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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 1

PHY 113 C General Physics I11 AM – 12:15 PM MWF Olin 101

Plan for Lecture 22:

Chapter 21: Ideal gas equations

1. Molecular view of ideal gas

2. Internal energy of ideal gas

3. Distribution of molecular speeds in ideal gas

4. Adiabatic processes

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 2

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 3

From Webassign (Assignment #19)

A combination of 0.250 kg of water at 20.0°C, 0.400 kg of aluminum at 26.0°C, and 0.100 kg of copper at 100°C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine the final temperature of the mixture.

iIiFii TTcmQ

Q0

0container insulatedThermally

387 J/(kg*oC) 1001.0264.02025.00 FCuFAlFwater TcTcTc

4186 J/(kg*oC) 900 J/(kg*oC)(From Table 20.1)

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 4

From Webassign (Assignment #19)

A thermodynamic system undergoes a process in which its internal energy decreases by 465 J. Over the same time interval, 236 J of work is done on the system. Find the energy transferred from it by heat.

JJJWEQWQE

701236465int

int

Note: Sign convention for Q : Q>0 system gains heat from environment

iclicker question:Assuming the system does not change phase, what can you say about TF versus TI for the system?

A. TF>TIB. TF<TI

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 5

From Webassign (Assignment #19)

A 2.20-mol sample of helium gas initially at 300 K, and 0.400 atm is compressed isothermally to 1.80 atm. Note that the helium behaves as an ideal gas. (a) Find the final volume of the gas.

(b) Find the work done on the gas.

(c) Find the energy transferred by heat.

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 6

From Webassign (Assignment #19)

A 2.20-mol sample of helium gas initially at 300 K, and 0.400 atm is compressed isothermally to 1.80 atm. Note that the helium behaves as an ideal gas. (a) Find the final volume of the gas.

2

1

1

11

2

112

2

1

1

2

11

22

11

22

22221111

PP

PRTn

PPVV

PP

VV

RTnRTn

VPVP

RTnVPRTnVP

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 7

From Webassign (Assignment #19)

A 2.20-mol sample of helium gas initially at 300 K, and 0.400 atm is compressed isothermally to 1.80 atm. Note that the helium behaves as an ideal gas. (b) Find the work done on the gas.(c) Find the energy transferred by heat.

QVV

nRTdVV

nRTPdVWi

fV

V

V

V

f

i

f

i

ln

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 8

From Webassign (Assignment #19)

One mole of an ideal gas does 2 900 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and volume of 28.0 L.

(a) Determine the initial volume of the gas.

(b) Determine the temperature of the gas.

KnRVP

TnRTVP offff 14.341

314472.81028.010013.1 5

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 9

From Webassign (Assignment #19)

One mole of an ideal gas does 2 900 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and volume of 28.0 L.

(a) Determine the initial volume of the gas.

(b) Determine the temperature of the gas.

JVV

nRTdVV

nRTPdVWi

fV

V

V

V

f

i

f

i

2900ln

:processisothermalFor

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 10

From Webassign (Assignment #19)

In the figure, the change in internal energy of a gas that is taken from A to C along the blue path is +795 J. The work done on the gas along the red path ABC is -530 J.

(a) How much energy must be added to the system by heat as it goes from A through B to C?(b) If the pressure at point A is five times that of point C, what is the work done on the system in going from C to D?(c) What is the energy exchanged with the surroundings by heat as the gas goes from C to A along the green path?(d) If the change in internal energy in going from point D to point A is +495 J, how much energy must be added to the system by heat as it goes from point C to point D?

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 11

Review:Consider the process described by ABCA

iclicker exercise:What is the net work done on the system in this cycle?

A. -12000 JB. 12000 JC. 0

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 12

Equation of “state” for ideal gas(from experiment)

nRTPV

pressure in Pascals

volume in m3 # of moles

temperature in K

8.314 J/(mol K)

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 13

Ideal gas -- continued

..............................diatomicfor

monoatomicfor gasidealofon typedependingparameter

11

11:energyInternal

:stateofEquation

5735

int

PVnRTE

nRTPV

Note that at this point, the above equation for Eintis completely unjustified…

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From The New Yorker Magazine, November 2003

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 15

Microscopic model of ideal gas:

Each atom is represented as a tiny hard sphere of mass m with velocity v. Collisions and forces between atoms are neglected. Collisions with the walls of the container are assumed to be elastic.

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 16

Proof:Force exerted on wall perpendicular to x-axis by an atom which

collides with it:

average over atoms

What we can show is the pressure exerted by the atoms by their collisions with the walls of the container is given by:

avgavgK

VNvm

VNP

32

32 2

21

tvm

tpF ixiix

ix

2

d

x

ixvdt /2

22

2

/22

xii

ixi

i

ix

ixi

ix

ixiix

vmVN

dAvm

AFP

dvm

vdvmF

vix

-vix

number of atoms

volume

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 17

22122

22222

2222

222

22

2

32

3

3

that noteAlso

:,,alongmovelikely toequally aremoleculesSince

/22

iiiiixi

ixiziyixi

iziyixi

iziiyiixi

ixii

ixi

i

ix

ixi

ix

ixiix

vmVNvm

VNvm

VNP

vvvvv

vvvv

vmvmvm

zyx

vmVN

dAvm

AFP

dvm

vdvmF

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 18

iclicker question:What should we call ?

A. Average kinetic energy of atom.B. We cannot use our macroscopic equations

at the atomic scale -- so this quantity will go unnamed.

