Absolute Entropy:
T2 C Because there is an absolute zero for
∆S = ∫T1 pdT
T temperature, the entropy for any state
can be calculated ... if we know S(0).
T
∫0
Cp(T´) dT´
S(T) = + S(0)
T´
We also defined entropy statistically:
W(T) ∝ number of
accessible states at T
SS==kkln
lnW
W or S(T) = k lnW(T)
W(T) depends on T through
Boltzmann population factors:
W ∝ probability of a given −∆E/kT
state being populated Pn ∝ e
Lets change the definition of W slightly:
Flip three coins W = W = #
S of states S
probability
HHH A 1/8 -2.08 k 1 0k
“isoenergetic”
HHT
HTH B 4 possible 3/8 -0.98 k 3 1.10 k
THH “states”
TTH 8 possible
THT C outcomes 3/8 -0.98 k 3 1.10 k
HTT
TTT D 1/8 -2.08 k 1 0k
If there is only one accessible microstate, the entropy of the
system is given by:
This is consistent with our view of
entropy as a measure of disorder.
SS==kkln W = k ln 1 = 0
ln W = k ln 1 = 0 No disorder (only one possible state)
⇒ S = 0.
Third Law of Thermodynamics:
SS==00for
forall
allperfectly
perfectlycrystalline
crystallinematerials
materialsat
atTT==00K
K
If there is only one possible arrangement of atoms: (i.e. a perfect
crystal):
SS==kkln
lnW
W==kkln
ln11==00
Every substance has a finite (nonzero) heat capacity.
T C (T´) dT´
S(T) =
∫
0
p
T´
+ S(0)
∴ except for a perfect crystal at absolute zero, every substance
has a finite positive entropy.
Standard absolute or “third law” molar entropies are tabulated,
generally at 298 K.
Third Law Entropy: Consider the absolute molar entropy of HCl
at 1 atm, 298.15K
98.36 K 158.91 K 188.07 K
Solid I Solid II Liquid
12.1 12.6 85.9
Gas
1.3 + 29.5 21.1 9.9
13.5
SS298.15
298.15
=
= 185.9
185.9 JK
JK -1mol
-1
mol
-1
-1
S = ∫ Cp dlnT from 0 K to 16 K using Debye formula: Cp = aT3
Solid I
S = ∫ Cp dlnT from 16 K to 98.36 K using experimental Cp
S = ∆H/T for each phase transition
Solid II S = ∫ Cp dlnT from 98.36 K to 158.91 K using experimental Cp
Liquid S = ∫ Cp dlnT from 158.91 K to 188.07 K using experimental Cp
Gas S = ∫ Cp dlnT from 188.07 K to 298.15 K using experimental Cp
Residual Entropy:
The entropy of a substance may not go to zero - even at 0 K.
Consider CO.
If there is no preferred orientation
O C µ = 0.12 D W = 2N
For one mole: W = 2NA
S = k ln W
Actual ca.
= k ln 2NA
4.2 J K-1 mol-1
= NAk ln 2
= R ln 2 = 5.8 J K-1 mol-1
Consider CH3D. No dipole moment, same number of electrons
No preferred orientation!
W = 4N or for one mole W = 4NA
S = R ln 4 = 11.5 J K-1 mol-1
Actual 11.7 J K-1 mol-1
The Three Laws of Thermodynamics:
Atkins Moore
The internal energy You can’t win,
∆U = q + W
1. of an isolated system the best you can
Energy Conservation
is constant do is break even
∆S ≥ 0
The entropy of the You can only
everything tends
2. universe tends to break even at
toward increasing
increase absolute zero
disorder
The entropy of a
S > 0 except for You can never
perfectly crystalline
3. perfect crystals reach absolute
substance is zero at
at T = 0 K zero
T = 0 K
Summary:
The energy of the universe is constant; the entropy
of the universe tends always toward a maximum.
Rudolf Julius Clausius (1822-1888)