PHY 2311

PHYSICS 231Lecture 26: Ideal gases

Remco ZegersWalk-in hour:Monday 9:15-10:15

Helproom

PHY 2312

Temperature scales

ConversionsTcelsius=Tkelvin-273.5Tfahrenheit=9/5*Tcelcius+32

We will use Tkelvin.

If Tkelvin=0, the atoms/moleculeshave no kinetic energy and everysubstance is a solid; it is called theAbsolute zero-point.

Kelvin

Celsius Fahrenheit

PHY 2313

Thermal expansion

L=LoT

L0

L

T=T0T=T0+T

A=AoT =2

V=VoT =3

length

surface

volume

Some examples:=24E-06 1/K Aluminum=1.2E-04 1/K Alcohol

: coefficient of linear expansion different for each material

PHY 2314

quiz (extra credit)

A substance is measured to have a temperatureof 00C (Celcius). Which of the following is not true?

a) The temperature in Kelvin is 273.5 K b) The temperature in Fahrenheit is 320F c) The atoms in the block are not moving at alld) The substance can be a gas, a liquid or a solid

PHY 2315

Ideal Gas: properties

Collection of atoms/molecules that• Exert no force upon each other

The energy of a system of two atoms/molecules cannot be

reduced by bringing them close to each other• Take no volume

The volume taken by the atoms/molecules

is negligible compared to the volume they

are sitting in

PHY 2316

R

Potential Energy

0-Emin

Rmin

Ideal gas: we are neglecting the potential energy betweenThe atoms/molecules

R

Potential Energy

0

Kinetic energy

PHY 2317

Number of particles: mol

1 mol of particles: 6.02 x 1023 particlesAvogadro’s number NA=6.02x1023 particles per mol

It doesn’t matter what kind of particles:1 mol is always NA particles

PHY 2318

What is the weight of 1 mol of atoms?

X

Z

A

Number of protons

molar mass

Name

PHY 2319

Weight of 1 mol of atoms

1 mol of atoms: A gram (A: mass number)

Example: 1 mol of Carbon = 12 g 1 mol of Zinc = 65.4 g

What about molecules?H2O 1 mol of water molecules:

2x 1 g (due to Hydrogen)1x 16 g (due to Oxygen)

Total: 18 g

PHY 23110

Example

A cube of Silicon (molar mass 28.1 g) is 250 g.A) How much Silicon atoms are in the cube? B) What would be the mass for the same number of gold atoms (molar mass 197 g)

A) Total number of mol: 250/28.1 = 8.90 mol 8.9 mol x 6.02x1023 particles = 5.4x1024 atoms

B) 8.90 mol x 197 g = 1.75x103 g

PHY 23111

Boyle’s Law

P0 V0 T0

2P0 ½V0 T0

½P0 2V0 T0

At constant temperature: P ~ 1/V

PHY 23112

Charles’ law

P0 V0 T0

P0 2V0 2T0

If you want to maintain a constant pressure, the temperature must be increased linearly with the volume V~T

PHY 23113

Gay-Lussac’s law

P0 V0 T0 2P0 V0 2T0

If, at constant volume, the temperature is increased,the pressure will increase by the same factor

P ~ T

PHY 23114

Boyle & Charles & Gay-LussacIDEAL GAS LAW

PV/T = nR

n: number of particles in the gas (mol)R: universal gas constant 8.31 J/mol·K

If no molecules are extracted from or added to a system:

2

22

1

11 constant T

VP

T

VP

T

PV

PHY 23115

ExampleAn ideal gas occupies a volume of 1.0cm3 at 200C at1 atm. A) How many molecules are in the volume?B) If the pressure is reduced to 1.0x10-11 Pa, while thetemperature drops to 00C, how many molecules remainedin the volume?

