SL-1Spontaneous “Reactions”?

ΔH = 0

ΔH = 0

At one time scientists thought that for a reaction (or process) to be spontaneous, it must be exothermic (evolve energy)… like a ball rolling down a hill.

Ba(OH)2 (s) + 2 NH4 NO3 (s) Ba(NO3)2 (s) + 2 H2O (l) + 2 NH3 (aq) ΔH > 0

Let’s do a thought experiment to test this hypothesis…

These spontaneous reactions (processes) can’t be explained by 1st Law alone!

SL-2What’s missing?

Examples where the processes are either thermoneutral (ΔH = 0) or endothermic have in common an increase in the disorder or randomness of the system.

There is a competition between the tendency to minimize the energy and maximize the disorder.

We know how to calculate and consider the energy. How do we quantify the disorder?

We need a state function that describes disorder. We rule out heat because it is a path function…

SL-3A new state function

dVV

nRTdTTCwdUq Vrevrev +=−= )(δδ

dVVnR

TdTTC

Tq Vrev +=

)(δ

TqdS revδ

≡

Consider the heat transfer associated with small reversible change in T and V of an ideal gas:

We know q is not a state function… but q/T is!(Why?)

Let q/T be the new state function; entropy…

SL-4An old friend shows us S is a state function…

1

21

1, ln

VVnRTdV

VnRTq Arev ==δ

dVV

nRTTT

qS

V

V

revAA

1

1

2

11

, 2

1

1∫∫ ==Δ

δ

Path A vs. Path B+C vs. Path D+EIs the change in entropy different for these different paths?

Path ARecall (from FL-10):

1

2lnVVnRSA =Δ

SL-5Path B + C

0=Δ BS

∫=Δ= 1

2

)(,

T

T VCCrev dTTCUq

dTT

TCT

qS

T

TVrevC

C ∫∫ −==Δ 2

1

)(2

1

,δ

Path BAdiabatic Expansion

∫∫ −= 2

1

2

1

1V

V

T

TV dV

VnRdT

TC

1

2lnVVnRSC =Δ

Path CRecall (from FL-12)

Recall (from FL-16)

So…

SL-6Path D + E

dVPqqq ErevDrevEDrev 1,,, =+=+ δδδ

∫ +=Δ2

1

,

Tq

S EDrevδ

∫=Δ +2

11

V

VED TdVPS ∫=Δ +

2

1

V

VED VdVnRS

1

2lnVVnRS ED =Δ +

Path D+E

So…VP

nRT 1

1=

The entropy changes for all three different paths are exactly the same!

See FL-13

SL-7Entropy and disorder

∫=Δ2

1 TqS revδ

Entropy is related to the disorder of the system. If you add energy, as heat, to a system then its entropy increases because the thermal disorder increases.

Energy delivered at lower T contributes more to an entropy increase than energy delivered at higher T.

SL-8

(i.e., we can learn about spontaneity due to S alone… and not worry about U or H

Entropy and Spontaneity … One Case

TA VA TB VB

Energy as heat will flow spontaneously from a region of high temperature to a region of low temperature. What is the role of entropy?

Consider an isolated two compartment system

Both compartments are in equilibrium but not with each other

UA + UB = constant

VA = constant

VB = constant

S = SA + SB

In an isolated system:

The system can do no work and can have no work done to it.There is no exchange of energy with the surroundings.

SL-9Evaluate the energy and entropy

0== BA dVdV AArevA dSTqdU == δ

BBrevB dSTqdU == δ

B

B

A

ABA T

dUT

dUdSdSdS +=+= BA dUdU −=

No work

Why?

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

ABB TT

dUdS 11 TB > TA , dUB ?, dS ?

TB < TA , dUB ?, dS ?

SL-10What does this tell us about spontaneity?

The spontaneous flow of energy as heat from a body at higher temperature to a body at lower temperature is governed by the condition dS > 0.

Entropy, unlike energy, is not conserved; it increases whenever a spontaneous process takes place. When the system reaches equilibrium dS = 0.

Figure 20.5

SL-11Isolated versus non-isolated…

0>dS0=dS

exchangeprod dSdSdS +=

For a spontaneous process

At equilibrium or for a reversible process

Isolated system

(Why?)General systemThere is exchange of heat with the surroundings. dS becomes the sum of the system (prod) and the exchange:

TqdSdS prodδ

+=

0, == prodrev dSqq δδTqdS revδ

=

or

TqdSdS irr

exchprodδ

=> ,0TqdS irrδ

>

Reversible Process:

Irreversible/Spontaneous Process:

SL-12The second law of thermodynamics…

TqdS revδ

=TqdS irrδ

>TqS δ

∫≥Δ

TqdS δ

≥

There is a thermodynamic function of a system called entropy,S, such that for any change in the state of the system:

where the equality sign applies if the change is carried out reversibly and the inequality sign applies if the change is carried out irreversibly at any stage.

Other ways to write the 2nd law mathematically:

SL-13First and Second Law of Themodynamics

wqU +=ΔThe energy of the universe is a constant…

∫≥ΔTqS δ

The First Law

The entropy is tending toward a maximum…The Second Law

Rudolph Clausius(see page 816!)

SL-14Entropy and Statistical Mechanics

WkS B ln=

)(EΩ

The most famous equation of statistical thermodynamics:

Entropy is a state function related to disorder. Disorder can be expressed in a number of ways. We’ll discuss one (statistical) way here…

Recall: The last time we had each hot chocolate (system) in the collection of hot chocolates (ensemble) with the same N, V, and T (canonical ensemble).

Now: Let’s consider an ensemble with the same N, V, and E (microcanonical ensemble).

