phy 113 c general physics i 11 am – 12:15 p m mwf olin 101 plan for lecture 22: chapter 21: ...

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 gas2. Internal energy of ideal gas3. Distribution of molecular speeds in ideal

gas4. 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

Q0

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.

JJJWEQWQE

701236465int

int

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

PP

PRTn

PPVV

PP

VV

RTnRTn

VPVP

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.

QVV

nRTdVV

nRTPdVWi

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.

KnRVP

TnRTVP offff 14.341

314472.81028.010013.1

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.

JVV

nRTdVV

nRTPdVWi

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

11

1 :energy Internal

:state ofEquation

5735

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

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


22122

22222

2222

222

22

2

32

3

3

that note Also

:,, along move likely toequally are molecules Since

/22

iiiiixi

ixiziyixi

iziyixi

iziiyiixi

ixii

ixi

i

ix

ixi

ix

ixiix

vmVNvm

VNvm

VNP

vvvvv

vvvv

vmvmvm

zyx

vmVN

dAvm

AFP

dvm

vdvmF


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 :Note32

27-

221

nnMNm

m

N

vmVNP

i

i

ii

atoms gas ideal of mole ofenergy kinetic average

23or

32

32

:law gas ideal toConnection

221

2212

21

221

i

ii

i

Mv

RTMvRTMv

nRTMvnPV

nRTE23

int for mono atomic ideal gas


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

MRTv

RTMv

i

i

323

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



nRTE23

int

For monoatomic ideal gas:

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

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

monoatomicfor 1

1

5735

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-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 1.41

H2O 4/3 1.30

Big leap!


Comment on “big leap” – case of diatomic molecule

vCM

w

22

int

21

21 wIMv

EEE

CM

rotCM

RTI

RTMvCM

22

:guess Educated23

:shown have we,Previously

221

221

w

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


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 system Note: = CP/CV

PHY 113 C Fall 2013 -- Lecture 22 29

1γ

1γ

-RC

TnCTR-nQ

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-nE 1γ

int


Example: Isobaric process (P=constant):

In terms of “heat capacity”:

Note: = CP/CV

fifififi WQTR-nE 1γ

int

1

1γγ

1γ

1γ1γ

γ-γR C

-RR

-RC

TnCTnRTR-nVVPTR

-nQ

PP

fiPfifiifififi


Summary

XRRC

CR

CC

CC

RCCXRC

XXnRTE

V

VV

P

V

P

VP

V

1

1 :algebra From

:Define

constant a is ere wh :Suppose int

1

1

int

nRTE

RX


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


More examples:

Isothermal process (T=0)

T=0 DEint = 0 Q=-W

WQTR-nE

RT-nE

1γ

1γ

int

int

i

fV

V

V

V VV

nRTVdVnRTPdVW

f

i

f

i

ln


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


VVPP

VVPP

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

VVTT

VTVT

VTnRPnRTVP

VPVP


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 ;

1γ int -

RCTnCTR-nE VV

1γγ-RCP

i

fii

i

fV

V VV

VPVV

nRTPdVWf

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

iV

V

ii

VVVPPdVW

f

i


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


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?

phy 113 c general physics i 11 am – 12:15 p m mwf olin 101 plan for lecture 22: chapter 21: ...

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webassign assignment

insulated container