A) PV/T=nR, so n=PV/(TR) R=8.31 J/molK T=200C=293K P=1atm=1.013x105 Pa V=1.0cm3=1x10-6m3

n=4.2x10-5 mol n=4.2x10-5*NA=2.5x1019 molecules

B) T=00C=273K P=1.0x10-11 Pa V=1x10-6 m3

n=4.4x10-21 mol n=2.6x103 particles (almost vacuum)

PHY 23116

And another!An airbubble has a volume of 1.50 cm3at 950 m depth (T=7oC). What is its volume when it reaches the surface(water=1.0x103 kg/m3)?

P950m=P0+watergh =1.013x105+1.0x103x950x9.81 =9.42x106 Pa

293

10013.1

280

1050.11042.9

566

2

22

1

11

surfaceV

T

VP

T

VP

Vsurface=1.46x10-4 m3=146 cm3

PHY 23117

Correlations

A volume with dimensions LxWxH is kept underpressure P at temperature T. A) If the temperature isRaised by a factor of 2, and the height of the volume made5 times smaller, by what factor does the pressure change?

Use the fact PV/T is constant if no gas is added/leakedP1V1/T1= P2V2/T2

P1V1/T1= P2(V1/5)/(2T1)P2=5*2*P1=10P1

A factor of 10.

Next quiz is something like this…

PHY 23118

Diving BellA cylindrical diving bell (diameter 3m and 4m tall, with anopen bottom is submerged to a depth of 220m in the sea.The surface temperature is 250C and at 220m, T=50C. Thedensity of sea water is 1025 kg/m3. How high does the sea water rise in the bell when it is submerged?

Consider the air in the bell.Psurf=1.0x105Pa Vsurf=r2h=28.3m3 Tsurf=25+273=298KPsub=P0+wg*depth=2.3x106Pa Vsub=? Tsub=5+273=278KNext, use PV/T=constantPsurfVsurf/Tsurf=PsubVsub/Tsub plug in the numbers and find:Vsub=1.15m3 (this is the amount of volume taken by the air left)Vtaken by water=28.3-1.15=27.15m3= r2hh=27.15/r2=3.8m rise of water level in bell.

PHY 23119

A small matter of definition

Ideal gas law: PV/T=nR PV/T=(N/NA)R

n (number of mols)=N (number of molecules)

NA (number of molecules in 1 mol)

Rewrite ideal gas law: PV/T = NkB

where kB=R/NA=1.38x10-23 J/K Boltzmann’s constant

PHY 23120

From macroscopic to microscopic descriptions: kinetic theory of gases

For derivations of the next equation, see the textbook

1) The number of molecules is large (statistical model)2) Their average separation is large (take no volume)3) Molecules follow Newton’s laws4) Any particular molecule can move in any direction with a large distribution of velocities5) Molecules undergo elastic collision with each other6) Molecules make elastic collisions with the walls7) All molecules are of the same type

PHY 23121

Pressure

2

2

1

3

2vm

V

NP

Number of Molecules

Volume

Mass of 1 molecule

Averaged squared velocity

Average translation kinetic energy

Number of moleculesper unit volume

PHY 23122

M

RT

m

Tkvv

nRTTNkE

Tkvm

vmk

T

TNkPV

vmNPV

brms

Bkin

B

B

B

33

2

3

2

32

3

2

1

)2

1(

3

2

2

1

3

2

2

2

2

2

Microscopic

Macroscopic

Temperature ~ average molecular kinetic energy

Average molecular kinetic energy

Total kinetic energy

rms speed of a moleculeM=Molar mass (kg/mol)

PHY 23123

example

What is the rms speed of air at 1atm and room temperature? Assume it consist ofmolecular Nitrogen only (N2)?

M

RT

m

Tkvv b

rms

332

R=8.31 J/molK T=293 K M=2*14x10-3kg/molvrms=511 m/s !!!!!

PHY 23124

And another...

What is the total kinetic energy of the air molecules in thelecture room (assume only molecular nitrogen is present N2)?

1) find the total number of molecules in the room

PV/T= Nkb P=1.015x105 Pa V=10*4*25=1000 m3

kb=1.38x10-23 J/K T=293 KN=2.5x1028 molecules (4.2x104 mol)

2) Ekin=(3/2)NkBT=1.5x108J (same as driving a 1000kg car at 547.7 m/s)