1. Entropy = S
Entropy is
disorder
randomness
dispersal of energy

2. 2nd Law of Thermodynamics
Suniverse
> 0
for spontaneous processes
spontaneous =
no external intervention
Ssystem
Ssurroundings
positional disorder
energetic disorder

3. Energetic Disorder
a) 
b) 
P.E K.E.
ordered
random
a) endothermic reaction
b) exothermic reaction
reactants
P.E.
Ssurr=
- qsys
(J/K)
products T
qsystem 0
<
qsurroundings 0
>
Ssurr
depends on T
heat  surroundings
Ssurroundings > 0
high T
small effect
low T
relatively larger effect

4. = energy and position of atoms in state
Positional Disorder
2 dice
microstates
S = kB
ln W kB
= R/NA 2 3 4 5 6 7 8 9 10 11 12
distribution =
state
microstates
= W
= energy and position of atoms in state

5. S = kB ln W W 2 W kB = R/NA ∆S = S2 – S1 = kB ln W2/W1 = kB ln 2 x 2
for 1 mole gas
ln x 10 23
∆S = kB
= 6.02 x 1023 kB ln 2
= R ln 2
∆SV → V = 1 2
R ln (V2/V1)

6. Positional Disorder Boltzman S = kB ln W = R ln (V2/V1) ∆S
W = microstates
ordered states
low probability
low S
disordered states
high probability
high S
 Ssystem  Positional disorder
Increases with number of possible positions
(energy states)
Ssolids
Sliquids
Sgases
<
<<

7. Entropy
(J/K)
[heat entering system at given T]
convert q
to S
System 1
Pext = 1.5 atm
E = 0
w = -182 J
q = +182 J
T = 298 K
E = 0
System 2
Pext = 0 atm
w = 0
q = 0

8. System 3 P1 = 6.0 atm P2 = 1.5 atm V1 = 0.4 L V2 = 1.6 L
T1 = 298 K = T2
Pext =
Pint
+ dP
reversible process
- infinitely slow
n = .10 V2 =
- nRT
ln (V2/V1)
wr=
- Pext  dV
= - nRT dV V  V1
- nRT ln(1.6/4.0)
= J
wr=

9. Ssystem ∆S = n R ln (V2/V1) qr = n R T ln (V2/V1) Ssystem = qr
Pext = 1.5 atm
w = -182 J
q = +182 J
S =
System 2
Pext = 0 atm
w = 0
q = 0
S =
System 3
Pext = Pint + dP
wr =
qr =
S =
-nRT ln (V2/V1)
= J J
1.15 J/K
1.15 J/K
1.15 J/K
∆S =
n R ln (V2/V1)
qr =
n R T ln (V2/V1)
Ssystem = qr
= J
298 K T
∆S = n CP ln (T2/T1)
∆S = n CV ln (T2/T1)

10. 3rd Law of Thermodynamics
Entropy
of a perfect crystalline
substance
at 0 K
= 0

11. Entropy curve S Temperature (K) gas liquid solid vaporization qr T
fusion
Temperature (K)

12. Entropy At 0K, S = 0 Entropy is absolute S  0
for elements in standard states
S is a State Function
Sorxn =
nSoproducts
- nSoreactants
S is extensive

13. Increases in Entropy 1. Melting (fusion) Sliquid > Ssolid
2. Vaporization Sgas >> Sliquid
3. Increasing ngas in a reaction
Heating ST2 > ST1 if T2 > T1.
Dissolution Ssolution > (Ssolvent + Ssolute) ?
6. Molecular complexity number of bonds
7. Atomic complexity e-, protons and neutrons