•  Main points of today’s lecture: •  Ideal gas law:

Physics 231 Lecture 30

2B

2 N 1 2 N 3P m v KE ; KE KE k T3 V 2 3 V 2⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = ≡ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

BPV nRT Nk T= =

Phases of Matter

Slide 12-16

Ideal Gas: properties

Approximatedescrip/onofagasasacollec/onofatomsormoleculesorboththat• Exertnoforceuponeachother

Theenergyofasystemoftwoatoms/moleculescannotbereducedbybringingthemclosetoeachother• Atomsthemselvestakeupveryli?leofthetotalvolume

Thevolumetakenbytheatoms/moleculesisnegligiblecomparedtothevolumethatthegasoccupies.

R

Poten/alEnergy

0-Emin

Rmin

Idealgas:weareneglec/ngthepoten/alenergybetweenTheatoms/moleculesbecausetheyonlyrarelygetthatclosetoeachother.

Important Properties of gases

V=volumeP=pressureT=temperatureinK(Kelvin)N=numberofatomsormolecules

Amount of substance

•  If there are N particles in a gas then the number of moles: NA = 6.02*1023 Avogadro’s number

NA is not defined randomly, it is defined such that there is exactly NA atoms of

carbon in 12 g of 12C

n = NNA

number of moles = mass

mass of one mole

N = total number of atoms (or molecules)

n = NN A

= number of moles or N = n N A

so one mole contains N A atoms (or molecules)

M = m N = total mass of all atoms (or molecules) , each having mass m

M = n Mmolar

Number of atoms and moles

Molar mass

m = mass of one atom (or molecule)

check that it works: 12 x 1.66x10-27 kg =1.99 x10-26 kg

Another way to calculate m(carbon): Approximate mass: m(carbon)=12 Nucleons x atomic mass unit 1 atomic mass unit =1µ = 1.66x10-27 kg.

N A = 6.02×1023 Avagodro's number

Mmolar = m N A molar mass

for example Mmolar (carbon) = 12.0 g = 0.0120 kg

m(carbon) =

Mmolar

N A

0.0126.02×1023 kg = 2.00×10-26 kg

X

Z

A

Numberofelectrons

molarmassingrams

Name

Weight of 1 mol of atoms

1molofatomsweighsAgrams(Aisthemolarmass)

Examples:1molofHydrogenweighs1.0g1molofCarbonweighs12.0g1molofOxygenweighs16.0g1molofZincweighs65.4g

Whataboutmolecules?H2O1molofwatermolecules:

2x1.0g(duetoHydrogen) 1x16.0g(duetoOxygen)

Total:18.0g

Example: Silicon cube

AcubeofSilicon(molarmass28.1g)is250g.A)  Mmolar=28.1g.HowmuchSiliconatomsareinthecube?

B)Whatwouldbethemassforthesamenumberofgoldatoms(molarmass197g)a)197gb)394gc)875gd)1750g

Totalnumberofmolesn=M/Mmolar=250/28.1=8.90N=nNA=(8.9)(6.02x1023)=5.4x1024atoms

M=nMmolar=(8.90)(197g)=1750g

Clicker Quiz

1)  1molofCO2hasalargermassthan1molofCH22)  1molofCO2containsmoremoleculesthan1molofCH2

a)  1)true2)true

b)  1)true2)false

c)  1)false2)true

d)  1)false2)false

Clicker Quiz

1)  1molofCO2hasalargermassthan1molofCH22)  1molofCO2containsmoremoleculesthan1molofCH2

a)  1)true2)true

b)  1)true2)false

c)  1)false2)true

d)  1)false2)false

2) 1 mole of has NA molecules

1) 1 mole CO2 weighs 12g +2*16 g =44 g 1 mole of CH2 weighs 12g + 2*1g = 14g

Boyle’s Law

P0V0T0

2P0½V0T0

½P02V0T0

Atconstanttemperatureandconstantnumberofcons/tuents:P~1/V

Charles’ law

P0V0T0

P02V02T0

Ifyouwanttomaintainaconstantpressurewithconstantnumberofcons/tuents,thetemperaturemustbeincreasedlinearlywiththevolume V~T

Gay-Lussac’s law

P0V0T0 2P0V02T0

If,atconstantvolume,thetemperatureisincreased,thepressurewillincreasebythesamefactor

P~T

•  Ideal gas pressure depends linearly on temperature. •  Here n=number of moles of the gas. There are NA = 6.02x1023

molecules or atoms (if the atoms don’t combine into molecules) per mole. One mole = .0224 m3 of gas at T=0oC and P=1 atm.

