Lecture 25 Phases & Phase Changes Thermal Processes

Lecture 25Phases & Phase Changes

Thermal Processes

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Solids

Solids and Elastic Deformation

Solids have definite shapes (unlike fluids), but they can be deformed. Pulling on opposite ends of a rod can cause it to stretch:

Stretching / Compression of a SolidThe amount of stretching will depend on the force; Y is Young’s modulus and is a property of the material:

The stretch is proportional to the force, and also to the original length

The same formula works for stretching or compression (but sometimes with a different Young’s modulus)

Shear ForcesAnother type of deformation is called a shear deformation, where opposite sides of the object are pulled laterally in opposite directions.

The “lean” is proportional to the force, and also to the original height

Shear Modulus

S is the shear modulus.

Uniform Compression

Here, the proportionality constant, B, is called the bulk modulus.

Under uniform pressure, an object will shrink in volume

Stress and Strain

The applied force per unit area is called the stress, and the resulting deformation is the strain. They are proportional to each other until the stress becomes too large; permanent deformation will then occur.

Phase Changes

EvaporationMolecules in a liquid can sometimes

escape the binding forces and become vapor (gas)

Phase Equilibrium

If a liquid is put into a sealed container so that there is a vacuum above it, some of the molecules in the liquid will vaporize. Once a sufficient number have done so, some will begin to condense back into the liquid. Equilibrium is reached when the numbers in each phase remain constant.

Vapor PressureThe pressure of the gas when it is in equilibrium with the liquid is called the equilibrium vapor pressure, and will depend on the temperature.

A liquid boils at the temperature at which its vapor pressure equals the external pressure.

a) Charlottesville

b) Denver (the “mile high” city)

c) the same in both places

d) I’ve never cooked in Denver, so I really don’t know

e) you can boil potatoes?

Boiling PotatoesBoiling Potatoes

Will boiled potatoes cook faster in Charlottesville or in Denver?

a) Charlottesville

b) Denver (the “mile high” city)

c) the same in both places

d) I’ve never cooked in Denver, so I really don’t know

e) you can boil potatoes?


Will boiled potatoes cook faster in Charlottesville or in Denver?

The lower air pressure in Denver means that the water will boil at a lower temperature... and your potatoes will take longer to cook.

Phase Diagram

The vapor pressure curve is only a part of the phase diagram.

When the liquid reaches the critical point, there is no longer a distinction between

liquid and gas; there is only a “fluid” phase.

There are similar curves describing the pressure/temperature of transition from solid to liquid, and solid to gas

Fusion CurveThe fusion curve is the boundary between the solid and liquid phases; along that curve they exist in equilibrium with each other.

One of these two fusion curves has a shape that is typical for most materials, but the other has shape specific to water.

Which is which?

(a) Curve 1 is the fusion curve for water

(b) Curve 2 is the fusion curve for water

(c) Trick question: there is no fusion curve for water!

Curve 1

Curve 2

Fusion CurveThe fusion curve is the boundary between the solid and liquid phases; along that curve they exist in equilibrium with each other.

One of these two fusion curves has a shape that is typical for most materials, but the other has shape specific to water.

Which is which?

(a) Curve 1 is the fusion curve for water

(b) Curve 2 is the fusion curve for water

(c) Trick question: there is no fusion curve for water!

Curve 1

Curve 2

Fusion curve for water

Ice melts under pressure!This is how an ice skate works

Phase EquilibriumThe sublimation curve marks the boundary between the solid and gas phases.

The triple point is where all three phases are in equilibrium.

Heat and Phase ChangeWhen two phases coexist, the temperature remains the same even if a small amount of heat is added. Instead of raising the temperature, the heat goes into changing the phase of the material – melting ice, for example.

Latent HeatThe heat required to convert from one phase to another is called the latent heat.

The latent heat, L, is the heat that must be added to or removed from one kilogram of a substance to convert it from one phase to another. During the conversion process, the temperature of the system remains constant.

