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Lecture 8: Thermo and Entropy

Reading: Zumdahl 10.2, 10.3

Outline

Answers to yesterdays questions

Isothermal processes

Isothermal gas expansion and work Reversible and Irreversible processes

Thermodynamic definition of entropy


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Answers

Not terribly satisfying but

Heat death is only a hypothesisIf the expansion of theuniverse is slowing, then the increase in the entropydensity of the universe is also slowing.

Gravity leads to heavier objects. Eventually this leads to

black holes. Steven Hawkings originally proposed blackholes as maximum entropy per unit mass.

But now he recants

A simpler explanation will become apparent todaythat is that raising and releasing a body with regard to PEis not perfect hence an increase in entropy.

Entropy is also seen as the reason time flows forward


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AnswersBoltzmann distributions

0

1

2

3

4

5

6

7

0 2 4 6 8 10

Weighted Average Energy

Count

s


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Isothermal Processes

Recall: Isothermal means T = 0. Since E = nCvT, then E = 0 for an

isothermal process.

Since E = q + w:

q = -w (isothermal process)

We will use isothermal processes to

illustrate what is meant by reversible and

irreversible processes.


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Isothermal Expansion (cont.)

Initially, V = V1

P = P1

Pressure of gas is equal to that

created by mass:

P1

= force/area = M1

g/A

where A = piston area

g = gravitational acceleration (9.8 m/s2)

kg m-1 s-2 = 1 Pa


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Isothermal Expansion (cont.) One-Step Expansion. We change the

weight to M1/4, then

Pext = (M1/4)g/A = P1/4

The mass will be lifted until the internalpressure equals the external pressure. In

this case Vfinal = 4V1

w = -PextV = -P1/4 (4V1 - V1) = -3/4 P1V1


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Two Step Expansion

In this expansion we go in two steps:Step 1: M1 to M1/2

Step 2: M1/2 to M1/4

In first step:

Pext = P1/2, Vfinal = 2V1

w1 = -PextV = -P1/2 (2V1 - V1) = -1/2 P1V1


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Two Step Expansion (cont.)

In Step 2 (M1/2 to M1/4 ):

Pext = P1/4, Vfinal = 4V1

w2 = -PextV = - P1/4 (4V1 - 2V1) = -1/2 P1V1

wtotal = w1 + w2 = -P1V1/2 - P1V1/2 = -P1V1

wtotal,2 step > wtotal,1 step


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Two Step Expansion (cont.) Graphically, we can envision this two-step processon a PV diagram:

Work is given by the area under the PV curve.


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Two Step Compression Now we will do the opposite.take the gas and

compress:

Vinit = 4V1

Pinit = P1/4

Compression in two steps:

first place on mass = M1/2

second, replace mass with one = M1


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Compression/Expansion

We have undergonea cycle where thesystem returns to

the starting state.

In two step example:wexpan. = -P1V1

wcomp. = 2P1V1

wcycle

= P1V

1

qcycle = -P1V1

Now, E = 0 (state fcn)

But, q = -w 0


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Irreversible Processes Since q and w for this cycle are not zero, the

surroundings of the system have changed.

Though the system has been restored to its

starting state, the universe has changed. When it is not possible to restore a system to

its original state without leaving a change inthe universe we call the process irreversible.


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Infinite Step Expansion

Imagine that we perform a process in which we change

the weight infinitesimally between expansions.

(Perhaps we use very fine grains of sand.)

Instead of determining the sum of work performed ateach step to get wtotal, we integrate:

w = PexdVVinitial

Vfinal

= PdVVinitial

Vfinal

= nRT

VdV

Vinitial

Vfinal


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Infinite Step Expansion (cont.) Grapically:

Infinitely Many Steps

(reversible)Six Steps


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Infinite Step Expansion (cont.)

If we perform theintegration from V1 to

V2:

2 22

11 1

expan 2

1

(ln( )) ln|V V

V

totalVV V

VnRTw PdV dV nRT V nRT

V V

= = = =

expan expan2

1

lnrev rev

Vw nRT q

V

= =


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Infinite Step Compression

If we perform the

integration in the

backwards direction,

from V2 to V1 :

1 11

22 2

comp 1

2

(ln( )) ln|rev

V VV

VV V

VnRTw PdV dV nRT V nRT

V V

= = = =

comp comp 1

2

lnrev rev

Vw q nRT

V

= =

2

1

ln V

nRT

V

=


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Reversible Processes

Comparing the results for the expansionand the compression

So this cycle can in principle be performed

without changing the surroundings.

These types of processes are called

reversible.

comp expan comp expanandrev rev rev rev

q q w w= =


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Reversible and Irreversible

Processes

After a reversible process both the systemand its surroundings can be restored to

their original state.

After an irreversible process this is not

possible. It is not possible to restore the

system to its original state withoutchanging the surroundings.


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Reversible and Irreversible

Processes (cont.)

