IV. Kinetic theory (continued – see previous lecture)

5. Heat capacitance a) Monoatomic gas

nRTKE tr 23

dTnCdQ

nRdTdK

V

tr

23

} RCV 23

b) Equipartition principle. Degrees of freedom

nRTfE2

RfCV 2

olumeconstant vat heat specificmolar - dTdQ1

nCV

Diatomic gas, including rotations: RCV 25

Diatomic gas, including rotations and vibrations: RCV 27

Molar Specific Heats of Gases at Constant Volume CV At Room Temperature

CV (J/mol·K)Monatomic He 12.47

Ar 12.47

Diatomic H2 20.42N2 20.76 O2 21.10

Polyatomic CO2 28.46SO2 31.39

3R/2 = 12.47 J/mol·K

5R/2 = 20.79 J/mol·K

7R/2 = 29.10 J/mol·K

How many rotational degrees of freedom does the water molecule have?

Question

1. 12. 23. 34. 4

nRTE 3 RCV 3

5. Distribution of molecular speeds (Maxwell distribution)

Tkmv

B

BevTk

mvf 2/22

32

24)(

dvvNfdN )(

c) A hint of quantum theory

b) Heat capacitance of solids

Molar Specific Heats of Elemental Solids at Constant Volume due to Lattice Vibrations

The rule of Dulong and Petit only holds at high temperatures. At low T, quantum mechanical effects reduce CV.

V. The first low of Thermodynamics(conservation of energy)

dU = dQ - dW

1.•Macro- a micro- parameters•Equation of state

for monatomic ideal gas:

•Thermodynamic equilibrium

2. Internal energy (kinetic plus potential energy of particles)

dU = dQ + dWext dWext = -dW

dW = P dV dQ = T dS

dU = dQ - dW = T dS - P dV

RTmPV

ΔU = ΔQ - ΔW

3. Work

V

P

V1 V2

1 2

ΔW = P(V2 - V1)>0

V

P

V2 V1

2 1

ΔW = P(V2 - V1)<0

ΔW12 = 0 ΔW23 > 0ΔW34 > 0ΔW42 < 0

•Work is path dependent•Heat is path dependent•Internal energy is path independent

V

P

V1 V2

1 2

ΔW

V

P

V1 V2

1

23

4ΔW

PdVPAdxFdxdW

2

1

)(V

VdVVPWPdVdW

Example1: A quantity of air is taken from state a to state b along a path that is a straight line in the PV-diagram, where Va=0.070 m3, Vb=0.110 m3, Pa=1.00*105 Pa, Pb=1.40*105 Pa. Assume that the gas is ideal. a) What is the work, W, done by the gas in this process?

V

P

a

b

Va Vb

W = ½ (Pa + Pb )(Vb - Va )

W = ½ (1.00 + 1.40)* 105 Pa*(0.1100 - 0.0700 ) m3

W = 4.8* 103 J

b) What happen with temperature and internal energy of this gas?For ideal gas:

2.20700.01000.11100.01040.1

35

35

mPamPa

VPVP

TT

aa

bb

a

b

b

bb

a

aa

TVP

TVP

TK ~ 2.2a

b

a

b

TT

KK

Example2: This P-V diagram represents a system consisting of a fixed amount of ideal gas that undergoes three different processes in going from state A to state B. Rank work, heat transfer, change in internal, kinetic, and potential energy for each process.

V

State A

I

ΔW1 < ΔW2 < ΔW3

ΔU1 = ΔU2 = ΔU3 = UB - UA

ΔT1 = ΔT2 = ΔT3 = TB - TA

ΔU = ΔQ - ΔW ΔQ1 <ΔQ2 <Δ Q3

For ideal gas:

PE=0 and Δ PE=0

T~ K=U ΔK1 = ΔK2 = ΔK3 = KB - KA

P

State B

23

Example3: A system consisting of a quantity of ideal gas is in equilibrium state A. It is slowly heated and as it expands, its pressure varies. It ends up in equilibrium state B. Now suppose that the same quantity of ideal gas again starts in state A,but undergoes a different thermodynamic process (i.e., follows a different path on a P-V diagram), only to end up again in the same state B as before. Consider the net work done by the system and the net heat absorbed by the system during these two different processes. Which of these statements is true?

1. The work done may be different in the two processes, but the heat absorbed must be the same.2. The work done must be the same in the two processes, but the heat absorbed may be different.3. The work done may be different in the two processes, and the heat absorbed may be different in the two processes.4. Both the work done and the heat absorbed must be the same in the two processes, but are not equal to zero. 5. Both the work done and the heat absorbed by the system must be equal to zero in both processes.

Note: See example 2

~1/3 of questions test understanding of concepts/principles~2/3 of questions test numerical application of concepts/principlesBring soft (#2) pencils, erasers, and your scientific calculator

Studying:• Review during lecture: Friday, Dec. 12• Practice problems and solutions for Ch. 20 are posted• Compare your homework solutions with the posted solutions• Review the lecture questions and the text supplements • Review and solve the example problems in the text• Solve problems on the practice exams -- Seven Phys 221 practice Final Exams are posted plus practice

problems on waves and thermodynamics (Ch. 15-20)• The formula sheet that will be provided with the exam will be posted• Meet with me before the exam to clear up any serious problems you

are having with the course material

Final Exam: Tues, Dec. 16, 2008, 7:00-9:00 p.m.

Physics 221 Final Exam ConflictsIf you are enrolled in any of the three following courses you are entitled to take a Physics 221 Final Exam Make-up at a time and place to be arranged. Math 195, Math 196, Acct 215

Prior to final exam week, students must request to take the make-up exam.

Physics 221 Final ExamTuesday, Dec. 16, 7:00-9:00 p.m.

