Thermo & Stat Mech - Spring 2006 Class 6

1

Thermodynamics and Statistical Mechanics

Entropy and the Second Law of Thermodynamics

Thermo & Stat Mech - Spring 2006 Class 6

2

Second Law

Kelvin-Planck statement:

No series of processes is possible whose sole result is the absorption of heat from a thermal reservoir and the complete conversion of this energy to work.

There are no perfect engines!

Thermo & Stat Mech - Spring 2006 Class 6

3

Second Law

Clausius statement:

No series of processes is possible whose sole result is the transfer of heat from a reservoir at a given temperature to a reservoir at a higher temperature.

There are no perfect refrigerators!

Thermo & Stat Mech - Spring 2006 Class 6

4

Carnot Cycle is Best

A Carnot cycle is the most efficient possible, operating between two reservoirs at temperatures T1 and T2. Proof: Assume there is a more efficient engine. Let it produce work, and use that work to run a Carnot refrigerator between the same two reservoirs.

Thermo & Stat Mech - Spring 2006 Class 6

5

Carnot Cycle is Best

Thermo & Stat Mech - Spring 2006 Class 6

6

Carnot Cycle is Best

and '

' where,' 22 Q

W

Q

W

Then, |Q2 < |Q2|. Also, |W| = |Q2 – |Q1 = |Q2| – |Q1|.So, |Q2| – |Q2 = |Q1| – |Q1, and |Q1 < |Q1| also.

Heat has been taken out of the low temperature reservoir and put into the high temperature reservoir with no expenditure of work!

Not possible.

Thermo & Stat Mech - Spring 2006 Class 6

7

For a Carnot Engine

TQTT

Q

Q

1

2

1

2

1

2

1

2

TT

QQ or 0

2

2

1

1 TQ

TQ

Thermo & Stat Mech - Spring 2006 Class 6

8

For infinitesimal cycles

02

2

1

1 TQd

TQd

0i i

i

TQd

Any cycle can be represented as a sum of infinitesimalCarnot cycles. Then,

Thermo & Stat Mech - Spring 2006 Class 6

9

Carnot Cycles

Thermo & Stat Mech - Spring 2006 Class 6

10

Entropy

0 TQd

TQd

i i

i

dSTQd

For reversible processes.Entropy is a state variable.

Thermo & Stat Mech - Spring 2006 Class 6

11

Carnot Cycle

Thermo & Stat Mech - Spring 2006 Class 6

12

Carnot Cycle on T-S Plot

Thermo & Stat Mech - Spring 2006 Class 6

13

Carnot Cycle

The area enclosed by the cycle on a P-V plot is the net work done per cycle. (đW = PdV)

The area enclosed by the cycle on a T-S plot is the net heat added per cycle. (đQ = TdS for any reversible process.)

These two quantities are equal.

Thermo & Stat Mech - Spring 2006 Class 6

14

Irreversible “Carnot” Cycle

0

or Then

11 and

2

2

1

1

2

2

1

1

2

1

2

1

2

1

2

1

2

1

T

Q

T

Q

T

Q

T

Q

T

T

Q

Q

Q

Q

Q

Q

Q

Q

Thermo & Stat Mech - Spring 2006 Class 6

15

Clausius Inequality

0 so 02

2

1

1

T

Qd

T

Qd

T

Qd

Irreversible cycle

In general 0T

Qd

Thermo & Stat Mech - Spring 2006 Class 6

16

Entropy Change

T

QddS

The equal sign applies forreversible processes.

Thermo & Stat Mech - Spring 2006 Class 6

17

Free Expansion of a Gas

Thermo & Stat Mech - Spring 2006 Class 6

18

Free Expansion

Thermo & Stat Mech - Spring 2006 Class 6

19

Isothermal Expansion

Thermo & Stat Mech - Spring 2006 Class 6

20

Isothermal Expansion

Reversible route between same states.

f

iTQd

SđQ = đW + dU Since T is constant, dU = 0. Then, đQ = đW.

dVV

nRTPdVWd

VdV

nRVdV

TnRT

TWd

TQd

dS

Thermo & Stat Mech - Spring 2006 Class 6

21

Entropy Change

2ln2

ln2

nRVV

nRVdV

nRSV

V

The entropy of the gas increased.For the isothermal expansion, the entropy of theReservoir decreased by the same amount. So for the system plus reservoir, S = 0

For the free expansion, there was no reservoir.

Thermo & Stat Mech - Spring 2006 Class 6

22

Second Law of Thermodynamics

The entropy of an isolated system increases in any irreversible process and is unaltered in any reversible process. This is the principle of increasing entropy.

S 0

Thermo & Stat Mech - Spring 2006 Class 6

23

First and Second Laws

First Law: dU = đQ – đW

First law, combined with the second law:

(for reversible processes)

dU = TdS – PdV