Entropy = S

Entropy is disorderrandomnessdispersal of energy

2nd Law of Thermodynamics

Suniverse

spontaneous = no external intervention

positional disorder energetic disorder

for spontaneous processes

Ssystem Ssurroundings

> 0

Energetic Disorder

orderedreactants

products

P.E.

a) endothermic reactionb) exothermic reaction

qsystem 0qsurroundings 0

Ssurroundings > 0

a) b)

<>

random

P.E. K.E.

Ssurr= - qsys

Ssurr depends on T

high T small effectheat surroundings

low T relatively larger effect

T(J/K)

2 dice

2 3 4 5 6 7 8 9 10 11 12

distribution = state

microstates

microstates = W= energy and position of atoms in state

S = kB ln W

Positional Disorder

kB = R/NA

W 2 W

∆S = S2 – S1 = kBln W2/W1

= kB ln 2 x 2 x 2for 1 mole gas

∆S = kB ln 2 6.02 x 10 23= 6.02 x 1023 kB ln 2= R ln 2

∆SV → V =1 2 R ln (V2/V1)

S = kB ln WkB = R/NA

Positional Disorder

ordered states low probabilitydisordered states high probability

low Shigh S

Ssystem Positional disorder

Increases with number of possible positions

Ssolids < Sgases

(energy states)

Boltzman

W = microstatesS = kB ln W = R ln (V2/V1)∆S

Sliquids <<

Entropy

System 1

System 2

w = -182 J

w = 0

E = 0

E = 0

q = +182 J

q = 0

(J/K)[heat entering system at given T]

T = 298 K

Pext = 1.5 atm

Pext = 0 atm

convert q to S

System 3

P1 = 6.0 atm P2 = 1.5 atmV1 = 0.4 L V2 = 1.6 L T1 = 298 K = T2

Pext =

reversible process

wr=

Pint + dP

- Pext= dV = - nRT dV

V - nRT ln (V2/V1)

V2

V1

- infinitely slow

wr= - nRT ln(1.6/4.0) = - 343.5 J

n = .10

Ssystem

System 1Pext = 1.5 atmw = -182 Jq = +182 JS =

System 2Pext = 0 atmw = 0q = 0S =

System 3Pext = Pint + dPwr =qr =S =

= - 343.5 J+ 343.5 J

= 343.5 J 298 K

1.15 J/K 1.15 J/K 1.15 J/K

Ssystem =

-nRT ln (V2/V1)

∆S = n R ln (V2/V1) qr = n R T ln (V2/V1)

qr

T

∆S = n CP ln (T2/T1) ∆S = n CV ln (T2/T1)

3rd Law of Thermodynamics

Entropyof a perfect crystallinesubstance

at 0 K= 0

Entropy curve

Sqr

T

Temperature (K)

solid liquid gas

00

fusion

vaporization

Entropy

At 0K, S = 0 Entropy is absolute

S 0 for elements in standard states

S is a State Function

Sorxn = nSo

products - nSoreactants

S is extensive

Increases in Entropy

1. Melting (fusion) Sliquid > Ssolid

2. Vaporization Sgas >> Sliquid

3. Increasing ngas in a reaction

4. Heating ST2 > ST1 if T2 > T1.

5. Dissolution Ssolution > (Ssolvent + Ssolute)?

6. Molecular complexity number of bonds7. Atomic complexity e-, protons and neutrons