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Unformatted text preview: ∆S = T2 C ò T1 pdT T T1 = 0K Because there is an absolute zero
for temperature, the entropy for
any state can be calculated ...
if we know S(0K). S(T) = T ò0 Cp(T) dT
+ S(0K)
T We also defined entropy statistically: S(T) = k lnW(T) p The third law: Absolute entropies W(T) = number of
accessible
microstates at T W(T) depends on T through the
Boltzmann distribution (fewer
states accessible at low T):
∆E/kT Pn ∝ e Nils Walter: Chem 260 This definition gives us a useful S(T = 0K)
“isoenergetic” Flip three coins HHH
HHT
HTH
THH
TTH
THT
HTT
TTT A W = # of
P=
probability microstates S = k ln W 1/8 1 0k 3/8 3 1.10 k C 8 possible
microstates 3/8
(distributions) 3 1.10 k D 1/8 1 0k B 4 possible
macrostates If there is only one accessible microstate, the entropy of the system (macrostate)
is given by:
This is consistent with our view of entropy
S = k ln W = k ln 1 = 0
as a measure of disorder. No disorder
S = k ln W = k ln 1 = 0
(only one possible state) Þ S = 0 @ T = 0 K
Nils Walter: Chem 260 The third law of thermodynamics
S = 0 fforall perfectly crystalline materials at T = 0 K
S = 0 or all perfectly crystalline materials at T = 0 K S = k ln W = k ln 1 = 0
S = k ln W = k ln 1 = 0 p If there is only one possible arrangement of atoms
(i.e., in a perfect crystal): Since every substance has a finite (nonzero) heat capacity
=0
T C (T) dT
p
and S(T) =
+ S(0K)
T
0 ò Þ 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 1 bar and 298 K Chem 260
Nils Walter: ( ) One can calculate the molar entropy of HCl
98.36 K
Solid I
1.3 + 29.5 12.1 158.91 K
Solid II
21.1 12.6 Liquid
9.9 188.07 K
85.9 Gas
13.5 Sum: S298.15 = 185.9 J K1 mol1
Sum: S298.15 = 185.9 J K1 mol1
Solid I S = ò Cp dlnT from 0 K to 16 K using Debye T3law: Cp = aT3
S = ò Cp dlnT from 16 K to 98.36 K using experimental Cp
S = ∆H/T for each phase transition Solid II S = ò Cp dlnT from 98.36 K to 158.91 K using experimental Cp Liquid S = ò Cp dlnT from 158.91 K to 188.07 K using experimental Cp Gas S = ò Cp dlnT from 188.07 K to 298.15Nilsusing experimental Cp
K Walter: Chem 260 The exception:
An imperfect crystal still has residual entropy
a) Consider CO
O
C µ = 0.12 D If there were no preferred
orientation W = 2N
For one mole: W = 2NA
S = k ln W
Actual value:
= k ln 2NA
4.2 J K1 mol1
= NAk ln 2
= R ln 2 = 5.8 J K1 mol1 b) 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 K1 mol1 Actual value:
11.7 J K1 mol1 Nils Walter: Chem 260 The three laws of thermodynamics
Atkins Moore
You can’t construct
a perpetuum mobile
that generates
energy from nothing
You can’t construct a
machine that does
nothing but convert
heat into work 1. ∆U = q + w
Energy Conservation The internal energy of
an isolated system is
constant 2. ∆S ≥ 0
everything tends
toward increasing
disorder The entropy of the
universe tends to
increase 3. S > 0 except for
perfect crystals at
T=0K The entropy of a
perfectly crystalline
substance is zero at
T=0K You can never
reach absolute
zero Summary:
The energy of the universe is constant; the entropy
of the universe tends always toward a maximum.
Nils Walter: Chem
Rudolf Julius Clausius (18221888) 260 ...
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This note was uploaded on 12/26/2011 for the course CHEM 260 taught by Professor Staff during the Spring '08 term at University of Michigan.
 Spring '08
 STAFF

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