Even though every system has the same energy, each system can be in different quantum states due to degeneracy.

is the degeneracy associated with energy E.typically huge #

SL-15Application of statistics

∏==

jja

Aaaa

AaaaW!

!!...!!

!,...),,(321

321

Let W be the number of ways of having a1 systems in state 1, a2 in state 2, etc. (the systems are distinguishable).

Imagine we have A = 4 (i.e., 4 systems) and 5 states…

1 2 3 4 5

Wmin =Wmax =

W = W = W = W = W =

= 1 system

SL-16Entropy and W

WkS B ln=

BAtotal SSS += BAAB WWW =

BABBABBABABBtotal SSWkWkWWkWkS +=+=== lnlnlnln

Ω= lnBkS

For a perfectly ordered arrangement:

For a “completely” disordered arrangement:

This is the form using degeneracy. (Derived in M&S: pg. 831)

Why ln W? We know: &

SL-17Partition Functions and Entropy

∏==

jj

BBensemble aAkWkS

!!lnln

∑−=j

jjBB aakAAk lnln∑=

=N

n

nN1

ln!ln NNNN −= ln!ln

Aa

p jj =A

SS ensemblesystem =

∑−=j

jjBsystem ppkS ln

From Math Chapter J (Eq 7 and 8)

Stirling’s approximation

Using: and

∑−=j

jjBBensemble ApApkAAkS lnln

∑∑ −−=j

jBj

jjBB AApkpApkAAk lnlnln

If all probabilities are 0 except one: If all probabilities are equal :

See pg 840-841

SL-18S in terms of Q

),,(

),(

β

β

VNQep

VNE

j

j−

= ∑−=j

jjBsystem ppkS ln

( )∑ −−−=−

jj

E

B QEQ

ekj

lnββ

∑ ∑ −−

+=j j

EBE

jB

j

j

eQ

QkQeE

k ββ

β lnNow we have S in

terms of Q…

QkTUS B ln+=

SL-19S for an ideal gas…

QkTQTkS B

VNB lnln

,

+⎟⎠⎞

⎜⎝⎛

∂∂

=

1

2/3

2 !2),,( e

NNB g

NV

hTmkTVNQ ⎟⎠⎞

⎜⎝⎛=π

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+= 1

2/3

2

2ln25

eA

B gNV

hTmkRRS π

EX-SL1

SL-20Section 20-9

• Proof of:

• And proof that:

WkS B ln=TqdS revδ

=

TkB

1=β

and

are equivalent.

As promised on BZ-6

SL-21Entropy and Applications

revrev wdUq δδ −=

0=dU

The next few slides discuss some physical processes and the role of entropy in each process. (Chemical processes will be discussed in the next chapter.)

Start with: An isothermal expansion of a gas into a vacuum…

Ideal gas Vacuum

Since entropy is a state function, we can use the following equation (even though the process isn’t reversible!).

dVV

nRTwrev −=δ

∫=Δ2

1 TqS revδ

∫∫ −==Δ2

1

2

1 Tw

TqS revrev δδ

1

2lnVVnR=

Is ΔS + or - ?

(Why?)

SL-22Entropy of Mixing

22

2

2

2

22

22lnln

BrN

NN

N

BrNNN VV

VRn

VVV

RnS+

−=+

=Δ

22

2

2

2

22

22lnln

BrN

BrN

Br

BrNNBr VV

VRn

VVV

RnS+

−=+

=Δ

Isothermal mixing of two ideal gases

22 BrN SSS Δ+Δ=Δ

2222lnln BrBrNN

mix yyyyR

S−−=

Δ

total

ii n

ny =

22

2

2

22

2

2lnln

BrN

BrBr

BrN

NN nn

nRn

nnn

RnS+

−+

−=Δ

Since these are ideal gases, n is proportional to V and ΔS becomes:

Mole fraction:

+ or - ?

SL-23Heat Transfer

)()( cVhV TTCTTC −=−

2ch TTT +

=

∫=Δ 2

1

T

TV

TdTCS

Heat transfer between two pieces of the same metal

Tc Th

Heat lost by hotter piece = heat gained by colder piece

ch SSS Δ+Δ=Δ

ch

chV TT

TTCS4

)(ln2+

=Δ

Uh = Uc

What assumption made here?Solve for T:

+ or - ?

SL-24Carnot Cycle and Heat Pumps/Engines

Sadi Carnot

Carnot used the following cycle to show that a heat engine can not convert all the heat energy supplied to it into mechanical work.

1. Isothermal expansion at Th from V1 to V22. Adiabatic expansion from V2 to V3 , T drops to Tc3. Isothermal compression at Tc from V3 to V44. Adiabatic compression from V4 to V1 , T rises to Th

(You should be able to draw the Carnot cycle on a P-V diagram and solve for w, q, ΔU and ΔS for each step and the cycle as a whole).

What is a heat engine?

Heat ReservoirTh

Heat engine

Cold ReservoirTcqh

wqc

What are signs of qh , qc , and w?

SL-25Carnot Cycle and Heat Pumps/Engines 2

0,, =++=Δ crevhrevengine qqwU

0,, =+=Δc

crev

h

hrevengine T

qTq

Sδδ

What is a heat pump?

Heat ReservoirTh

Heat engine

Cold ReservoirTcqh

wqc

What are signs of qh , qc , and w?

Consider the thermodynamics of the heat engine (or pump)…

SL-26Engine/Pump Efficiency

hrevqw

,

−=η

h

c

hrev

crevhrev

TT

qqq

−=+

= 1,

,,η

h

ch

TTT −

=η

Sadi Carnot (and we) want to know how efficient this hypothetical heat engine (or pump) could be.

Maximum efficiency = work output / heat input

Using equations on bottom of SL-26

EX-SL2