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Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

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Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008
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Page 1: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Lecture #8

Cassandra PaulPhysics 7A

Summer Session II 2008

Page 2: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Announcements• No Lecture or DL on Monday! (Labor Day)• There is DL on Tuesday and Wednesday.• I will be out of town next Tuesday and Wednesday.

– Substitute for my office hours.• Lecture next Wednesday

– Sub– Evaluations– Short Lecture– Quiz 5

• Review Sessions posted over the Weekend• Final Lecture Monday September 8th

Page 3: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Today

• Intro Model to Thermodynamics• Ideal Gas Model• Practice with both• Enthalpy• Microstates• Entropy

Page 4: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Intro Model of Thermodynamics

A tool we use to talk about the internal energy of a system.

Page 5: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What is Thermodynamics?

In a nutshell, it is the study of the transfer of energy between systems and how the energy instills movement, i.e., how the system responds.

Ex. If we heat something, it expands.

Open box of gas

Page 6: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

But wait… does it have to expand?

Closed box of gas Open box of gas

No, we can constrain the system.

Page 7: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Does it take more or less energy to change the temperature of the constant volume system?

Closed box of gas Open box of gas

A) MoreB) LessC) The SameD) Depends on

Substance

Page 8: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Does it take more or less energy to change the temperature of the constant volume system?

Closed box of gas Open box of gas

Closed box: all heat goes into the gas’s internal energy

Open box: Some heat goes into pushing air out of the way (work is leaving the system)

CV measurement

CP measurement

Page 9: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What is ‘work’ in the Thermodynamic Model?

Final Initial

EEthth

T

Work

Volume is decreasing, positive work is being done on the system.

ΔEth = W

Page 10: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What is ‘work’ in the Thermodynamic Model?

Initial Final

EEthth

T

Work

Volume is increasing, negative work is being done by the system.

ΔEth = W

Page 11: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

U = Internal Energy

When we add heat or work to a substance we are changing the internal energy, but what is internal energy anyway?

It is the energy of the system associated with the particles in the system.

ΔEth + ΔEbond = ΔU = Q + W

Eth + Ebond = U

Page 12: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Then how is U different than Etot?Then how is U different than Etot?

It does not include external energies of the system. For example…

Page 13: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

If we throw a coconut off the roof…

The Internal Energy of the coconut:

• Does not account for the height of the coconut

• Does not account for the velocity of the coconut

• Does account for the Temperature of the coconut

• Does account for the bond energy of the coconut

• Does account for the average KE of the particles in the coconut

• Does account for the average PE of the particles in the coconut

Page 14: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

∆U : Internal energyEnergy associated with the atoms/molecules inside the body Of material

∆Etotal must include all changes of energy associated with the system…

∆Etotal = ∆Ethermal + ∆Ebond + ∆Eatomic + ∆Enuclear + ∆Emechanical

Energy associated with the motion of a body as a whole (KE and PE of totalSystem)

ΔU is a part of ΔEtot

Remember conservation of energy?

…if there’s no change in ∆Emechanical, ∆ U

First law of Thermodynamics

Page 15: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

External Energies vs. U

External• KE of the SYSTEM• PE of the SYSTEM

Internal (U)• Average KE of the

Particles in the system• Average PE of the

Particles in the system• Eth of the System

• Ebond of the System

Page 16: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

State Functions

A State Function is a macroscopic property of a physical system of matter that has a definite value that depends only on certain observable parameters of a system.

A State Function does not depend on how the system evolved, but only on the value of the current parameters. (Initial and Final values)

Page 17: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Let’s bring back a bucket of water…

What are some observable (measurable) properties of the water in this bucket?

Temperature

Volume

Pressure

Eth

Eb KEPEAll State Functions!!!!!! Entropy

Enthalpy

Page 18: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

But can I measure how much heat was added and work was done on this

bucket since the beginning of time?

Only If you were there since the beginning of time!

Page 19: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

State Functions can have an Instantaneous Value,

Processes can not!

Page 20: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Processes

• Processes are quantities that depend on HOW a system ends up at it’s final state.

• The have no ‘instantaneous values.’ They can only be measured over some period of time.

• They are not considered properties of the system.

Can you think of any processes? (Hint: they cause change, but aren’t changes themselves.)

Heat and Work!

Page 21: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

ΔU = Q + W

This is the “First Law of Thermodynamics”It is often EXTREMELY useful to use it like this:ΔEth + ΔEbond = ΔU = Q + W

And to also use it in conjunction with the Ideal gas law, so let’s pause a moment and (officially) introduce the ideal gas model…

Page 22: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Ideal Gas Model

A tool we use to describe gases in situations where we can ignore inter-

particle interactions.

Page 23: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

PV=nRT or PV=NkBT

• n= number of moles• N=number of atoms• kB =1.381x10-23 J/K

• R=8.314 J/(K mole)There are other constructs of this model, but

most we have introduced in other places, check the blue pages for more information.

