Honors Physics
CHAPTER 17
Phases and Phase Changes
Teacher: Luiz Izola
Chapter Preview
1. Ideal Gases
2. Kinetic Theory
3. Solids and Elastic Deformation
4. Phase Equilibrium and Evaporation
5. Latent Heats
6. Phase Changes and Energy Conservation
Learning Objectives
 Discuss the phases of matter (solid, liquid, gas) in detail.
 Show that “ideal gas” (one that has no interactions between
its molecules) is a good approximation to real gases.
 Show the relation amongst kinetic theory of gases,
temperature of a substance, and kinetic energy of its molecules.
 Explore the relationship between a force applied to a solid
and the resulting deformation.
 Finally, the behavior of a substance when it changes phases.
Ideal Gases
 Ideal gases are gases where intermolecular interactions are
almost non-existent.
 The pressure of an ideal gas depends on the following:
•
•
•
Temperature (T)
Number of Molecules (N)
Volume (V)
There three combinations:
Constant V, N
Constant V, T
Constant N, T
 P = (constant)T
 P = (constant)N
 P = (constant) / V
Ideal Gases
 Combining the three previous observations, the equation of
state, a relationship between the thermal properties of a
substance, for an ideal gas was defined.
Equation of State for an Ideal Gas
PV = kNT
 K is known as Boltzmann Constant: k = 1.38 x 10-23 J/K
Ex: A person’s lungs might hold 6.0L (1L = 10-3 m3) of air at
body temperature (310K) and atmospheric pressure of (101
kPa). Given that the air is 21% oxygen, find the number of
oxygen molecules in the lung.
Ideal Gases
 Another way to write the ideal gas equation of state is in
terms of the number of moles in a gas – opposed to the number
of molecules, N.
 A mole is the amount of substance that contains as many
elementary entities as there are atoms in 12g of carbon-12.
 The number of atoms in a mole of carbon-12 is know as
Avogadro’s Number (NA).
Avogadro’s Number (NA)
NA = 6.022 x 1023 molecules/mol
Now, if we call n = number of moles in a gas. We can
calculate the number of molecules as:
N = n x NA
Ideal Gases
 Then, we can rewrite the ideal-gas equation of state as:
PV = nNAkT
 The constants NA, k are combined to form the universal gas
constant R.
Universal Gas Constant (R)
R = NA k = (6.022 x 1023 )(1.38 x 10-23 ) = 8.31 J/ (mol.K)
 Thus, another way to write the ideal-gas equation is:
PV = nRT
Ideal Gases
Ex:
How many moles of air are in an inflated basketball?
Assume that the P = 171kPa, the temperature is 293K,
and the diameter of the ball is 30cm.
 A mole has precisely the same number (NA) of particles.
What differs from substance to substance is the mass in each
mole. Therefore, we define atomic or molecular mass (M) as:
“The mass in grams of 1 mole of that substance.”
Notice that, if we measure a mass of copper equal to 63.546g,
we have, in effect, counted out NA atoms of copper. Therefore,
the mass of an individual copper atom, m, is:
m = M / NA
Ideal Gases
Ex: Find the mass of (a) a copper atom, (b) a molecule of
oxygen, O2. Atomic masses are listed in Appendix E.
ISOTHERMS
Robert Boyle established the fact that the pressure of a gas
varies inversely with volume – as long as the temperature and
the number of molecules are held constant. This is known as
Boyle’s Law.
Boyle’s Law
PiVi = PfVf
Ideal Gases
Ex: A cylindrical flask of cross-sectional area A is fitted with an
airtight piston. Contained within the flask is an ideal gas. The
initial pressure is 130kPa and the piston height above the base
of the flask is 25cm. When more mass is added, the pressure
goes to 170kPa. Assuming the temperature is always 290K,
what is the new height of the piston?
Ideal Gases
 Charles’s Law states that the volume of a gas divided by its
temperature is constant, as long as the pressure and the
number of molecules are constant.
Charles’s Law
Vi / Ti = Vf / Tf
Ex: Consider the example from previous slide. In this case the
temperature is changed from an initial value of 290K to a final
value of 330K. The pressure remains constant at 130kPa, and
the initial height of the piston is 25cm. Find the final height of
the piston.
Kinetic Theory of Gases
In kinetic theory, we imagine a gas to be made up of a collection
of molecules moving inside a container of volume V. We
assume the following:
• Container holds a very large number N of identical molecules.
Each molecule has a mass m. and behaves as a point particle.
• The molecules obey Newton’s laws of motion at all times.
• When the molecules hit the walls or collide with one another,
they bounce elastically.
Kinetic Theory of Gases
Imagine a cube container with side L. Its volume is L3. Consider
a molecule moving in the negative direction x towards the wall.
If its initial speed is vx, its initial momentum is pi,x = -mvx. After
bouncing off the wall (elastically), it moves with the same spped
in the + direction, therefore pf,x = mvx. As a result, the
molecule’s change in momentum is:
Δpx = pf,x - pf,x = mvx – (- mvx) = 2mvx
The time required for a 2L roundtrip is: Δt = 2L/vx
By 2nd Newton’s Law, we have:
F = Δpx / Δt = 2mvx / (2L/vx) = mvx2 / L
F = mvx2 / L
Kinetic Theory of Gases
The average pressure exerted by the this wall is simply the
force divided by the area. Since the area is A = L2, we have:
P=F/A
P = mvx2 / V
Kinetic Theory of Gases
Kinetic Energy and Pressure
 The pressure of a gas is proportional to the number of
molecules and inversely proportional to the volume.
