
dz=(δz/δx)ydx+(δz/δy)xdy
The y and x outside are subscript to say those variables remain constant for that partial derivative.

Question 2

Prove that the second partial derivative of z for z=z(x,y) is the same whether you do with respect to x or y first and the other second.

Answer 2

dz=Mdx+Ndy 
M=(δz/δx)y
N=(δz/δy)x
Partially differentiate M with respect to y and N with respect to x and you get δ^2z/δyδx for both
All δ symbols are actually curly ds

Question 3

What is the equation for the Gibbs function, g?

Answer 3

g=h-Ts

Question 4

What is the equation for the Helmholtz function, a?

Answer 4

a=u-Ts

Question 5

How to derive the last two Gibbs relations

Answer 5

Start with Gibbs function and Helmholtz function equations. Differentiate them (not with respect to anything in particular). Have to use the product rule. Use the previous two Gibbs relations to sub in for Tds and cancel out the du or dh as appropriate. End with
dg=-sdT+vdP
da=-sdT-Pdv

Question 6

How to derive the four Maxwell relations for the four Gibbs relations

Answer 6

The Gibbs relations are in the form dz=Mdx+Ndy
Use the fact that (δM/δy)x=(δN/δx)y. End up with
(δT/δv)s=-(δP/δs)v
(δT/δP)s=(δv/δs)P
(δs/δv)T=(δP/δT)v
(δs/δP)T=-(δv/δT)P
All δ symbols are curly ds

Question 7

When can Maxwell relations be used?

Answer 7

For simple compressible systems

Question 8

What is the Clapeyron equation and what is it used for?

Answer 8

Used to determine enthalpy change associated with a phase change from knowledge of P,v and T data alone. Makes use of the slope of the saturation curve on a P-T diagram.
(dP/dT)sat=h12/Tv12
Subscripts 1 and 2 indicate the two phases

Question 9

Deriving the Clapeyron equation part 1

Answer 9

Start with 3rd Maxwell relation (δs/δv)T=(δP/δT)v.
During phase change, the pressure is the saturation pressure which only depends on temp. So partial derivative (δP/δT)v can be expressed as total derivative (dP/dT)sat which is the slope of the saturation curve on a P-T diagram at a specified saturation state. The slope is independent of v so can be treated as constant for integration between the two saturation states at the same temperature. You get
s2-s1=(dP/dT)sat(v2-v1)

Question 10

Deriving the Clapeyron equation part 2

Answer 10

Then get (dP/dT)sat=s12/v12. During this process the pressure also remain constant so can use Gibbs relation dh=Tds+vdP and say dP tends to 0. Integrate what remains between the two saturation states to get h12=Ts12. Sub s12 into above equation to get
(dP/dT)sat=h12/Tv12

Question 11

How to simplify the Clapeyron equation for liquid to vapour changes

Answer 11

At low pressures, vg»vf so vfg is about vg. If you treat the vapour as an ideal gas vg=RT/P. Substituting these into Clapeyron equation gives (dP/dT)sat=Phfg/RT^2. Times by dT and divide by P gives (dP/P)sat=(hfg/R)(dT/T^2)sat. For small temperature intervals hfg can be considered constant at som average value. Integrating between two saturation states gives
ln(P2/P1)sat=(hfg/R)(1/T1-1/T2)sat
Can also be used for enthalpy of sublimation, hig

Question 12

What is the volume expansivity of a gas, β, and what does it mean?

Answer 12

β=(1/v)(δv/δT)P
All δ symbols are curly ds
Measures the change in volume with temperature at a constant pressure

Question 13

What is the isothermal compressibility of a gas, α?

Answer 13

α=-(1/v)(δv/δP)T
All δ symbols are curly ds
Always gives a positive answer

Question 14

What is the Mayer relation?

Answer 14

cp-cv=vTβ^2/α