Physics 231 Topic 12: Temperature, Thermal Expansion, and Ideal Gases Wade Fisher

Physics 231
Topic 12: Temperature, Thermal Expansion,
and Ideal Gases
Wade Fisher
Nov
16-26
2012
MSU Physics
231 Fall
2012
1
Key Concepts: Temperature,
Thermal Expansion, and Ideal Gases
Temperature and Thermometers
Thermal Energy & Temperature
Thermal Expansion
Coefficient of thermal expansion
Ideal Gases
State Variables
Ideal gas law
Kinetic Theory of Gases
Kinetic & thermal energy
Maxwell distribution
Covers chapter 12 in Rex & Wolfson
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Binding Forces
Potential Energy
Kinetic energy ~ T
0
-Emin
R
The curve depends on
the material, e.g. Emin is
different for water and
iron
R
2 atom/molecules
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Solid (low T)
Potential Energy
0
Kinetic energy ~ T
Rmin
R
-Emin
The temperature (and thus kinetic energy)
is so small that the atoms/molecules can only
oscillate around a fixed position Rmin
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Liquid (medium T)
Potential Energy
Kinetic energy ~ T
Rmin
0
R
-Emin
On average, the atoms/molecules like to
stick together but sometimes escape and
can travel far.
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Gas (high T)
Kinetic energy ~ T
Potential
Energy
Rmin
0
-Emin
R
The kinetic energy is much larger than
Emin and the atoms/molecules move around
randomly.
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What happens if the temperature
of a substance is increased?
Rmin=Rave(T=0)
Kinetic energy ~ T
Rave(T>0) > Rmin
0
-Emin
R
T=0: Average distance between
atoms/molecules: Rmin
T>To: The average distance
between atoms/molecules is
larger than Rmin:
the substance expands
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Temperature scales
Conversions
Tcelsius=Tkelvin-273.5
Tfahrenheit=9/5*Tcelcius+32
We will use Tkelvin.
If Tkelvin=0, the atoms/molecules
have no kinetic energy and every
substance is a solid; it is called the
Absolute zero-point.
Celsius
Fahrenheit
Kelvin
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Clicker Question!
Which is the largest unit:
one Celsius degree, one
Kelvin degree, or one
Fahrenheit degree?
a) one Celsius degree
b) one Kelvin degree
c) one Fahrenheit degree
d) both one Celsius degree and
one Kelvin degree
e) both one Fahrenheit degree
and one Celsius degree
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Thermal expansion
length
L=LoT
surface
A=AoT =2
volume
V=VoT =3 
L
L0
: coefficient of linear expansion
different for each material
Some examples:
=24E-06 1/K Aluminum
=1.2E-04 1/K Alcohol
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T=T0
T=T0+T
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Thermal equilibrium
Thermal contact
Low temperature
Low kinetic energy
Particles move slowly
High temperature
High kinetic energy
Particles move fast
Transfer of kinetic energy
Thermal equilibrium: temperature is the same everywhere
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Zeroth law of thermodynamics
If objects A and B are both in thermal equilibrium
with an object C, than A and B are also in thermal
equilibrium.
There is no transfer of energy between A, B and C
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Thermal expansion: an example
In the early morning (T=30oF=272.4K) a person is asked to measure the
length of a football field with an aluminum measuring stick and finds
109.600 m. Another person does the same in the afternoon
(T=60oF=289.1K) using the same ruler and finds 109.556 m. What is
the coefficient of linear expansion of the ruler?
L=LoT so = L/(L0T)
T=16.7K L0=109.60 L=109.644-109.600=0.044
So: =24E-06 1/K
The ruler has become larger by 0.044 m, so the
person measures a length smaller by the same amount
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A Heated Ring
A metal ring is heated. What is true:
a) The inside and outside radii become larger
b) The inside radius becomes larger, the outside radius
becomes smaller
c) The inside radius becomes smaller, the outside radius
becomes larger
d) The inside and outside radii become smaller
PHY
231
MSU
Physics 231 Fall 2012
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14
Demo: Bimetallic Strips
top
bottom
Application: contact in a refrigerator
top<bottom if the temperature increases,
The strip curls upward, makes contact and switches
on the cooling.
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Water: a special case
Coef. of expansion is
negative: If T drops
the volume becomes
larger
Coef. Of expansion is
positive: if T drops the
volume becomes smaller
Ice is formed (it floats on water)
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Ice
 (g/cm3)
liquid
1
Phase transformation
0.917
ice
Ice takes a larger volume than water!
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Ideal Gas: properties
Collection of atoms/molecules that:
1) 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
2) Take up no volume.
-- The volume taken by the atoms/molecules
is negligible compared to the volume they
are sitting in
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Potential
Energy
Rmin
0
-Emin
R
Ideal gas: we are neglecting the potential energy between
The atoms/molecules
Potential
Energy
Kinetic energy
0
R
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Number of particles: mole
1 mole of particles: 6.02 x 1023 particles
Avogadro’s number NA=6.02x1023 particles per mole
It doesn’t matter what kind of particles:
1 mole is always NA particles
1 mole = 1 mol
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What is the weight of 1 mol of
atoms?
Number of protons
Name
Z
X
A
molar
mass
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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
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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
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Question
1) 1 mol of C02 has a larger mass than 1 mol of CH2
2) 1 mol of CO2 contains more particles than 1 mol of CH2
a) 1) is true and 2) is true
b) 1) is true and 2) is not true
c) 1) is not true and 2) is not true
d) 1) is not true and 2) is not true
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Boyle’s Law
½P0 2V0 T0
P0 V0 T0
2P0 ½V0 T0
At constant temperature: P ~ 1/V
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Charles’ law
P0 2V0 2T0
P0 V0 T0
If you want to maintain a constant pressure, the
temperature must be increased linearly with the volume
V~T
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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
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Boyle & Charles & Gay-Lussac
IDEAL 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:
PV
 constant
T
P1V1 P2V2

