Honors Physics CHAPTER 17 Phases and Phase Changes Teacher: Luiz Izola

Honors Physics

CHAPTER 17

Phases and Phase Changes

Teacher: Luiz Izola

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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 P = (constant)T

Constant V, T P = (constant)N

Constant N, T 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.


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


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 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


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.


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



Based on the previous observations, the calculation of such force and the resulting change in length is given by:

F = Υ (ΔL/ L0)A



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?


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.



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?



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?


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.



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.



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?

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Honors Physics CHAPTER 17 Phases and Phase Changes Teacher: Luiz Izola