Entropy

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# Entropy - PowerPoint PPT Presentation

Entropy. heat engine. T h =T 2. T c =T 1. Let’s consider the efficiency of a Carnot engine again. Taking into account . and. Clausius’ Theorem. Generalization of the above considerations. W D. T 2. T 1. T N. C N. C 2. C 1. T R. Clausius’ Theorem:.

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Presentation Transcript
Entropy

heat engine

Th=T2

Tc=T1

Let’s consider the efficiency of a Carnot engine again

Taking into account

and

Clausius’ Theorem

Generalization of the above considerations

WD

T2

T1

TN

CN

C2

C1

TR

Clausius’ Theorem:

The heats Qi exchanged at absolute temperatures Ti

by a cyclic process of N steps satisfy

.

Note: the Qi might be irreversibly exchanged

Theorem applies to any real engine cycle

Consider a general cyclic device

in thermal contact with a single heat reservoir

Proof of the Clausius theorem:

General cyclic Device

Ti

Ci

TR

General cyclic Device

WD

T2

T1

TN

CN

C2

C1

TR

With a particular sign convention for work and heat we obtain:

Consider convention for the case Ci is a Carnot engine:

Flowing from the Carnot engine into the device

at the lower temperature Ti

Heat

Work

done by the engine on the surrounding

Heat

Flowing from the hot reservoir to the engine

Consider our combined system as a black box

The net effect is:

Exchange work

Exchange heat

TR

Kelvin statement:

It is impossible to take heat from a single reservoir and

change it entirely into work

(we are allowed to do nothing or to do work on the

device and transfer it entirely into heat)

Heat exchanged by

Carnot cycles

According to our above sign convention we have

Proof of Clausius’ Theorem

Holds in general also for irreversible devices

What happens in the particular case of a reversible device

Corollary (of Clausius’ theorem): For an arbitrary reversible process

.

Proof:

Let device operate in reversed direction and apply the Clausius theorem

while

Let’s consider now a reversible cyclic process

(represented as closed contour in state space)

isotherms

P

V

Reversible cyclic process can be represented by subsequent Carnot cycles

Approximation of the closed contour by N steps with

In the limit N

From the exact differential theorem we conclude

is exact

dS is the differential of a state function called entropy S.

Consequences of Existence of Entropy

Entropy determined up to an additive constant ( which will be specified later by the 3rd law)

Since

is extensive, S and dS are extensive

In some cases an inexact differential can be changed into an exact one by

multiplying with an integrating factor

is the integrating factor which makes

Exact differentials and theory of differential equations)

Relationship between the internal energy and the equation of state

=

dS exact

Calculation of the derivatives yields

P=P(T,V)

Obviously:

U=U(T,V)

and

not independent

Example:

We show for the ideal gas:

is a consequence of U=U(T)

1

U=U(T)

From

is derived from

U=U(T)

2

U=U(T)

U independent of V

CV

Heat capacity in terms of entropy

and

With

0

Let’s take advantage of the new notation using differential forms

@ V=const.

dV=0

V=const.

P=const.

Q

m

Q

Let’s summarize the various representations of heat capacities for PVT systems

Constant volume

Constant pressure

where H=U+PV

Explicit expression for the entropy of an ideal gas

for the ideal gas

in general

for the ideal gas

Comparison

(in order to obtain dimensionless argument of

the logarithm introduce a reference state (Tr,Vr) )

(T,P)

Change from (T,V)

(P,V)

Easily done with

PV=nRT

Isentropic processes

isentropic

Reversible adiabatic processes are isentropic processes that is, processes

in which the entropy is constant.

S=const.

Isentropic processes in ideal gases

A

x

We know already the isothermal bulk modulus

Reminder:

x

P2=P1+P

0

x

Proportionality

constant

defines BT

T=const.

We now define the adiabatic bulk modulus

Let’s calculate

Since

derived for

Summary of important measurable properties of PVT systems

Thermal expansion

volume coefficient

Isothermal bulk

modulus

modulus

Constant pressure

heat capacity

Constant volume

heat capacity

Easier to measure than CV

measured in experiment from speed of sound

Remember:

Existence of entropy provides relationships between material properties

Consider

with

with

derived from general considerations

X

S

S

Y

P

V

Z

T

T

Second relationship looking for CP/CV expressed in terms of BS and BT

Consider again

and let’s calculate

Applying

for

and

Entropy and the Second Law

Entropy never decreases

We are going to show the meaning of the sloppy phrase:

Therefore:

Reminder:

Clausius´ theorem

Heat exchanged

In a cyclic process of N steps

@

absolute temperatures

How to calculate the entropy change involved in a real (irreversible) process

Find a reversible equilibrium process that takes the system between

the initial and the final states of the actual processes.

