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Lec-10

Content

Entropy and 2nd Law of thermo

Lost work

Entropy a state function

Entropy is not conserved

Entropy

The first law relates the energy, its conversion into heat and work, but

places no limits on the amounts that can be converted, second law is

concerned with the limits on the conversion of heat into work by heat

engines.

For a closed system, the reversible heat flow divided by the absolute

temperature of the system is a sate function known as Entropy S.

Qrev/T=dSChanges in entropy to transfer of heat and to irreversibilities in a

macroscopic sense, without implications to the atomic or molecular

structure or randomness in a system.

for a closed system that undergoes a change in U, as per 1st law

dU=Q+W

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Entropy

As U is independent of the path, we may choose a reversible pathbetween the same starting and ending points U and U+dU, so that

dU = Qrev + Wrev As U is independent of the path, the two values of U and U+dU are equal

dU = Qrev + Wrev =Q+ W

Qrev = Q- Wrev + W

Where (-Wrev + W) is called the lost work

and is denoted as lw. It is the difference between

The irreversible work done and the actual work. It is the measure of theirreversibilities involved in the transfer of energy between system and

surroundings

Qrev = Qlw

Path 1

Path 2 reversible

U U+dU

Lost work through P-V diagram

Example

The gas is initially at P1 and V1. If gas

expands reversibly the work done is the

negative of the integral of PV curve. If

Resisting force during expansion is constant

and equal to final pressure, the expansion is

Irreversible. Work done is P2(V2-V1)

So

dS=Qactual/T - lw/T

Volume

pressure

Rev. work

2

1

v

v

PdVW

V1 V2

P1

P2

Volume

pressure

Actual work, W

V1 V2

P1

P2

lost work, -lw

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Consider the isothermal expansion of one mole of

Ideal gas from volume V to double the volume 2V.

This expansion can be conducted in two-ways. First

By reversible process, i.e. resisting pressure= pressure of gas,

By first law

U=Q+W

The change in U is zero as the U of an ideal gas is only function of temperature. The work done is

Considering Q+W=0, Q=RT In 2, which is Qrev. The entropy change is

Second the gas will be expanded under no pressure adiabatically as shown in figure.

In this case there is no work done by the gas as the resisting pressure is 0. so U is zero.

S=Qactual/T - lw/T = -lw/T, (as heat transfer is zero) but lost work is the work done if the

process was reversible

2V,

P=0

P1,

V1

2)2

(

)(22

RTInV

VRTInW

gasidealdVV

RTPdVW

v

v

v

v

S=Qrev/T = RT In 2/T=R In 2


So,

Hence the calculated entropy in both cases is the same.

Because the initial and final states of the gas are the

Same in the two cases.

2V,

P=0

P1,

V1

S = -(-RT In 2/T) =R In 2

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Entropy not conserved

Entropy unlike energy is not conserved in natural processes.

Two bodies connected via a thermal conductor (Cu-Bar), if TII>TII,

Heat will flow from I to II. If no heat losses then QI=-QII

Taking the whole apparatus (bodies I, II and connector) as a system,

The change in U is zero. Energy is conserved, work done is 0.

UI+II=UI + UII =QI+QII=0

If heat transfer at the boundaries of reservoir I and II is reversible. i.e. the T of the Cu-bar

adjacent to the reservoir is at the T of reservoir.

SI+II=SI+ SII= QI/TI + QII/TII

As QII=-QI

SI+II = - QI/TI + QII/TIISI+II = - QII [TI TII/TITII]

Which shows that the entropy is not conserved for this isolated system. The heat flow is irreversible,

as seen by the TI and TII difference. That irreversibility gave rise to the entropy increase.

Therefore entropy of this isolate system can never decrease, it can only increase or remain

constant.

I

TICopper

II

TII

Q

TI>TII

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