Transcript Slide 1

Entropy = S
Entropy is
disorder
randomness
dispersal of energy
2nd Law of Thermodynamics
Suniverse > 0
spontaneous =
for spontaneous processes
no external intervention
Ssystem
positional disorder
Ssurroundings
energetic disorder
Energetic Disorder
P.E. a)  K.E.
b) 
ordered
reactants
random
a) endothermic reaction
b) exothermic reaction
P.E.
products
qsystem < 0
qsurroundings > 0
Ssurroundings > 0
Ssurr= - qsys
T
(J/K)
Ssurr depends on T
heat  surroundings
high T
small effect
low T
relatively larger effect
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
W
2W
∆S = S2 – S1 = kBln W2/W1
S = kB ln W
kB = R/NA
= kB ln 2 x 2 x 2
for 1 mole gas
∆S = kB ln 2
6.02 x 10 23
= 6.02 x 1023 kB ln 2
= R ln 2
∆SV 1→ V2 = R ln (V2/V1)
Positional Disorder
Boltzman
S = kB ln W
∆S = R ln (V2/V1)
W = microstates
ordered states
disordered states
low probability
high probability
low S
high S
 Ssystem  Positional disorder
Increases with number of possible positions
(energy states)
Ssolids < Sliquids << Sgases
Entropy (J/K)
[heat entering system at given T]
convert q to S
System 1 Pext = 1.5 atm
w = -182 J
q = +182 J
T = 298 K
System 2 Pext = 0 atm
w=0
q=0
E = 0
E = 0
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
wr= -  Pext dV = -  nRT dV = - nRT ln (V2/V1)
V
V
wr= - nRT ln(1.6/4.0) = - 343.5 J
1
Ssystem
System 1
Pext = 1.5 atm
w = -182 J
q = +182 J
S = 1.15 J/K
System 2
Pext = 0 atm
w=0
q=0
S = 1.15 J/K
∆S = n R ln (V2/V1)
System 3
Pext = Pint + dP
wr = -nRT ln (V2/V1) = - 343.5 J
qr = + 343.5 J
S = 1.15 J/K
qr = n R T ln (V2/V1)
Ssystem = qr = 343.5 J
298 K
T
∆S = n CP ln (T2/T1)
∆S = n CV ln (T2/T1)
3rd Law of Thermodynamics
Entropy
of a perfect crystalline
substance at 0 K= 0
Entropy curve
solid
liquid
gas
vaporization
S
qr
T
fusion
0
0
Temperature (K)
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
Increases in Entropy
1.
2.
3.
4.
5.
6.
7.
Melting (fusion) Sliquid > Ssolid
Vaporization
Sgas >> Sliquid
Increasing ngas in a reaction
Heating
ST2 > ST1 if T2 > T1.
Dissolution ?
Ssolution > (Ssolvent +
S
Molecular
complexity number of bonds
solute)
Atomic complexity e-, protons and neutrons