(hence we can later take advantage of Clausius´theorem )

Consider a cyclic process

State space spanned by (P,V) for instance but not necessarily

P

cannot be represented by a line

In state space

real process

1

2

reversible

process

V

Now let´s apply Clausius´ theorem

Clausius

Consider the reversible equilibrium part of the total cyclic process

start “0” of real

final “f” of real

P

1

2

reversible

process

final

of rev

start

of rev

reversible heat exchange at T

V

System with single temperature

Qi=0

Consider the real process

Entropy statement of the second law:

The total entropy of an adiabatically isolated system never decreases.

The idea that entropy will increase, if it can is emphasized in

Clausius summary of the second law:

The entropy of the universe* tends toward a maximum.

*Universe is a synonym for a completely isolated system (in contrast to

adiabatically isolated systems of the above statement )

Clausius statement and

the principle of increase of entropy

-Entropy change of the high temperature bath.

-Entropy change of the low temperature bath.

Heat is leaving the low temperature bath

Total entropy change

because

Existence of a non-Clausius device violates entropy statement of 2nd law

Kelvin statement and the entropy statement of the 2nd law

-Entropy change of the high temperature bath:

-Device: only effect is to accept heat from the reservoir

and transform it entirely into work

Device in the same state at the start and finish

of the process

Total entropy change

Existence of a non-Kelvin device violates entropy statement of 2nd law

Time’s arrow

S1

S2

Entropy and Irreversibility

Consider a process in an adiabatically isolated system where

process allow by 2nd law

Reversed process:

Start and final state become exchanged

Reversed process forbidden by 2nd law

Entropy concept allows to quantify irreversibility

S1 < S2

t1

t2

Two examples for irreversible processes with corresponding entropy increase

Free expansion of an ideal gas

1

V0 ,T0

Vf ,Tf

Thermal insulation

gas

no heat flow

fixed walls

no work done by the container

W=0

U=U(Tf)-U(T0)=0

With

and

ln1=0

ln2

Let’s calculate S via an integral in state space using a path, L, of reversible equilibrium processes taking the system from the initial to the final states of the actual process.

Fast free expansion is certainly not a sequence of reversible equilibrium processes

only initial and final states can be represented in state space

We know that free expansion of the ideal gas takes place at

T=const.=T0=Tf

We can therefore bring the system from (T0,V0) to (Tf,Vf)

via an isothermal expansion

T=const.

Cu

1kg

2kg

Temperature equalization

2

In order to have a specific example consider:

No heat exchange with

surrounding

T0Cu=80C

=353.15K

T0W=10C

=283.15K

Final equilibrium temperature

no heat exchange with surrounding

heat flow into the water>0

heat flow out of the Cu block<0

with

Total entropy change

Entropy of the Cu block decreases

Total entropy of the isolated system increases

Stability of thermodynamic systems

According to Clausius: entropy is at maximum in systems in equilibrium at

Constant U and V.

Consider two systems in equilibrium with

and

entropy of combined system

(not in equilibrium) is

+

Internal energy: 2U

volume: 2V

Compare with

entropy of combined system is

Internal energy: 2U

volume: 2V

We know entropy is extensive

would evolve into 2 systems with

otherwise system with

Combined system with internal energy 2U and volume 2V,

has entropy

in equilibrium which is at maximum

Alternative formal approach:

We know:

Absolute temperature T>0

Moreover, we want S=S(U,V) be a concave function of U

S

S(U0,V)

Equation of the linear function

U

U0+U

U0

U0-U

Consider the limit U->0

With

and for U->0

Let’s apply the same consideration for the volume

Consider the limit V->0

With

For V->0

Finally let’s consider the small changes U and V together

(both inequalities yield the same result)

Consider the limit U->0,V->0

With e.g.,

For all U,V

in the vicinity of (U,V)=(0,0)

Remember from math: sufficient condition for a maximum of f(x,y) at (x0,y0)

For more background

Hesse-matrix

& determinant of Hesse-matrix>0

Our problem: in order to make sure

(U,V)=(0,0)

has to be a maximum of

since f(0,0)=0

with

> 0

> 0

> 0

What are the physical implications

> 0

see textbook

1

< 0

From

2

A

x

Intuitive meaning of

>0

Consider a volume fluctuation V>0

x

with

x

Pressure decreases

0

External pressure drives system back to equilibrium

otherwise

T=const.