p = (N/3)
mv 2
l3
(12.15)
This equation rearranges to
pV = N mv 2 /3
(12.16)
because volume V is the cube of the length l. The form of the ideal gas law given above
shows the pressurevolume product is directly proportional to the meansquare velocity of
L
L

+

Figure 12.3: Left end of bar is xed as length changes
Figure 12.4: dashed line represents plastic recovery *incomplete*
stretches a the material by constant amounts and the corresponding stress is measured and
plotted. Typical solid metal bars w
12.2.1
Solids, liquids, gasses
12.2.2
Pressure in uids
12.2.3
change of phase
12.3
Deformation of solids
12.3.1
strain, stress
Stress () and strain ( ) is one of the most fundamental concepts used in the mechanics of
materials. The concept can be easily i
Chapter 12
Heat and Properties of Matter
12.1
Phases of matter
12.1.1
Density
Matter is a substance which has mass and occupies space. The density of matter refers to how
much mass is in a given volume. Said dierently, you can imagine the density to be th
M ass =
1.35N
= .138kg
9.8m/s2
Using the mass and the volume we determine the density of the soap:
density =
138g
= .92g/cm3
150cm3
Water has a density of 1g/cm3 , therefore the soap is less dense than water, allowing
it to oat.
12.2
Phases of matter
Alth
5cm
2cm
5cm
3cm
3cm
3cm
Now that you know the density of aluminum and lead, which object would be bigger
(larger volume): 1kg of Lead or 1kg of Aluminum.
Solution:
1kg of aluminum will be much larger in volume than 1kg of lead. Aluminum has a
smaller dens
Sometimes the design of the support is such that the support needs to withstand the rock
mass without the force breaking the roof.
Therefore hydraulic supports (hydro = water) use the principles of force and pressure such
that as a force is exerted on the
Metastable states may sometimes be considered as phases, although strictly speaking they
arent because they are unstable. For example, each polymorph of a given substance is usually
only stable over a specic range of conditions. For example, diamond is on
Essay 3 : Pressure and Forces
Author: Asogan Moodaly
Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering
from the University of Natal, Durban in South Africa. For his nal year design project he worked on
a 3axi
Get a plastic container lid (or anything waterproof) swing it around in air and then try to
swing it around under water. The density of the water is much larger than the air, making you
have to work harder at swinging the lid in water. This is why boats a
Material
Aluminium
Water
Ethyl alcohol
Methyl ether
Nitrogen
Helium
Temperature in Celsius
2467 C
100 C
78.5 C
25 C
195.8 C
268.9 C
Temperature in Kelvin
2740 K
373.15 K
351.6 K
248 K
77.3 K
4.2 K
Table 12.1: Boiling points for various materials in deg
+


Figure 12.5: dashed line represents plastic recovery *incomplete*
12.4
Ideal gasses
Author: Grald Wigger
e
Grald Wigger started his Physics studies at ETH in Zuerich, Switzerland. He moved to Cape
e
Town, South Africa, for his Bachelor of Science de
12.4.3
Pressure of a gas
In the kineticmolecular theory of gases, pressure is the force exerted against the wall of a
container by the continual collision of molecules against it. From Newtons second law of motion,
the force exerted on a wall by a single
According to this expression, the mean free path of the molecules should get longer as the
temperature increases; as the pressure decreases; and as the size of the molecules decreases.
Worked Example 69
mean free path
Calculate the length of the mean free
Calculate the rootmeansquare velocity of oxygen molecules at room temperature,
25 C.
Solution:
Using
vrms =
3RT /Mmol ,
the molar mass of molecular oxygen is 31.9998 g/mol; the molar gas constant has
the value 8.3143 J/mol K, and the temperature is 298.
V = 850mL = 0.850L = 0.850dm3
P = 746torr/760torr = 0.982atm
T = 25.0 C + 273.15 = 298.15K
pV = nRT
(0.982atm)(0.850L) = (n)(0.0821Latmmol 1K 1)(298.15K)
n = 0.0341mol
molecular mass = g/mol = 1.090 g/ 0.0341 mol = 31.96 g/mol. The gas is oxygen.
Or the e
The kineticmolecular theory of gases is a theory of great explanatory power. We shall see
how it explains the ideal gas law, which includes the laws of Boyle and of Charles; Daltons law
of partial pressures; and the law of combining volumes.
The kinetic
(p2 ) before it bursts, causing a buildingattening explosion. What is the maximum
temperature the cylinder can withstand before bursting?
Plugging in the known variables into the expression for the GayLussac law yields
T2 = p2 /p1 T1 = 500atm/50atm 300K
p1 /T1 = p2 /T2 = constant
p1,2
T1,2
: pressures (P a)
: Temperatures (K)
That means, that pressure divided by temperature is a constant. On the other hand, if we
plot pressure versus temperature, the graph crosses 0 pressure for T = 0 K = 273.15 C as sh
where p = pressure, V = volume, n = number of mols, T = kelvin temperature and R the
ideal gas constant.
The ideal gas constant R in this equation is known as the universal gas constant. It arises from
a combination of the proportionality constants in the
Figure 12.6: PressureVolume diagram for the ideal gas at constant temperature.
Worked Example 62
compressed Helium gas
A sample of Helium gas at 25 C is compressed from 200 cm3 to 0.240 cm3 . Its
pressure is now 3.00 cm Hg. What was the original pressure
3.60 103 Hg 1 atm/76.0 cm Hg = 4.74 105 atm
One way to experience this is to dive under water. There is air in your middle ear, which
is normally at one atmosphere of pressure to balance the air outside your ear drum. The water
will put pressure on the ea
Note:
The surface is taken to mean the cloud tops of the gas giants (Jupiter, Saturn, Uranus
and Neptune) in the above table.
Worked Example 59
Comparative Problem 2
Question: On Earth a man weighs 70kg. On the planet Beeble how much will he
weigh if Beeb
Step 1 : We start with the situation on Earth
mE m
r2
Step 2 : Now we consider the situation on Zirgon
W = mg = G
(10.17)
mZ m
(10.18)
2
rZ
Step 3 : Relation between conditions on Earth and Zirgon
but we know that mZ = 2mE and we know that rZ = r so we co
E = 800 J
1
, E
p1 k1
2
, E
p2 k2
Step 2 : N
ow we know that in an explosion, total kinetic energy is not conserved. There is a
denite change in shape of the exploding object! But we can always use momentum
conservation to solve the problem. So:
pBef ore
After the collision:
2
, E
p2 k2
1
, E
p1 k1
Step 2 : Convert all units into S.I. units
m1
=
5000 g
m1
=
5 kg
Step 3 : Decide which equations to use in the problem
Now we know that in an explosion, total kinetic energy is not conserved. There is a
denite
After the explosion, the can is completely destroyed. But momentum is always conserved, so:
pBef ore
pBef ore
0
= ter
pAf
= +
p1 p2 p3 p4
+
= p
p
p
p
1
2
3
4
However, the kinetic energy of the system is not conserved. The cans shape was changed in
knocked people o their feet and broke windows up to 650 km away (the same as the
distance from Bloemfontein to Durban!). The explosion registered on seismic stations
across Europe and Asia, and produced uctuations in atmospheric pressure strong enough
to
T he Zeeman Effect
R obert DeSerio
P h 461
O bjective: I n t his e xperiment, you will m easure t he s mall energy shifts i n
t he m agnetic sublevels o f a toms i n "weak" m agnetic fields. T he visible light
f rom t ransitions b etween various multiplet