•  R=8.31 J/(mole�K) •  Example: A molecular gas is contained in an 8.0-L vessel at a

temperature of 20°C and a pressure of 9.0 atm. (a) Determine the number of moles of gas in the vessel. (b) How many molecules are in the vessel?

All the dependence of pressure P on T, V, and number of moles is contained in the Ideal gas law

•  Ideal gas pressure depends linearly on temperature.

P = nRV

TK

When T is given in Kelvin

PVa) nRT

= ( ) ( )( ) ( )

5 39 1.01x10 Pa .008m8.31J / K 293K

= 3.0 moles=

Ab) N nN= 233 6.02x10= ⋅ 241.8x10 molecules=

Reading Quiz 2.  A sample of nitrogen gas is in a sealed container with a constant

volume. The temperature of the gas is increase by heating it. The pressure

A.  increases B.  stays the same C.  decreases D.  can’t be determined with the information given

Slide 12-8

P = nRT

V

Example

•  Gas is confined in a tank at a pressure of 10.0 atm and a temperature of 15.0°C. If half of the gas is removed and the temperature is raised to 65.0°C, what is the ratio of the final density over the initial density?

•  What is the new pressure in the tank?

00

N m V

ρ = ff

N m V

ρ =0

1 N m2

V= 0

12ρ=

00 B 0

NP k T V

= ff B f

N P k T V

=

fB f

f

00B 0

N k TP VNP k TV

= f f

0 0

N TN T

= 1 338 0.592 288

= =

fP 0.59 10atm= ⋅ 5.9atm=

\

\

\

\

Reading Quiz What is the mass, in u, of a molecule of carbon dioxide, CO2?

A.  12 B.  24 C.  32 D.  36 E.  44

Slide 12-12

mO = 16µmC = 12µmCO2

= 12+ 2 ⋅16 = 44µ

Kinetic properties of an ideal gas •  Consider the gas volume at shown on the right. A

molecule with mass m is elastically scattered by the wall which supplies an impulse force F1.

tmvFmvvmmvtF x

xxx Δ=⇒=−−=Δ 22)( 11

•  After a time Δt=2d/(vx), the mass bounce off the far wall and return to bounce of this wall again. If we use this value for Δt in the expression above, we can get the average force for much large time intervals.

( ) dmvvd

mvF xx

x //2

2 21 ==

•  Multiplying by the number of molecules in the gas and dividing by the area A, we get the pressure on the wall.

wall

wall

d

A

><=⎟⎠⎞⎜

⎝⎛== KE

VNvm

VNvm

VNP

32

21

323/ 22

•  This would be correct if each molecule was moving the x direction with exactly velocity vx. To correct for the differences in velocities, we use the average <vx

2> instead. Also <v2>= <vx2>+ <vy

2>+ <vz2>=3 <vx

2>. Thus”

2 2x x

N NP mv mvAd V

= =

Average kinetic energy of gas molecules •  If we combine the last expression with the ideal gas law equation of

state, we get a useful expression for the means kinetic energy of gas molecules:

•  Example: If the translational rms speed of water molecules (H2O) in a volume of air is 648 m/s, what is the translational rms speed of carbon dioxide (CO2)? Both gases are at the same temperature. (vrms=[<v2>]1/2)

2 NP KE 3 V

= < >

2

2

Brms,H O

H O

3k Tv m

=

B2 N NKE k T3 V V

⇒ < >= 2 2rms B

1 1 3 KE m v mv k T 2 2 2

⇒ < >= ≡ =

A BA

R N nN kN

= ≡B

NP k TV

⇒ =nP RT V

=

vrms,CO2

=3kBTmCO2

=mH2O

mCO2

3kBTmH2O

2

2 2

2

H Orms,CO rms,H O

CO

mv v

m⇒ = 2

2

2

A H Orms,H O

A CO

N mv

N m= ( )2 16 648m / s 414m / s

12 2 16+= =+ ⋅

note: increasing temperature increases thermal (KE) energy

12

mvrms2 = 3

2kBT ⇒

=

mH2O

mCO2

⋅3kBTmH2O

/ /

/ /

kB=1.38x10-23 J/K