Ch 13: Temperature and
Kinetic Theory
Kinetic Theory
 All matter is composed of atoms.
 Atoms are in continuous motion (“kinetic” is Greek for
“moving”)
 Temperature is thought of as how „hot‟ or „cold‟ a
material is.
 Each phase of matter has atoms with specific amounts
of Kinetic energy. (ie solids have slower moving atoms
than gases).
Thermometers
 Thermometers are
instruments used to measure
temperature. They are based
on a property of matter that
changes with temperature.
(Typically this is expansion of
the material.)
 We have 2 common scales
(°C, °F), that are based on the
freezing and boiling points of
water.
Zeroth Law of Thermal
Equilibrium
 Two objects at different temperatures placed in thermal
contact will eventually reach the same temperature.
 The Zeroth law says if two systems are in thermal
equilibrium with a third system, then they are in thermal
equilibrium with each other. (Consider an ice cube in a
glass of water).
 If TA= TB and TC= TB then TA = TC
Thermal Expansion
 Most materials expand when heated and contract when cooled.
The type of material determines the amount of expansion.
 Change in length, ΔL, is proportional to change in temperature,
ΔT, and also to original length, L0.
 ΔL = αL0ΔT ( α is a coefficient of linear expansion for the
particular material at 20°C)
 2m iron rod expands ½ as long as a 4m iron rod for the same
ΔT.
 If T-T0 is negative, then L-L0 is also negative and object shrinks
in length.
Thermal Expansion cont‟d
 Table 13-1 has some common coefficients of linear
expansion, α, and of Volume expansion, β.
 Units of α and β are C°-1
 If a cookie sheet had a hole cut out and then was heated,
what would happen to the hole? (Consider drawing a circle
on the sheet instead of cutting out the hole, what happens to
the circle?)
 In solid objects, ALL sections expand with increased
Temperature.
Example Problem 13-5
 An iron ring is to fit snugly on a 6.445cm diameter
cylindrical iron rod. If the ring has a diameter of
6.420cm and needs to be 0.008cm larger to fit
snugly, to what temp must the ring be brought so it
will slip over the rod?
 Solution: The hole in the ring must be increased
from a diameter of 6.420cm to 6.445cm + 0.008cm
= 6.453cm. The ring must be heated since the hole
diameter will increase linearly. Solve for ΔT.
 ΔT = ΔL / αL0 so
L
6.453 cm 6.420 cm
T
430 C
6
1
L0 (12 x10 C )( 6.420 cm)
 So it must be raised to T = (20°C + 430°C) = 450°C
Changing Volume
 During a temperature change, the Volume of a material
can change and this is given by the equation
which is similar to that of linear expansion.
 For solids, β≅ 3α.
 Linear expansion does not apply to liquids or gases as
they have no fixed shape.
Gas Laws
 The volume of a gas depends
very much on pressure and
temperature.
 For constant temperature, the
volume of a gas is inversely
proportional to the pressure
applied to it. (Absolute pressure,
not gauge)
 If a pressure on a gas is doubled,
the volume of it reduces to ½ V0.
This is Boyle’s Law:
PV=constant for a fixed
Temperature.
Charles‟ Law
 If pressure can be held constant, the volume of a gas
increases with temperature increases at a fairly constant
rate.
 If the slope of a V-T graph were extended below the
liquefaction point, it crossed the zero volume at -273°C.
All gases seemed to share this trend.
 This indicates that at -273°C it would have zero volume.
This became known as absolute zero or 0 Kelvin.
Charles‟ Law
Gay-Lussac‟s Law
 After Charles‟ and Boyle‟s Laws were established,
Joseph Gay-Lussac stated a 3rd law:
 At constant volume, the pressure of a gas is directly
proportional to the absolute temp.
 For example: a closed jar or aerosol can thrown into a
fire will explode due to increased pressure!
Ideal Gas Laws- at STP
 Each of the previous laws can be combined to come up
with a general relationship regarding a fixed quantity of
gas:


PV
T
PV
nRT
however the amount of the gas is also an
important variable…
where n is the number of moles of the
gas and R is the universal gas constant.
R=8.315J/molK or 1.99cal/molK
Example Problem
 Gas laws at STP assume Temp = 273 K (0°C) and
Pressure = 1 atm or 1.013x105 N/m2
 Determine the volume of 1.00 mol of any gas
 Solution:
V
V
nRT
P
(1.00 mol )(8.315 J / mol K )( 273 K )
(1.013 x10 5 N / m 2 )
3
22 .4 x10 m
3
 This is approximately a cube of one foot on each side.
Example 13-13 Cold Tires
 A car tire filled to gauge pressure of 200kPa at 10°C is
driven 100km and temp within the tire rises to 40°C. What is
the pressure within the tire now?
 Solution: Since volume remains relatively constant V1=V2,
therefore P1/T1 = P2/T2
 Recall pressures are in absolutes and temps are in Kelvin.
P2
P1
T2
T1
(3.01x105 Pa)(313K )
(283K )
333kPa
Subtracting atmospheric pressure we get Pg= 232kPa which is a 15%
increase.
Kinetic theory of Gases
 Basic Assumptions (postulates) of Kinetic theory:
 1. Large number of molecules, N, each of mass, m, moving in
random directions with a variety of speeds.
 2. Molecules are far apart from one another. (Relative)
 3. Molecules obey laws of mechanics and exert forces on each
other only when they collide. (KE)
 4. Collisions with walls of containers or other molecules are
assumed to be perfectly elastic.
Kinetic Theory applied
 Another way to write the ideal gas law relationship
considers the number of molecules in a mole of any gas,
NA, which is Avogadro‟s number. According to the # of
molecules present:
PV
nRT
PV
NkT
N
RT
NA
 k is called Boltzmann‟s constant.
 k = 1.38x10-23 J/K
Therefore…
The average translational kinetic energy
of molecules in a gas is directly
proportional to the absolute temperature.
KE
1 2
mv
2
3
kT
2
This equation also applies reasonably accurately to liquids and
gases.
We can also then use this to calculate the average velocity of
molecules at certain temperatures by:
v
3kT
m
Your turn to Practice
 Please do Ch 13 Review pgs 413-414 #s 9, 12, 15, 31,
32, 35, 36, & 41
 #46 = bonus 