Page 24: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

OK what I was saying…

• It is VERY useful to use the first law of thermodynamics with the sum of the internal energies and with ideal gas law:

ΔU = Q + WΔU =ΔEth + ΔEbond

PV=nRT

Let’s See How…

Page 25: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Here is a particular cycle that happens to one mole of a monatomic gas, no phase

change happens during this process.x105

x10-3

a

b

c

Let’s write this on the board so we can keep track of it.

Page 26: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

From abca, what is ΔU?

Let’s see:ΔU =ΔEth + ΔEbond

ΔU =ΔEth

ΔU = (# of modes per particle)(# of particles) ½ kBΔT

ΔU = 3(6.02x1023) ½ kB(Tf-Ti)

OK… let’s find the T at a and a…. wait…The final STATE is the same so:Tf=Ti! Therefore ΔU = 0

Note: if you don’t believe me, use PV=nRT

Page 27: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

OK well what is the Q for the process abca ?

ΔEth + ΔEbond = ΔU = Q + W

Does Q equal zero for this process? (Don’t calculate anything.)

A.Yes, Q = 0B.No, Q has a negative valueC.No, Q has a positive valueD.Can’t be determined.

Page 28: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What is the Q for the process abca ?

ΔEth + ΔEbond = ΔU = Q + W

ΔU = Q + W = 0Work is easier to find, let’s find that and then

get our Q.

Page 29: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Work for abc:Wabc= ½ (1x10-3 x 1x105) + (1x10-3 x 1x105)

Work for ca:Wca = (1x10-3 x 1x105)

Wabca = Wabc + Wca = -150J + 100J = -50J

x105

x10-3

a

b

c

Positive of Negative?

Negative W: V is increasing

Wabc = -150J

Positive of Negative?

Positive W: V is decreasing.

Wca = 100J

Page 30: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What is the Q for the process abca ?

ΔEth + ΔEbond = ΔU = Q + W

ΔU = Q + W = 0 = Q + -50JQ = 50J

Page 31: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Harder Problem…

What is the change in Eth from bc?

Start with: ΔEth + ΔEbond = ΔU = Q + W

ΔEth = Q + W

We could find W… but how do we find Q?Need to use something else…PV=nRT!

Page 32: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Harder Problem…

What is the change in Eth from bc?

ΔEth = Q + W

PbVb=nRTb

(2x105)(1.8x10-3) = (1)(8.31)Tb

Tb = 43.3°K

PcVc=nRTc

(1x105)(2.3x10-3) = (1)(8.31)Tc

Tc = 27.7°K

Page 33: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What is the change in Eth from bc?

ΔEth = Q + W

Tb = 43.3°K

Tc = 27.7°K

What equation do we use for Eth?

ΔEth = (# of modes per particle)(# of particles) ½ kBΔT

ΔEth = (3)(6.02x1023) ½ (1.38x10-23) (27.7 – 43.3)

ΔEth = -194.4 J

Page 34: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

So… use all three equations together!

ΔU = Q + WΔU =ΔEth + ΔEbond

PV=nRT

Page 35: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Enthalpy

Is a state function:- U depends only on state of system- P depends only on state of system- V depends only on state of system

=> H depends only on state of system

(Hess’s law)

Page 36: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

initial

final

P

V

initial

final

P

V

W = 0

Constant volume Constant pressure

Note: works for solids and liquids too!

Enthalpy

*Derivation in P.84

ΔEth + ΔEbond = ΔU = Q + W

Page 37: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Intro to Statistical Model of Thermodynamics

A tool we use discuss probability of particles or a system being in a

certain state.

Page 38: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

States and Microstates

• A state in the Thermodynamic Model is a combination of instantaneously measurable parameters. – Ex: Solid phase, 30C, at 1Atmosphere of Pressure.

• You can think of it is any point on any state diagram P

V

1

2

3

4

5

Page 39: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What the heck is a Microstate?

• A microstate is one of the different ‘ways’ a system can be in that state.

P

V

State 1

Microstate1

Page 40: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What the heck is a Microstate?

• A microstate is one of the different ‘ways’ a system can be in that state.

P

V

State 1

Microstate 2

Page 41: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What the heck is a Microstate?

• A microstate is one of the different ‘ways’ a system can be in that state.

P

V

State 1

Microstate 3

Page 42: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

Let’s try a real world example..

(Well sort 0f real world)

Page 43: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

After graduating from Davis, you decide you love it here so much

that you want to start a farm and live here forever..

Page 44: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

You go to the state Fair to get your first animals. There are only two kinds for sale and you can only buy 1.

You go back the next day, and pick another one. What are your possible combinations?

Page 45: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

What are the States? And what are the microstates of the system?

State = 1Cow and 1 Sheep State = 1Cow and 1Sheep

State = 2 Cows State = 2 Sheep

Microstate = CS Microstate = SC

Microstate = SSMicrostate = CC

Three States, but 4 microstates, which State is most probable?

Page 46: Lecture #8 Cassandra Paul Physics 7A Summer Session II 2008.

If you leave it up to chance…

You are most likely to have a state of one cow and one sheep, because there are more microstates in that state!


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