11/14/2013PHY 113 C Fall 2013 -- Lecture 221 PHY 113 C General Physics I 11 AM – 12:15 PM MWF Olin 101 Plan for Lecture 22: Chapter 21: Ideal gas equations

PHY 113 C Fall 2013 -- Lecture 22 111/14/2013

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



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

Q

0

0 container insulatedThermally

387 J/(kg*oC)

1001.0264.02025.00 FCuFAlFwater TcTcTc

4186 J/(kg*oC) 900 J/(kg*oC)

(From Table 20.1)



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.

JJJWEQ

WQE

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>TI

B. TF<TI



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.



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

P

P

P

RTn

P

PVV

P

P

V

V

RTn

RTn

VP

VP

RTnVPRTnVP



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.

QV

VnRTdV

V

nRTPdVW

i

fV

V

V

V

f

i

f

i

ln



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.

KnR

VPTnRTVP off

ff 14.341314472.81

028.010013.1

5



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.

JV

VnRTdV

V

nRTPdVW

i

fV

V

V

V

f

i

f

i

2900ln

:process isothermalFor



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?


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


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)


Ideal gas -- continued

..............................

diatomicfor

monoatomicfor

gas ideal of on type dependingparameter

1

1

1

1 :energy Internal

:state ofEquation

57

35

int

PVnRTE

nRTPV

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


From The New Yorker Magazine, November 2003


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.


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

V

Nvm

V

NP

3

2

3

2 22

1

t

vm

t

pF ixiix

ix

2

d

x

ixvdt /2

22

2

/22

xii

ixi

i

ix

ixi

ix

ixiix

vmVN

dAvm

AF

P

dvm

vdvm

F

vix

-vix

number of atoms

volume


22122

22222

2222

222

22

2

3

2

3

3

that note Also

:,, along move likely toequally are molecules Since

/2

2

iiiiixi

ixiziyixi

iziyixi

iziiyiixi

ixii

ixi

i

ix

ixi

ix

ixiix

vmV

Nvm

V

Nvm

V

NP

vvvvv

vvvv

vmvmvm

zyx

vmV

N

dA

vm

A

FP

d

vm

vd

vmF


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


atoms of moles ofnumber

atom) Hefor kg (0.004 massmolar thedenotes M where

)atom Hefor kg106.6( atom of mass

atoms ofnumber :Note3

2

27-

221

n

nMNm

m

N

vmV

NP

i

i

ii

atoms gas ideal of mole ofenergy kinetic average

2

3or

3

2

3

2

:law gas ideal toConnection

221

2212

21

221

i

ii

i

Mv

RTMvRTMv

nRTMvnPV

nRTE2

3int

for mono atomic ideal gas


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

M

RTv

RTMv

i

i

3

2

3

2

221

Relationship between average atomic velocities with T


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

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



Molecular mass




Molecular mass

kg/mole 0.001 of unitsin M



nRTE2

3int

For monoatomic ideal gas:

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

..............................

diatomicfor

monoatomicfor

1

1

57

35

int

nRTE


Macroscopic Microscopic

BNknR

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

molecules 10 6.022 mole 1 23


RT-n

Tk-

NTkNvmNE BB 1γ1γ2

321

2int

Internal energy of an ideal gas:

derived for monoatomic ideal gas more general relation for

polyatomic ideal gas

Gas g (theory) (g exp)

He 5/3 1.67

N2 7/5 1.41

H2O 4/3 1.30

Big leap!


Comment on “big leap” – case of diatomic molecule

vCM

w

22

int

2

1

2

1 IMv

EEE

CM

rotCM

RTI

RTMvCM

2

2

:guess Educated2

3

:shown have we,Previously

221

221

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


Comment on “big leap” – continued

RT-n

Tk-

NTkNvmNE BB 1γ1γ2

321

2int

Internal energy of an ideal gas:

derived for monoatomic ideal gas more general relation for

polyatomic ideal gas

Big leap!

g can be measured for each gaseous system Note: = CP/CV

PHY 113 C Fall 2013 -- Lecture 22 29

1γ

1γ

-R

C

TnCTR-n

Q

V

fiVfifi

11/14/2013

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-n

E 1γ

int


Example: Isobaric process (P=constant):

In terms of “heat capacity”:

Note: = CP/CV

fifififi WQTR-

nE

1γ int

1

1γ

γ

1γ

1γ1γ

γ-

γR C

-

RR

-

RC

TnCTnRTR-

nVVPTR

-

nQ

PP

fiPfifiifififi


Summary

XRR

C

C

R

C

C

C

C

RCC

XRC

XXnRTE

V

VV

P

V

P

VP

V

1

1 :algebra From

:Define

constant a is ere wh :Suppose int

1

1

int

nRT

E

RX


iclicker question:

The previous discussionA. Made me appreciate the g factor in thermo

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


More examples:

Isothermal process (T=0)

DT=0 DEint = 0 Q=-W

WQTR-

nE

RT-

nE

1γ

1γ

int

int

i

fV

V

V

V V

VnRT

V

dVnRTPdVW

f

i

f

i

ln


Even more examples:

Adiabatic process (Q=0)

TnRVPPV

nRTPV

VPTR-

n

WE

1γ

int

γγγ

γ

lnln

γ

1γ

ffiii

f

i

f VPVPP

P

V

V

PP

VV

VPPVVP-TnR


VVP

P

VVP

P

ii

ii

:Isotherm

:Adiabat


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

V

VTT

VTVT

V

TnRPnRTVP

VPVP


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

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

For an adiabatic process, Q = 0

1γ ;

1γ int -

RCTnCTR

-n

E VV

1γγ-R

CP

i

fii

i

fV

V V

VVP

V

VnRTPdVW

f

i

lnln

1-γ1-γ

γγ

ffii

ffii

VTVT

VPVP


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 adiabatically to a volume Vf is given by:

1γ

11γ f

i

V

V

ii

V

VVPPdVW

f

i


P (

1.01

3 x

105 )

Pa

Vi Vf

Pi

Pf

A

B C

D

Examples process by an ideal gas:

A®B B®C C®D D®A

Q

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

DEint

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


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 mol of gas initially at 0°C.(b) What is the net energy added by heat to the system per cycle?

Documents

11/14/2013PHY 113 C Fall 2013 -- Lecture 221 PHY 113 C General Physics I 11 AM – 12:15 PM MWF Olin 101 Plan for Lecture 22: Chapter 21: Ideal gas equations