T1
T2
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Example
An ideal gas occupies a volume of 1.0cm3 at 200C at
1 atm.
A) How many molecules are in the volume?
B) If the pressure is reduced to 1.0x10-11 Pa, while the
temperature drops to 00C, how many molecules remained
in 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)
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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
P1V1 P2V2

T1
T2
5
9.42  10 6  1.50  10 6 1.013  10  Vsurface

280
293
Vsurface=1.46x10-4 m3=146 cm3
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Correlations
A volume with dimensions LxWxH is kept under
pressure P at temperature T. A) If the temperature is
Raised by a factor of 2, and the height of the volume made
5 times smaller, by what factor does the pressure change,
i.e. what is P2/P1?
No gas leaks or is added.
a) 0.4
b) 1
c) 2.5
d) 5
e) 10
Use the fact PV/T is constant if no gas is added/leaked
P1V1/T1= P2V2/T2
P1V1/T1= P2(V1/5)/(2T1)
P2=5*2*P1=10P1
A factor of 10.
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Diving Bell
A cylindrical diving bell (diameter 3m and 4m tall, with an
open bottom is submerged to a depth of 220m in the sea.
The surface temperature is 250C and at 220m, T=50C. The
density 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=298K
Psub=P0+wg*depth=2.3x106Pa Vsub=? Tsub=5+273=278K
Next, use PV/T=constant
PsurfVsurf/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= r2h
h=27.15/r2=3.8m rise of water level in bell.
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A small matter of definition
Ideal gas law:
PV/T=nR
PV/T=(N/NA)R
N (number of molecules)
n (number of mols)=
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
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From macroscopic to
microscopic descriptions:
kinetic theory of gases
1) The number of molecules is large (statistical model)
2) Their average separation is large (take no volume)
3) Molecules follow Newton’s laws
4) Any particular molecule can move in any direction with a
large distribution of velocities
5) Molecules undergo elastic collision with each other
6) Molecules make elastic collisions with the walls
7) All molecules are of the same type
For derivations of the next equation, see the textbook
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Pressure
Number of Molecules
Mass of 1 molecule
Averaged squared velocity
2  N  1 2 
P    mv 
3  V  2

Volume
Number of molecules
per unit volume
Average translation kinetic energy
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2 1 2
PV  N  mv  Microscopic
3 2

Macroscopic
PV  Nk BT
2 1 2
T
( mv )
3k B 2
Temperature ~ average molecular
kinetic energy
1 2 3
mv  k BT Average molecular kinetic energy
2
2
3
3
Ekin  Nk BT  nRT Total kinetic energy
2
2
3kbT
3RT rms speed of a molecule
2
vrms  v 

M=Molar mass (kg/mol)
m
M
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An Example
What is the rms speed of air at 1atm and room temperature?
Assume it consist of
molecular Nitrogen only (N2)?
vrms
3k bT
3RT
 v 

m
M
2
R=8.31 J/molK T=293 K M=2*14x10-3kg/mol
vrms=511 m/s !!!!!
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And another...
What is the total kinetic energy of the air molecules in the
lecture 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 K
N=2.5x1028 molecules (4.2x104 mol)
2) Ekin=(3/2)NkBT=1.5x108J
(same as driving a 1000kg car at 547.7 m/s)
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Thermal Energy in a Gas
Based on the bulk properties of a gas we can predict the
average kinetic energy of each gas molecule.
Why average? Because each molecule of gas has a semirandom distribution of energy & velocity. We cannot
measure a single molecule easily, but we can observe the
bulk gas.
1 2 3
m v  k BT
2
2
3
3
Ekin  Nk BT  nRT
2
2
3kbT
3RT
2
vrms  v 

m
M
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The Maxwell Distribution
However we can model the distribution of the velocities
(& thus the kinetic energies) of the individual gas
molecules. The result is the Maxwell Distribution.
1 2 3
m v  k BT
2
2
 m 

F (v)  4 
 2k BT 
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3/ 2
mv 2

2 k BT
2
ve
40
Diffusion
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For Next Week
Chapter 13:
Heat
Homework Set 10 Due 11/23
Covers Chapter 12
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