20 2-point problems + 20 4-point problems (multiple choice)Total: 40 problems worth 120 course points~ 1/3 Ch 17-20 ~ 1/2 Comprehensive (prior to Ch. 17)~ 1/5 Laboratory Final Exam

One cm3 of liquid water at 100 ˚C is boiled off at 1 atm pressure and is converted to 1670 cm3 of steam at 100 ˚C. The “system” is the water.

The work done by the system when it is converted to steam is __ L·atm. (1 L = 1000 cm3)

Question 1

1. 0.172. 1.73. 174. 1700

1. 12. 23. 34. 3

Question 2: The work done by the system in going from point 1 to point 2 is___ L·atm.

4. Thermodynamic processes

•Adiabatic process: ΔQ = 0 ΔU = - ΔW

ΔU = ΔQ - ΔW

•Isochoric process: ΔV = 0 ΔW = 0 ΔU = ΔQ

ΔW = P ΔV

•Isobaric process: ΔP = 0 ΔW = PΔV

•Isothermal process: ΔT = 0

•For ideal gas (!) only: ΔT = 0 ΔU = 0 ΔQ = ΔW

00VPPV

1

20000 ln2

1

2

1 VV

VPVdVVPPdVW

V

V

V

V

•Closed cycle process: ΔT = ΔP = ΔV = ΔU = 0 ΔQ = ΔW

The processes on a PV diagram(ideal gas)

The work done by the gas during this process is ___ L·atm.

Question

1. 1.52. 2.73. 3.94. 5.4

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0

p (a

tm)

V (L)

1

2

An ideal gas undergoes an isothermal expansion from V1 = 1.00 L to V2 = 2.72 L and p2 = 1.00 atm as shown in the figure.

1

20000 ln2

1

2

1 VV

VPVdVVPPdVW

V

V

V

V

The net heat absorbed by the system during one cycle is ___ L·atm

Question

1. 22. -23. 44. -4

Example: A monatomic ideal gas undergoes an increase in pressure from p1 = 1.00 atm to p2 = 3.00 atm at V = 24.0 L as shown in the figure. The heat absorbed by the gas during this process is ??? L·atm.

For Ideal monatomic gas:

PVnRTKU 23

23

atmLLatmU

VPUQV

722420

23

23

Heat capacitance of an ideal gas

dTdQ1 /Vn

CV

at constant volume

/P

dTdQ1

nCP

at constant pressure

PdVdUdQdWdUdQdWdQdU

dTdU1

dTdQ1 /V nn

CV

RCnnn

C VP dT

PdV1 dTdU1

dTdQ1 /P

For an ideal gas:

dTdU

dTdU

dTdU // PV

RCC VP

V

P

CC

RfCV 2

1

RCV 1

RCP

40.1 ,C ,C 5f :re temperaturoomat gas diatomic67.1 ,C ,C 3f :gas monoatomic

57

27

P25

v

35

25

P23

v

RRRR

nRTPV

Relating heat capacities at constant volume and pressure

TUTU

~~

Adiabatic processes for an ideal gas

For an adiabatic process: dQ = 0

dU = dQ -PdV

For an ideal gas: PV=nRT

dT ndU dTdU1 VV C

nC

dVCV VnRT-dT n

)1(

R

V

VP

V CCC

C

VdV

T1--dT

constTV

constVT

constVT

1

1

ln

lnln

ln1ln

constTV 1 122

111

VTVT

constPV 2211 VPVP nRTPV

Example:

?0

260

12

2

21

2

1

1

1

PQb

TTalVCT

lVatmP

VV1

P

V2

1

2

(a)

21

2

22

1

11

TTTVP

TVP

2211 VPVP

2

112 VVPP

atmllatmP 1

2122

(b)

2211

2

22

1

11

VPVP

TVP

TVP

2

112 VVPP atmP

2122

1

2

112

VVTT

21602 CT

The final pressure of the air is ___ atm. ( = 1.40 for air)

Example

The volume of the air inside the cylinder of an engine decreases from 1.00 L to 0.100 L during adiabatic compression. The initial pressure of the air is 1.00 atm and initial temperature is 27 ˚C.

The final temperature of the air is ___ ˚C.

Given: p1 1.00 atm; V1 1.00 L;V2 0.100 L; r V1 /V2 10.0;T1 27 °C = 300 K.Adiabatic compression:

p1V1 p2V2

p2 p1V1

V2 p1

V1

V2

p1r

= (1.00 atm)(10.0)1.40 25.1 atm.

T1V1 1 T2V2

1

T2 T1V1

V2

1

T1r 1

= (300 K)(10.0)1.40 1 754 K = 481 °C.

Adiabatic processes for an ideal gas (2)

constTV 1 122

111

VTVT

constPV 2211 VPVP

dT ndU VCdW

)P(P 1

1)P(P

)T(T n

22112211

21

VVVVRC

W

CW

V

V

PV=nRT

Work

)P-( 1

11

1

111

211

11

1211

112

1

2

1

VVVP

VVVP

VdVVPPdVW

V

V

V

V

)P(P 1

12211 VVW

The work done by the gas during this process is ___ L·atm. ( = 1.40 for air)

Question

1. -82. -43. 24. 6

The volume of the air inside the cylinder of an engine decreases from 1.00 L to 0.100 L during adiabatic compression. The initial pressure of the air is 1.00 atm and the final pressure is 25.1 atm.

)P(P 1

12211 VVW

The heat absorbed by the gas during this process is ___ L·atm.

Question

Some air undergoes an isobaric expansion at 1.00 atm pressure from V1 = 5.00 L to V2 = 9.00 L.(CV = 5R/2)

RCC

VPRCTnCQnRTpV

Vp

pp

atmLQ

LatmPdVRCQ p

14

)00.4(00.1)2/7(