Section 4: Themodynamic Potentials and Maxwell's Relations

Question 1

we can introduce three new functions of state, all with units of energy, which together with internal energy form the

Answer 1

four thermodynamic potentials

Question 2

the 4 thermodynamic potentials

Answer 2

internal energy
enthalpy
helmholtz free energy
gibbs free energy

Question 3

the internal energy of a system refers to

Answer 3

all the energy it has independent of external fields and total kinetic energy

Question 4

mathematically, enthalpy is defined as

Answer 4

H=U+PV

a partial Legendre transform of the internal energy

Question 5

enthalpy, physically it represents the

Answer 5

potential of the system to do work at constant pressure

Question 6

alternatively, if the internal energy is viewed as the required energy to create the system, enthalpy

Answer 6

represents the required energy to create the system and the work required to make room for it

Question 7

the helmholtz free energy refers to the potential

Answer 7

F=U-TS

Question 8

helmholtz represents the

Answer 8

energy of the system which can be converted into work at constant temperature

the energy which is free to do work

Question 9

helmholtz free energy is the total energy minus

Answer 9

energy which is ‘thermally bound’ (-TS)

Question 10

gibbs free energy is defined as

Answer 10

G=U+PV-TS

Question 11

analogous to free energy, gibbs subtracts

Answer 11

the part of energy which is unavailable without heat exchange

Question 12

internal energy - differential form

Answer 12

dU=TdS-PdV

variables S,V

Question 13

enthalpy - differential form

Answer 13

dH=TdS+VdP

variables S,P

Question 14

helmholtz free energy - differential form

Answer 14

dF=-PdV-SdT

variables V,T

Question 15

gibbs - differential form

Answer 15

dG=VdP-SdT

variables P,T

Question 16

deriving differential form of F=U-TS

Answer 16

dF = dU - TdS - SdT
=TdS - PdV - TdS - SdT
= PdV - SdT

Question 17

deriving differential form of H=U+PV

Answer 17

dH=dU+PdV+VdP
=TdS-PdV+PdV+VdP
=TdS+VdP

Question 18

deriving differential form of G=U+PV-TS

Answer 18

dG=dU+PdV+VdP-TdS-SdT
=TdS-PdV+PdV+VdP-TdS - SdT
=VdP-SdT

Question 19

Legendre transformation

Answer 19

the process of deriving one thermodynamic potential from another

allows us to recast a funtion with its derivative as an independent variable

Question 20

maxwell relations relate

Answer 20

the variables P,V,S,T to each other ad there is one such relation for each of the four thermodynamic potentials

Question 21

deriving the first maxwell relation

Answer 21

compare differentia form of internal energy with the total derivative of U

identify T= and P=

derivative of T wrt V at const S and deriv of P wrt S at const V

equate derivatives

Question 22

maxwell relations are useful way to

Answer 22

convert from a term that is difficult to calculate or measure to one that is easier to work with

in particular, entropy is hard to measure but heat capacities are experimentally accessible

Question 23

four thermodynamic potentials and four variables gives

Answer 23

336 permutations of the form (dX/dY)z

even without repeats, still 56 combinations many not measurable or useful

Question 24

measurable quantities

Answer 24

heat capacities at const v and p
coefficients of expansion and compression
bulk moduli and bulk compressibility
latent heats

Question 25

heat capacity is a measure of

Answer 25

the temperature rise per unit input heat δQ=TdS either at const v or p

Question 26

Cv=T(dS/dT)v=

Answer 26

(dU/dT)v using dU=TdS-PdV

Question 27

Cp=T(dS/dT)p=

Answer 27

(dH/dT)p using dH=TdS+VdP

Question 28

measuring heat capacities in the way of their definitions requires

Answer 28

measuring the internal energy and the enthalpy respectively would be easier to measure quantities such as pressure and volume

Question 29

derivation of Cp-Cv = T(dP/dT)v(dV/dT)p

Answer 29

start with entropy form (dS/dT)p by taking the deriv of S wrt T at const p insert definition of entropy use maxwell relation

Question 30

Cp-Cr should reduce to

Answer 30

nR since for an ideal gas Cp-Cr=nR