Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 13: Temperature and Ideal
Gas
•What is Temperature?
•Temperature Scales
•Thermal Expansion
•Molecular Picture of a Gas
•The Ideal Gas Law
•Kinetic Theory of Ideal Gases
•Chemical Reaction Rates
•Collisions Between Molecules
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.1 Temperature
Heat is the flow of energy due to a temperature difference.
Heat always flows from objects at high temperature to
objects at low temperature.
When two objects have the same temperature, they are in
thermal equilibrium.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The Zeroth Law of Thermodynamics:
If two objects are each in thermal equilibrium with a
third object, then the two objects are in thermal
equilibrium with each other.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.2 Temperature Scales
Absolute or
Kelvin scale
Fahrenheit
scale
Celsius
scale
Water boils*
373.15 K
212 F
100 C
Water freezes*
273.15 K
32 F
0 C
Absolute zero
0K
-459.67 F
-273.15C
(*) Values given at 1 atmosphere of pressure.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The temperature scales are related by:
Fahrenheit/
Celsius
TF  1.8 F/CTC  32F
Absolute/
Celsius
T  TC  273.15
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.3): (a) At what temperature (if
any) does the numerical value of Celsius degrees equal the
numerical value of Fahrenheit degrees?
TF  1.8TC  32  TC
TC  40 C
(b) At what temperature (if any) does the numerical value
of Kelvin equal the numerical value of Fahrenheit
degrees?
TF  1.8TC  32
 1.8T  273  32
 1.8TF  273  32
TF  574 F
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.3 Thermal Expansion of Solids
and Liquids
Most objects expand when their temperature increases.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
An object’s length after its temperature has changed is
L  1  T L0
 is the coefficient of
thermal expansion
where T=T-T0 and L0 is the length of the object at a
temperature T0.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.84): An iron bridge girder
(Y = 2.01011 N/m2) is constrained between two rock faces
whose spacing doesn’t change. At 20.0 C the girder is
relaxed. How large a stress develops in the iron if the sun
heats the girder to 40.0 C?
F
L
Y
Using Hooke’s Law:
A
L
 Y T 



 2.0 1011 N/m 2 12 10 6 K -1 20 K 
 4.8 107 N/m 2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
How does the area of an object change when its temperature
changes?
The blue square has an area of L02.
L0
L0+L
With a temperature change T each
side of the square will have a length
change of L = TL0.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The fractional change in area is:
new area  A  L0  TL0 L0  TL0 
 L  2TL   T L
2
0
2
0
2
2 2
0
 L20  2TL20
 A0 1  2T 
A
 2T
A0
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The fractional change in volume due to a temperature
change is:
V
  T
V0
For solids =3
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.4 Molecular Picture of a Gas
The number density of particles is N/V where N is the total
number of particles contained in a volume V.
If a sample contains a single element, the number of
particles in the sample is N = M/m. N is the total mass of
the sample (M) divided by the mass per particle (m).
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
One mole of a substance contains the same number of
particles as there are atoms in 12 grams of 12C. The number
of atoms in 12 grams of 12C is Avogadro’s number.
N A  6.022 1023 mol -1
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A carbon-12 atom by definition has a mass of exactly 12
atomic mass units (12 u).
 12 g  1 mole


23
12
mole
6.022

10


 1 kg 


 1000 g 
 1.66 10- 27 kg
This is the conversion factor between the atomic mass unit
and kg (1 u = 1.6610-27 kg). NA and the mole are defined
so that a 1 gram sample of a substance with an atomic
mass of 1 u contains exactly NA particles.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.39): Air at room temperature and
atmospheric pressure has a mass density of 1.2 kg/m3. The
average molecular mass of air is 29.0 u. How many air
molecules are there in 1.0 cm3 of air?
total mass of air in 1.0 cm3
number of particles 
average mass per air molecule
The total mass of air in the given volume is:
 1.2 kg  1.0 cm

m  V  
3 
 m  1
3
 1 m 


 100 cm 
3
 1.2  10 6 kg
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
total mass of air in 1.0 cm 3
number of particles 
average mass per air molecule
1.2 10 6 kg

