physics 231 lecture 26: ideal gases
DESCRIPTION
PHYSICS 231 Lecture 26: Ideal gases. Remco Zegers Walk-in hour:Monday 9:15-10:15 Helproom. Temperature scales. Conversions T celsius =T kelvin -273.5 T fahrenheit =9/5*T celcius +32. We will use T kelvin. If T kelvin =0 , the atoms/molecules have no kinetic energy and every - PowerPoint PPT PresentationTRANSCRIPT
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)