 The pressure of a gas is directly proportional to the average
kinetic energy of its molecules.
P = 1/3 (N/V)(2Kavg) = 2/3 (N/V) (1/2mv2)avg
Kinetic Energy and Temperature
Kavg = (1/2mv2) avg = (3/2)kT
Kinetic Theory of Gases
Ex: Find the average kinetic energy of air’s oxygen. Assume the
air is at temperature 210C.
Solving (1/2mv2)avg = (3/2)kT for v, we have a special kind of
velocity called “root mean square” :
Vrms = ( (3kT)/m)1/2
In terms of molecular mass, it becomes:
Vrms = ( (3RT)/M)1/2
Ex: The atmosphere is composed of N2 (78%), O2 (21%). (a) Is
the rms of N2 (28g/mol) >, <, or equal the rms of O2 (32g/mol)?
(b) Find the rms of N2 and O2 at 293K.
Kinetic Theory of Gases
The internal Energy of an Ideal Gas:
It is equals to the sum of all its potential and kinetic energies. As
we know, in an ideal gas, there is no molecular interactions,
other than perfectly elastic collisions; hence, there is no
potential energy.
The total energy of the system is the “sum of kinetic energy of
each one of its molecules”.
Internal Energy of a Monatomic Ideal Gas (U):
U = 3/2(nRT)
(Joules)
Ex: A basketball at 290K holds 0.95 mo; of air molecules. What
is the internal energy in the ball?
Solids and Elastic Deformation
 The shape of a solid can be changed if it is acted by a force.
Changing the Length of a Solid
 Pulling a solid rod with a force F, for example, will cause a
“intermolecular spring” in the direction of the force to expand by
an amount proportional to the force F.
 The stretch ΔL is proportional to the initial length L0. Also, the
amount of stretch for a given force F is inversely proportional to
the cross-sectional area A of the rod.
 The proportionality constant Υ, called Young’s Modulus, is
related to each specific material. See table 17.1, pg: 546
Solids and Elastic Deformation
Changing the Length of a Solid
 Based on the previous observations, the calculation of such
force and the resulting change in length is given by:
F = Υ (ΔL/ L0)A
Solids and Elastic Deformation
Changing the Length of a Solid
There is a straightforward connection between the previous
formula and Hooke’s Law. Notice that the force required to
cause a certain stretch is proportional to the stretch – just as in
Hooke’s Law. Therefore we can write their relation as:
F = (Υ.A / L0).ΔL = kx
Ex1: A person carries a 21-kg suitcase in one hand. Assuming
the upper arm bone supports the entire weight of the case, find
the amount by which it stretches. (The upper arm is about 33cm
with a cross-section = 5.2x10-4m2).
Ex2: A rock climber hangs freely from a nylon rope that is 12m
long and has a diameter of 5.5mm. If the rope stretches 4.7cm,
what is the mass of the climber?
Solids and Elastic Deformation
Changing the Shape of a Solid
 This kind of deformation is referred as shear deformation.
Consider the figure below. A force F is applied to the right of the
top cover and static friction applies a force to the left of the
bottom cover. The result is that the book remains at rest but
becomes slanted by the amount Δx. Table 17-2, page 548 has
the shear modulus of some materials.
Solids and Elastic Deformation
Changing the Shape of a Solid
 The force required to cause a given amount of slant is
proportional to Δx, inversely proportional to the thickness of the
book L0, and proportional to the surface area A of the book’s
cover. Defining the formula, we have:
F = S.(Δx/ L0).A
The constant of proportionality is the shear modulus (S).
Be aware of the differences between Young’s and Shear
modulus. The term L0 in Young’s refers to the solid’s length in
the direction of the applied force. In shear, refers to the
thickness of the of the solid measured in a direction
perpendicular to the applied force. What about the area?
Solids and Elastic Deformation
Changing the Shape of a Solid
Ex1: A horizontal force of 1.2N is applied to the top of a stack of
pancakes 13cm in diameter and 9cm high. The result is a shear
deformation of 2.5cm. What is the shear modulus of these
pancakes?
Ex2: A lead brick (see below) rests on a solid surface. A 2400N
force is applied. (a) What is the height change? (b) What is the
shear deformation?
Solids and Elastic Deformation
Changing the Volume of a Solid
Figure below shows a spherical solid whose volume decreases
by the amount Δx when the pressure acting on it increases by
ΔP. Experiments show that the pressure difference required to
cause a change in volume ,Δv, is proportional to and inversely
proportional to the initial volume V0.
Solids and Elastic Deformation
Changing the Volume of a Solid
ΔP = -B(ΔP/V0)
 The constant of proportionality is called bulk modulus (B).
 Since B is positive, we have to add a minus sign in front of
the equation. If the pressure increases (ΔP > 0), the volume will
decrease (ΔV < 0) and the final quantity will be positive.
 Table 17-3, page 549, gives you a list of Bulk Modulus for
some materials.
Solids and Elastic Deformation
Changing the Volume of a Solid
Ex1: A gold doubloon ( cylinder shape) 6.1cm in diameter and
2mm thick is dropped over the side of a pirate ship. When it
comes to rest on the ocean floor at a depth of 770m, how much
has its volume changed?
Ex2: The deepest place in all the oceans is the Marianas
Trench, where the depth is 10.9km and the pressure is
1.10x108Pa. If a copper ball 10.0cm in diameter is taken to the
bottom of the trench, by how much does its volume decrease?
Ex3: In 1934, Charles Beebe broke the diving record. He went
923m below the surface. He dove in a bathysphere, a steel
sphere 4.75ft in diameter. How much did the volume of the
sphere change at the record depth?