General Physics (PHY 2130)
Lecture 30
•  Thermal physics
  Thermal expansion
  Gases. Absolute temperature
  Ideal Gas law
•  Exam 3 review
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Sound
  Pipes. Doppler’s effect
2.  Thermal physics
  Temperature. Temperature scales. 0th law of thermodynamics.
Review Problem: When a hole is made in the side of a container holding water, water
flows out and follows a parabolic trajectory. I the container is dropped in free fall, the
water flow
1. diminishes
2. stops altogether
3. goes out in a straight line
4. curves upward
Recall:
pressure ~ ρ gh
in a container frame of reference
Comparing Temperature Scales
TC = TK − 273.15
9
TF = TC + 32
5
9
ΔTF = ΔTC
5
4
Example: On a warm summer day, the air temperature is 84°F. Express this
temperature in (a) °C and (b) kelvins.
Fahrenheit/Celsius
TF = (1.8 °F/ °C )TC + 32°F
TF − 32 0 F
TC =
1.8 0 F/ 0 C
Absolute/Celsius
T = TC + 273.15
5
Example: (a) At what temperature (if any) does the numerical value of
Celsius degrees equal the numerical value of Fahrenheit degrees?
Idea: use relations among temperature scales:
TF = 1.8TC + 32 = TC
TC = −40 °C
(b) At what temperature (if any) does the numerical value of kelvins equal the
numerical value of Fahrenheit degrees?
TF = 1.8TC + 32
= 1.8(T − 273) + 32
= 1.8(TF − 273) + 32
TF = 574 °F
Thermal Expansion
•  The thermal expansion of an object is a
consequence of the change in the average
separation between its constituent atoms or
molecules
•  At ordinary temperatures, molecules vibrate with a
small amplitude
•  As temperature increases, the amplitude increases
•  This causes the overall object as a whole to expand
Linear (area, volume) Expansion
•  For small changes in temperature
ΔL = α Lo ΔT
•  The coefficient of linear expansion, α , depends on the material
•  Similar in two dimensions (area expansion)
ΔA = γ Ao ΔT, γ = 2α
•  … and in three dimensions (volume expansion)
ΔV = β Vo ΔT for solids, β = 3α
8
Area Expansion, once again
How does the area of an object change when its temperature
changes?
The blue square has an area of L02.
With a temperature change ΔT each
side of the square will have a length
change of ΔL = αΔTL0.
L0
L0+ΔL
Thus,
A = ( L0 + αΔTL0 ) ( L0 + αΔTL0 )
= L20 + 2αΔTL20 + α 2 ΔT 2 L20
≈ L20 + 2αΔTL20
= A0 (1+ 2αΔT )
In other words,
ΔA = A − A0 = γ Ao ΔT, with
γ = 2α
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Expansion joints permit the roadbed of a bridge to expand and contract as the
temperature changes
Example
A copper telephone wire has essentially no sag between poles 35.0
m apart on a winter day when the temperature is –20.0°C. How
much longer is the wire on a summer day when TC = 35.0°C?
Assume that the thermal coefficient of copper is constant
throughout this range at its room temperature value.
Applications of Thermal Expansion
1. Thermostats
•  Use a bimetallic strip
•  Two metals expand differently
2. Pyrex Glass
•  Thermal stresses are smaller than for ordinary glass
3. Sea levels
•  Warming the oceans will increase the volume of the oceans
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Molecular Picture of a Gas
The number density of particles
= the number of particles in a unit volume
= 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).
Ideal Gas
•  Properties of gases
•  A gas does not have a fixed volume or pressure
•  In a container, the gas expands to fill the container
•  Ideal gas:
•  Collection of atoms or molecules that move randomly
•  Molecules exert no long-range force on one another
•  Molecules occupy a negligible fraction of the volume of their container
•  Most gases at room temperature and pressure behave
approximately as an ideal gas
Moles
•  It’s convenient to express the amount of gas in a
given volume in terms of the number of moles, n
mass
n=
molar mass
•  One mole is the amount of the substance that
contains as many particles as there are atoms in
12 g of carbon-12
Avogadro’s Hypothesis
•  Equal volumes of gas at the same temperature and
pressure contain the same numbers of molecules
•  Corollary: At standard temperature and pressure, one mole
quantities of all gases contain the same number of molecules
•  This number is called NA
•  Can also look at the total number of particles: N = n NA
Avogadro’s Number
•  The number of particles in a mole is called Avogadro’s
Number
•  NA=6.02 x 1023 particles / mole
•  The mass of an individual atom can be calculated:
matom
molar mass
=
NA
Equation of State for an Ideal Gas
•  Boyle’s Law
•  At a constant temperature, pressure is
inversely proportional to the volume
•  Charles’ Law
•  At a constant pressure, the temperature is
directly proportional to the volume
•  Gay-Lussac’s Law
•  At a constant volume, the pressure is
directly proportional to the temperature
Ideal Gas Law
•  Summarizes Boyle’s Law, Charles’ Law, and
Guy-Lussac’s Law
•
PV = n R T
•  R is the Universal Gas Constant
•  R = 8.31 J / mole K
•  R = 0.0821 L atm / mole K
•  P V = N kB T
•  kB is Boltzmann’s Constant
•  kB = R / NA = 1.38 x 10-23 J/ K
Question
An ideal gas is confined to a container with constant volume.
The number of moles is constant. By what factor will the
pressure change if the absolute temperature triples?
a. 1/9
b. 1/3
c. 3.0
d. 9.0
Question
An ideal gas is confined to a container with adjustable volume.
The number of moles and temperature are constant. By what
factor will the volume change if pressure triples?
a. 1/9
b. 1/3
c. 3.0
d. 9.0
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Example: Incandescent lightbulbs are filled with an inert gas to lengthen the
filament life. With the current off (at T = 20.0°C), the gas inside a lightbulb
has a pressure of 115 kPa. When the bulb is burning, the temperature rises
to 70.0°C. What is the pressure at the higher temperature?
Exam 3 Review
Exam 3 Review
•  Chapter 9: Solids and fluids
•  density and pressure
•  buoyant force
•  Archimedes’ principle
•  Fluids in motion
•  Note: no problems on viscosity
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Example: A flat-bottomed barge loaded with coal has a mass of 3.0×105 kg. The
barge is 20.0 m long and 10.0 m wide. It floats in fresh water. What is the depth
of the barge below the waterline?
y
FBD for
the barge
Apply Newton’s 2nd Law to the barge:
FB
∑F = F
B
−w=0
FB = w
mw g = (ρ wVw )g = mb g
x
w
ρ wVw = mb
ρ w ( Ad ) = mb
mb
3.0 ×10 5 kg
d=
=
= 1.5 m
3
ρ w A 1000 kg/m (20.0 m *10.0 m )
(
)