29.0 u/particle  1.66 1027 kg/u


 2.5 1019 particles
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.5 Absolute Temperature and
the Ideal Gas Law
Experiments done on dilute gases (a gas where interactions
between molecules can be ignored) show that:
For constant pressure
V T
Charles’ Law
For constant volume
P T
Gay-Lussac’s Law
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
For constant
temperature
For constant pressure
and temperature
1
P
V
V N
Boyle’s Law
Avogadro’s
Law
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Putting all of these statements together gives the ideal gas
law (microscopic form):
PV  NkT
k = 1.3810-23 J/K is
Boltzmann’s constant
The ideal gas law can also be written as (macroscopic
form):
PV  nRT
R = NAk= 8.31 J/K/mole is the
universal gas constant and n is
the number of moles.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.41): A cylinder in a car engine
takes Vi = 4.5010-2 m3 of air into the chamber at 30 C and
at atmospheric pressure. The piston then compresses the
air to one-ninth of the original volume and to 20.0 times the
original pressure. What is the new temperature of the air?
Here, Vf = Vi/9, Pf = 20.0Pi, and Ti = 30 C = 303 K.
PiVi  NkTi
Pf V f  NkTf
The ideal gas law holds
for each set of parameters
(before compression and
after compression).
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Take the ratio:
Pf V f
PiVi

NkTf
NkTi

Tf
Ti
 Pf
The final temperature is T f  
 Pi
 V f 
 Ti
 Vi 
 Vi 
 20.0 Pi  9 

 
303 K   673 K
 Pi  Vi 


The final temperature is 673 K = 400 C.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.6 Kinetic Theory of the Ideal
Gas
An ideal gas is a dilute gas where the particles act as point
particles with no interactions except for elastic collisions.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Gas particles have random motions. Each time a particle
collides with the walls of its container there is a force exerted
on the wall. The force per unit area on the wall is equal to
the pressure in the gas.
The pressure will depend on:
•The number of gas particles
•Frequency of collisions with the walls
•Amount of momentum transferred during each collision
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The pressure in the gas is
2N
P
K tr
3V
Where <Ktr> is the average translational kinetic energy of
the gas particles; it depends on the temperature of the gas.
K tr
3
 kT
2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The average kinetic energy also depends on the rms speed
of the gas
K tr
1
1 2
2
 m v  mvrms
2
2
where the rms speed is
3
1 2
kT  mvrms
2
2
3kT
vrms 
m
K tr 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The distribution of speeds in a gas is given by the MaxwellBoltzmann Distribution.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.60): What is the temperature of
an ideal gas whose molecules have an average
translational kinetic energy of 3.2010-20 J?
3
K tr  kT
2
2 K tr
T
 1550 K
3k
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.70): What are the rms speeds of
helium atoms, and nitrogen, hydrogen, and oxygen
molecules at 25 C?
vrms 
3kT
m
On the Kelvin scale T = 25 C = 298 K.
Element Mass (kg)
rms speed (m/s)
He
6.6410-27
1360
H2
3.32 10-27
1930
N2
4.64 10-26
515
O2
5.32 10-26
482
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.7 Temperature and Reaction
Rates
For a chemical reaction to proceed, the reactants must
have a minimum amount of kinetic energy called activation
energy (Ea).
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
If
3
Ea  kT
2
then only molecules in the high speed tail of MaxwellBoltzmann distribution can react. When this is the situation,
the reaction rates are an exponential function of T.
reaction rates  e


Ea

kT 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.76): The reaction rate for the
hydrolysis of benzoyl-l-arginine amide by trypsin at 10.0 C
is 1.878 times faster than at 5.0 C. Assuming that the
reaction rate is exponential, what is the activation energy?
r1  e


Ea


Ea
r2  e

kT 1 

kT 2 
where T1 = 10.0 C = 283 K and T2
= 5 C = 278 K; and r1 = 2.878 r2.
 Ea
Ea 
r1

The ratio of the reaction rates is
 exp  

r2
 kT1 kT2 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Solving for the activation energy gives:
 r1 
k ln  
r2 

Ea 
1 1
  
 T2 T1 

1.38  10


J/K ln 1.878
 1.37  10 19 J
1 
 1



 278 K 283 K 
 23
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§13.8 Collisions Between Gas
Molecules
On average, a gas particle will be able to travel a distance

1
2d 2  N / V 
before colliding with another particle. This is the mean free
path. The quantity d2 is the cross-sectional area of the
particle.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
After a collision, the molecules involved will have their
direction of travel changed. Successive collisions
produce a random walk trajectory.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Substances will move from areas of high concentration to
areas of lower concentration. This process is called
diffusion.
In a time t, the rms displacement in one direction is:
xrms  2Dt
D is the diffusion constant
(see table 13.3).
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 13.81): Estimate the time it takes a
sucrose molecule to move 5.00 mm in one direction by
diffusion in water. Assume there is no current in the water.
xrms  2Dt
Solve for t


3

2
x
5.00 10 m
t

 25000 s
10
2
2D 2 5.0 10 m /s
2
rms

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Summary
•Definition of Temperature
•Temperature Scales (Celsius, Fahrenheit, Absolute)
•Thermal Expansion
•Origin of Pressure in a Gas
•Ideal Gas Law
•Exponential Reaction Rates
•Mean Free Path
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