Any process taking place quickly in thereal world will be irreversible.

To be reversible, the changes in the system

must be slow and the system must be at

equilibrium at all times.

Real processes can approach beingreversible if they are performed slowly

enough.


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Thermodynamic Definition of Entropy

For a reversible isothermal process, the change

in the entropy of a system is

If the temperature changes during the process

In both cases, the qrev is the heat transferred in a

reversible process which converts the initial state

into the final state.

finalrev

initial

dqS

T =

r e vqST

=


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Irreversible process

S for a process is defined in terms of the heat for areversible process between those two states.

For an irreversible isothermal process

If the temperature changes during the irreversible

process

In both cases, the qirr is the heat transferred in an

irreversibleprocess.

final

irr

initial

dq

S T >

irrqST

>


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Change in Entropy for an Isothermal

Volume Change of an Ideal Gas

Initial State (T, Vinitial

)

Final State (T. Vfinal)

Now T = 0; therefore, E = 0 and q = -w

Do expansion reversibly.

ln

ln

finalrev rev

initial

finalrev

initial

Vq w nRT

VVq

S nR

T V

= =

= =


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Connecting with Lecture 6 From this lecture:

RTln R ln

T T

final finalrev

initial initial

V Vq nS n

V V

= = =

Exactly the same as derived in the previous lecture!

N ln R lninal final

initial initial

V VS k nV V

= =

The number of atoms, N = nNa.

F th i h i h ti f id l hi h t t t i t ?


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For the isochoric heating of an ideal gas, which statement is true?

Initial Final

V = 0

C. w = 0A. E = 0

D. q = 0B. H = 0


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Entropy Change for Isochoric

Temperature Change of an Ideal Gas

Initial State (Tinitial, V) Final State (Tfinal. V)

Now

V = 0; therefore, w = 0 qrev = E = nCvT

S (ln( ))

ln

|

final finalinal

initial

initial initial

T TT

rev V V T

T T

finalV

initial

dq nC dT nC TT T

T

nC T

= = =

=


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Calculating EntropyS=

dqrev

Tinitial

final

S= dqrev

Tinitial

final

=nR lnVfinal

Vinitial

dqrev =nRT

dV

VT = 0

S= dqrev

Tinitial

final

=nCv lnTfinal

Tinitial

V = 0 dqrev =nCvdT

S= dqrev

Tinitial

final

=nCp lnTfinal

Tinitial

P = 0 dqrev =nCPdT


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Calculating Entropy Example: What is S for the heating of a mole of a monatomic

gas isochorically from 298 K to 350 K?

ln

finalfinalrev

vinitialinitial

TdqS nC

T T

= =

3/2R

( ) 3503(1 ) ln2 298K

S mol RK

=

2JSK

=


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Four-step Cycle

Lets consider a four-step cycle: 1: Isothermal expansion (TH)

2: Isochoric cooling (V2)

3: Isothermal compression (TL)

4: Isochoric heating (V1)

Volume

V1 V2

1

2

3

4

Since the entropy is a state function, S for the cycle must be 0.


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Scycle 1. Isothermal Expansion at T = TH from V1 to V2

2. Isochoric Cooling at V = V2 from TH to TL

3. Isothermal Compression at T = TL from V2 to V1

4. Isochoric Heating at V = V1 from TL to TH

2 L Hln( )vS nC T T =

( )1 2 1lnS nR V V =

( ) ( )3 1 2 2 1 1ln lnS nR V V nR V V S = = =

4 H L L H 2ln ln( ) ( )v vS nC T T nC T T S = = =


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ScycleSum up the four entropy changes

As expected, S for a cycle is zero.

cycle 1 2 3 4

0

S S S S S = + + +

=


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Four-step Cycle

Lets consider a four-step cycle: 1: Isothermal expansion (TH)

2: Isochoric cooling (V2)

3: Isothermal compression (TL)

4: Isochoric heating (V1)

Volume

V1 V2

1

2

3

4

q and w are path dependent and are not zero when

summed around this cycle.


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q and w are not state functions

1. Isothermal Expansion at T = TH from V1 to V2

2. Isochoric Cooling at V = V2 from TH to TL

3. Isothermal Compression at T = TL from V2 to V1

4. Isochoric Heating at V = V1 from TL to TH

( )1 H 1 H 2 1T S T lnq nR V V = =

( ) ( )3 L 1 L 1 2 L 2 1T S T ln T lnq nR V V nR V V = = =

2 2 2 2 L HE E C (T T )vq w n= = =

4 4 4 4 H L 2E E C (T T )vq w n q= = = =

d f i


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q and w are not state functions

So heat transferred to system during this cycle

is

Since E = 0

This cycle can be thought of as a very simple

engine. Run in reverse it is a refrigerator.

1 2 3 4

H L 2 1(T T )ln( ) 0q q q q q

nR V V = + + += >

H L 2 1(T T )ln( )

0

w q nR V V = =

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