Lecture 34 – The Third Law of Thermodynamics
Chapter 10 – Wednesday November 21st
•Review of the 3rd law of thermodynamics
•Consequences of the third law
•Example Problems from chapter 10
Reading:
Note:
Chapter 10 (pages 169 - 180)
Homework 10 is be due on Mon. 26th
I will assign homework 10 on Mon. 26th
Exam 3 will be on Monday Dec. 3rd in class
Evening review next week (TBD)
Statements of the third law
lim G = lim H
T →0
T →0
lim ΔG = lim ΔH
T →0
T →0
•The Nernst formulation of the Third Law:
‘All reactions in a liquid or solid in thermal equilibrium
take place with no change of entropy in the neighborhood
of absolute zero.’
Statements of the third law
lim G = lim H
T →0
T →0
lim ΔG = lim ΔH
T →0
T →0
•Planck later postulated that:
⎛ ∂G ⎞
lim ⎜
= 0,
⎟
T →0 ⎝ ∂T ⎠
P
⎛ ∂H ⎞
lim ⎜
=0
⎟
T →0 ⎝ ∂T ⎠
P
⇒
lim S = 0
T →0
Statements of the third law
lim G = lim H
T →0
T →0
lim ΔG = lim ΔH
T →0
T →0
•Planck’s statement of the Third Law:
‘The entropy of a true equilibrium state of a system at
absolute zero is zero.’
The Tds equations
Specific heat of a solid
Copper
Statements of the third law
lim G = lim H
T →0
T →0
lim ΔG = lim ΔH
T →0
T →0
•Another statement of the Third Law is:
‘It is impossible to reduce the temperature of a system to
absolute zero using a finite number of processes.’
Adiabatic cooling
S
B
B1
M
B2 > B1
T3 T2
T1
T
dU = TdS – PdV = TdS – MdB = đQ – MdB
Adiabatic cooling
S
B1
M
B2 > B1
T3 T2
T1
T
dU = TdS – PdV = TdS – MdB = đQ – MdB
Equivalence of the 3rd law statements
S
B1
B2 > B1
0 T3
T2
T1
T
dU = TdS – PdV = TdS – MdB = đQ – MdB