Chapter 10: Thermal Physics
Thermal physics is the study of Temperature, Heat,
and how these affect matter.
Thermal equilibrium exists when two objects in thermal
contact with each other cease to exchange energy.
Thermometers and Temperature Scales
constant volume
gas thermometer
Zeroth Law of Thermodynamics (law of equilibrium):
If objects A and B are separately in thermal equilibrium with a third
object C, then A and B are in thermal equilibrium with each other.
Objects in thermal equilibrium with each other are at the same
temperature.
Any physical property which changes sensitively with the temperature
can be used as thermometer. For example, the pressure of gas at a
constant volume is an indicator of temperature. Celsius and Fahrenheit
are artificially defined temperature scales. The absolute temperature
(Kelvin) scale defines an absolute zero as a lower bound for physical
processes.
T = TC + 273.15
Thermal Expansion
Temperature Scales
TC = T − 273.15
TF = 95 TC + 32
Thermal expansion is due to the change of average spacing between
atoms as temperature changes.
The length of a solid object (with an initial length L0 at temperature T0)
changes by ΔL for a change ΔT in temperature:
ΔL = α L0 ΔT
L - L0 = ΔL = α L0 ( T - T0 )
α is the coefficient of linear expansion. The volume change of an object
with a change in temperature is
ΔV = β V0 ΔT
β is the coefficient of volume expansion. Usually, β = 3 α .
Thermal Expansion
Example Problems
18. A construction worker uses a steel tape to measure the
length of an aluminum support column. If the measured length
is 18.700 m when the temperature is 21.2°C, what is the
measured length when the temperature rises to 29.4°C? (Note:
Don’t neglect the expansion of the steel tape.)
6. A constant-volume gas thermometer is calibrated in dry ice (–
80.0°C) and in boiling ethyl alcohol (78.0°C). The two
pressures are 0.900 atm and 1.635 atm. (a) What value of
absolute zero does the calibration yield? (b) What pressures
would be found at the freezing and boiling points of water?
(Note that we have a linear relationship between P and T as P =
A + BT, where A and B are constants.)
Macroscopic Description of An Ideal Gas
Macroscopic Description of An Ideal Gas
PV = nRT
One mole of any substance contains as many particles as there are
atoms in 12 g of the carbon-12 isotope. (Avogadro’s number NA =
6.02 x 1023)
An ideal gas is a collection of atoms or molecules that move
randomly, exert no long-range forces on one another, and occupy a
negligible fraction of the volume of their container. For any ideal
gas:
The volume occupied by 1 mole of any gas at atmospheric pressure
and at 0oC is 22.4 L. (Equal volumes of gas at the same
temperature and pressure contain the same number of molecules.)
R = NA kB
PV = nRT = N kBT
PV = nRT ,
where n is the number of moles, T is the absolute temperature, the
universal gas constant R = 8.31 J/mol-K = 0.0821 L*atm / mol-K
N: number of molecules; kB: Boltzmann’s constant
k B = R / N A = 1.38 × 10 −23 J / K
The Kinetic Theory of Gases
Assumptions:
1. The number of molecules in the gas is large. However, the
average separation of molecules is large compared with their
dimensions.
2. The molecules obey Newton’s law of motion, but as a whole they
move randomly.
3. Molecules interact only through short-range forces during elastic
collisions.
4. The molecules make elastic collisions with the walls.
The Kinetic Theory of Gases
Assume N molecules of an ideal gas are inside a cubic
container of volume V (d3=V). Let’s figure out the impulse
exerted by one particular molecule in one collision with the
wall at x=d.
Δp x = mv x − (− mv x ) = 2 mv x
How often does this molecule collide with this wall?
Δt = 2 d / v x
In unit time, the total impulse received by the wall from this
molecule is
Δp x
2mv x
=
= mv x2 / d
Δt
2d / v x
5. All molecules in the gas are identical.
In unit time, the total impulse received by the wall from ALL
THE MOLECULES is
Nm v x2 / d
The Kinetic Theory of Gases
The average force received by the wall from the molecules is
F=
I
= Nm v x2 / d
Δt
1
3
3
KEtotal = N ( m v 2 ) = Nk BT = nRT
2
2
2
F
Nm v x2 / d N
P= =
= m v x2
A
d2
V
Because space is isotropic,
PV =
v x2 = v z2 = v y2 = 13 v 2
2
1
N ( mv2 )
3
2
U=
3
nRT
2
vrms =
3 kB T
=
m
3 RT
M
Maxwell velocity distribution N2
⎛ m ⎞
⎟⎟
f (v) = 4π ⎜⎜
⎝ 2πk BT ⎠
But we also know that PV=NkBT. Therefore,
1
3
m v 2 = k BT
2
2
Molecular Interpretation of Temperature
1
3
m v 2 = k BT
2
2
Temperature is a measure of
average molecular speed
3/ 2
v 2e
−
mv 2
2 k BT
For hydrogen molecules at RT,
the rms velocity ~ 1.9 km/s!
vesc =
2GM E
≈ 11.2 km / s
RE
Problems
50. A vertical cylinder of crosssectional area 0.050 m2 is fitted with a
tight-fitting, frictionless piston of mass
5.0 kg (Fig. P10.50). If there is 3.0 mol
of an ideal gas in the cylinder at 500 K,
determine the height h at which the
piston will be in equilibrium under its
own weight.
60. Two small containers of equal volume, 100 cm3, each contain
helium gas at 0°C and 1.00 atm pressure. The two containers are
joined by a small open tube of negligible volume, allowing gas to
flow from one container to the other. What common pressure will
exist in the two containers if the temperature of one container is
raised to 100°C while the other container is kept at 0°C?
Chapter 10 Summary
Zeroth law of thermodynamics.
Temperature scales
Thermal expansion
Ideal gas
Average (translational) kinetic energy per molecule:
proportional to T