The second law of thermodynamics

A cup coffee does not get hotter in a cooler room

HOT COFFEE

Heat

The first law of thermodynamics places no restriction on the direction of a process, but satisfying the 1st law alone does not ensure that a process will actually take place. This inadequacy is remedied by introducing the

second law of thermodynamics.

The 1st law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality.

The second law helps us to:

determine the quality as well as the degree of degradation of energy during a process. determine the theoretical limits for the

performance of engineering systems such as power plants, heat engines and refrigerators. predict the degree of completion of chemical

reactions.

Statement of the second law:Lord Kelvin and Planck(1) No apparatus can operate in such a way that its only effect (in system and surroundings) is to convert heat absorbed by the system completely into work done by the system.

[It is impossible by a cyclic process to convert the heat absorbed by a system completely into work done by the system]

[It is impossible to build a heat engine that produces a net work output by exchanging heat with a single

fixed-temperature region]

Statement of the second law (Cont.)

(2) No process is possible that consists solely in the transfer of heat from one temperature level to a higher one.

[It is impossible to build a device (a refrigerator or heat pump) that has as its only effect the transfer of heat from a low- to a

high-temperature region]

For reversible process:

CHC

H

TTTT CH

C

H QQ QQ ==

If both QH and QC refer to the heat engine itself (rather than to heat reservoirs) then QH is positive and QC is negative.

)9.5(..........0QQ

QQ

CH

CH

=+

=

CH

CH

TT

TT

For a complete cycle of a Carnot engine, the working fluid periodically returns to its initial state; i.e. its T, P, and U return to their initial values; i.e. the summation of changes in the properties of the working fluid is zero for any complete cycle.

Equation (5.9) suggests that a new property (Q/T) should exist!

When isothermal steps becomes infinitesimal, then:

only cycles reversiblefor 0 T

dQ

0T

dQ T

dQ

rev

C

C

H

H

=

=+

i.e. sum of changes of dQrev/T for a cyclic process = 0

dQrev/T must be a property ( like T, P, U)

This property is called entropy (S)

(5.12) trev TdSdQ =

The entropy change of a heat reservoir is always Q/T for heat source or sink whether transfer is reversible or irreversible. For an adiabatic and reversible process dQrev = 0 and dSt = 0i.e. the entropy is constant during an isentropic process.

Summary

(1) The change in entropy of any system undergoing a reversible process is

(2) For an irreversible (real) process applied to an arbitrarily reversible process that achieves the same change of state as the actual process.Entropy is a state function entropy changes for reversible and irreversible processes are identical.

(3) Entropy is a property derived from the second law just as U was derived from first law.

= TdQS revt = TdQS revt

Entropy changes of an ideal gas

First law of thermodynamics dU = dQ + dWReversible Process dU = dQrev PdVBut H = U + PV dH = dU + PdV + VdP

dH =(dQrev PdV) + PdV + VdPor dQrev = dH - VdP

Ideal gas dH = CPdT and V = RT / PdQrev = CPdT RT dP/P

This is a general equation for calculation of entropy changes of an ideal gas, since it relates properties only, i.e. independent of the process causing the changes (reversible/irreversible) [T1, P1T2, P2]

1

2

T

rev

lnS

TdQ

2

1PPR

TdTCgIntegratin

PdPR

TdTCdSor

T

P

P

=

==

=

+=+=

==

CH

CHtotal

CH

tC

tHtotal

C

tC

H

tH

TTTTQS

TQ

TQ

SSS

TQ

STQ

S and

If heat Q is transferred between two reservoirs TH and TC

Since TH > TC Stotal is positive For this irreversible process

reversible becomes process when i.e. T when 0 )(T as

H

H

Ctotal

Ctotal

TSTS

Stotal is always positive for an irreversible(actual) process, and approaches zero as the process becomes reversible.

What happens if Q=0 (Adiabatic process)?

Consider an irreversible adiabatic expansion from A to B.Suppose fluid is restored to its initial state by a reversible process. If the irreversible results in an entropy change of fluid, there must be heat transfer during the reversible process such that:

Both processes constitute a cycle dU = 0, and

== AB

revtB

tA

t

TdQSSS

=== revrevrevirr dQQWWWSince impossible by cyclic process to convert heat completely into work Qrev cannot directed into system

is negative is also negative

0)&( = tAtBtotal SSSSStB

tA SS revdQ

tA

tB SS

Stotal is always positive, approaching zero as a limit when process becomes reversible. True for any process.

Stotal 0Consider our cyclic heat engine working between TH and T

No net changes in its properties Total entropy change of process (S&S) is the sum of

entropy changes of reservoirs

)1(W

gives Q elminating , C

H

CHtotalC

CH

C

C

H

Htotal

TTQST

QQWTQ

TQ

S

+==

+=

This is the general equation for work done by heat engines working between TH and TC.

Minimum work output = 0 equivalent to irreversible heat transfer between two reservoirs

C

C

H

Htotal T

QTQ

S +=

Maximum work output obtained when engine is reversible

revH

total

QS

==

W and 0

dVPdW A=

TdSdQ =

The Second Law can be stated for a processas follows:

"It is impossible to have a process in which the sum of the change in the total entropy of the system and the change in the total entropy of the surroundings is negative."

That is,

(S2 - S1)system + (S2 - S1)surroundings > = 0