C. We made too many approximations, so it is not worth naming/discussion.

D. Very boring.

221

iivm

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 19

atomsofmolesofnumber atom)Hefor kg(0.004massmolar thedenotesM where

)atomHefor kg106.6(atomofmassatomsofnumber :Note

32

27-

221

nnMNm

mN

vmVNP

i

i

ii

atomsgasidealofmoleofenergy kineticaverage

23or

32

32

:lawgasideal toConnection

221

2212

21

221

i

ii

i

Mv

RTMvRTMv

nRTMvnPV

nRTE23

int for mono atomic ideal gas

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Average atomic velocities:(note <vi>=0)

MRTv

RTMv

i

i

323

2

221

Relationship between average atomic velocities with T

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Periodic table: http://www.nist.gov/pml/data/images/PT-2013-Large_2.jpg

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 22

Periodic table: http://www.nist.gov/pml/data/images/PT-2013-Large_2.jpg

Molecular mass

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Periodic table: http://www.nist.gov/pml/data/images/PT-2013-Large_2.jpg

Molecular mass

kg/mole0.001ofunitsin M

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nRTE23

int

For monoatomic ideal gas:

General form for ideal gas (including mono-, di-, poly-atomic ideal gases):

..............................diatomicfor monoatomicfor

11

5735

int

nRTE

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 25

Macroscopic Microscopic

BNknR

8.314 J/mole oK 1.38 x 10-23 J/molecule oK

molecules106.022mole1 23

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RT-nTk

-NTkNvmNE BB 1γ1γ2

321 2

int

Internal energy of an ideal gas:

derived for monoatomic ideal gas more general relation

for polyatomic ideal gas

Gas (theory) exp)

He 5/3 1.67N2 7/5 1.41H2O 4/3 1.30

Big leap!

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Comment on “big leap” – case of diatomic molecule

vCM

w

22

int

21

21 wIMv

EEE

CM

rotCM

RTI

RTMv CM

22

:guessEducated23

:shownhave we,Previously

221

221

w

Note: We are assuming that molecular vibrations are not taking much energy

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 28

Comment on “big leap” – continued

RT-nTk

-NTkNvmNE BB 1γ1γ2

321 2

int

Internal energy of an ideal gas:

derived for monoatomic ideal gas more general relation

for polyatomic ideal gas

Big leap!

can be measured for each gaseous systemNote: = CP/CV

1γ

1γ

-RC

TnCTR-nQ

V

fiVfifi

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 29

Determination of Q for various processes in an ideal gas:

Example: Isovolumetric process – (V=constant W=0)

In terms of “heat capacity”:

WQTR-nE

RT-nE

1γ

1γ

int

int

fififi QTR-nE 1γint

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Example: Isobaric process (P=constant):

In terms of “heat capacity”:

Note: = CP/CV

fifififi WQTR-nE 1γint

11γγ

1γ

1γ1γ

γ-γR C

-RR

-RC

TnCTnRTR-nVVPTR

-nQ

PP

fiPfifiifififi

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 31

Summary

XRRC

CR

CC

CC

RCCXRC

XXnRTE

V

VV

P

V

P

VP

V

1

1:algebraFrom

:Define

constantaisere wh:Suppose int

1

1

int

nRTE

RX

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 32

iclicker question:

The previous discussionA. Made me appreciate the factor in thermo

analysesB. Made me want to screamC. Put me to sleepD. No problem – as long as this is not on the test

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More examples:

Isothermal process (T=0)

T=0 Eint = 0 Q=-W

WQTR-nE

RT-nE

1γ

1γ

int

int

i

fV

V

V

V VV

nRTVdVnRTPdVW

f

i

f

i

ln

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 34

Even more examples:

Adiabatic process (Q=0)

TnRVPPVnRTPV

VPTR-n

WE

1γ

int

γγγ

γ

lnln

γ

1γ

ffiii

f

i

f VPVPPP

VV

PP

VV

VPPVVP-TnR

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 35

VVPP

VVPP

ii

ii

:Isotherm

:Adiabat

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 36

iclicker question:

Suppose that an ideal gas expands adiabatically. Does the temperature

(A) Increase (B) Decrease (C) Remain the same

1-γ

1-γ1-γ

γγ

f

iif

ffii

i

iiiii

ffii

VVTT

VTVT

VTnRPnRTVP

VPVP

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 37

Review of results from ideal gas analysis in terms of the specific heat ratio CP/CV:

For an isothermal process, Eint = 0 Q=-W

For an adiabatic process, Q = 0

1γ;

1γint -RCTnCTR

-nE VV

1γγ-RCP

i

fii

i

fV

V VV

VPVV

nRTPdVWf

i

lnln

1-γ1-γ

γγ

ffii

ffii

VTVT

VPVP

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 38

Note:

It can be shown that the work done by an ideal gas which has an initial pressure Pi and initial volume Vi when it expands adiabaticallyto a volume Vf is given by:

1γ

11γ f

iV

V

ii

VVVPPdVW

f

i

11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 39

P (1

.013

x 1

05 ) P

a

Vi Vf

Pi

Pf

A

B C

D

Examples process by an ideal gas:

AB BC CD DA

Q

W 0 -Pf(Vf-Vi) 0 Pi(Vf-Vi)

Eint

1-γ)( ifi PPV

1-γ)(γ iff VVP

1-γ)( iff PPV

1-γ)(γ ifi VVP-

1-γ)( ifi PPV

1-γ)( iff PPV

1-γ)( iff VVP

1-γ)( ifi VV-P

Efficiency as an engine:

e = |Wnet/ |/Qinput

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11/14/2013 PHY 113 C Fall 2013 -- Lecture 22 40

From Webassign (#19)

An ideal gas initially at Pi, Vi, and Ti is taken through a cycle as shown below. (Let the factor n = 2.6.)

netiiififnet QVPnVVPPW 21

(a) Find the net work done on the gas per cycle for 2.60 molof gas initially at 0°C.(b) What is the net energy added by heat to the system per cycle?