Latent HeatThe latent heat of fusion is the heat needed to go from solid to liquid;

the latent heat of vaporization from liquid to gas.

You’re in Hot Water!You’re in Hot Water!

Which will cause more severe

burns to your skin: 100°C water or

100°C steam?

a) water

b) steam

c) both the same

d) it depends...

Although the water is indeed hot, it releases only 1 cal/1 cal/gg of

heat as it cools. The steam, however, first has to undergo a

phase changephase change into water and that process releases 540 cal/g540 cal/g,

which is a very large amount of heat. That immense release of

heat is what makes steam burns so dangerous.

You’re in Hot Water!You’re in Hot Water!

Which will cause more severe

burns to your skin: 100°C water or

100°C steam?

a) water

b) steam

c) both the same

d) it depends...

a) Yes

b) No

c) Wait, I’m confused. Am I still in Denver?


Will potatoes cook faster if the water is boiling faster?


Will potatoes cook faster if the water is boiling faster?

The water boils at 100°C and remains at that temperature until allof the water has been changed into steam. Only then will the steam increase in temperature. Because the water stays at the same temperature, regardless of how fast it is boiling, thepotatoes will not cook any faster.

Follow-upFollow-up: : How can you cook the potatoes faster?

a) Yes

b) No

c) Wait, I’m confused. Am I still in Denver?

Phase Changes and Energy Conservation

Solving problems involving phase changes is similar to solving problems involving heat transfer, except that the latent heat must be included as well.

Water and Ice Water and Ice

You put 1 kg of ice at 0°C

together with 1 kg of water

at 50°C. What is the final

temperature?

LF = 80 cal/gcwater = 1 cal/g °C

a) 0°C

b) between 0°C and 50°C

c) 50°C

d) greater than 50°C

How much heat is needed to melt the ice?

QQ = = mLmLff = (1000 = (1000 gg) ) (80 cal/ (80 cal/gg) = 80,000 cal) = 80,000 cal

How much heat can the water deliver by cooling from 50°°C to 0°°C?

QQ = = ccwaterwater m xm xTT = (1 cal/ = (1 cal/gg °°C) C) (1000 (1000 gg) ) (50 (50°°C) = 50,000 calC) = 50,000 cal

Thus, there is not enough heat available to melt all the ice!!

Water and Ice Water and Ice

You put 1 kg of ice at 0°C

together with 1 kg of water

at 50°C. What is the final

temperature?

LF = 80 cal/gcwater = 1 cal/g °C

a) 0°C

b) between 0°C and 50°C

c) 50°C

d) greater than 50°C

Follow-upFollow-up: : How much more water at 50°C would you need?

a) Add more ice to the icewater

b) add salt to the icewater

c) hold the icewater in an evacuated chamber (vacuum)

d) Jump in the car and drive to a nearby convenience store

Ice Cold Root BeerIce Cold Root Beer

You have neglected to chill root beer for your son’s 5th-birthday party. You submerge the cans in a bath of ice and water as you start dinner. How can you hurry the cooling process?

a) Add more ice to the icewater

b) add salt to the icewater

c) hold the icewater in an evacuated chamber (vacuum)

d) Jump in the car and drive to a nearby convenience store

Ice Cold Root BeerIce Cold Root BeerYou have neglected to chill root beer for your son’s 5th-birthday party. You submerge the cans in a bath of ice and water as you start dinner. How can you hurry the cooling process?

Not a), because ice water at 1 atm is zero degrees, no matter the proportion of water and ice

Not c), because ice is less dense than water so you will raise the melting point when you reduce the pressure. This will allow the water to get a little warmer than 0o

Not d), because you’ll forget your wallet and it will end up taking more time

b) because salt interferes with the formation of ice. This barrier to the solid phase lowers the fusion temperature, and so reduces the temperature of the ice water

As pressure goes lower, the ice/water mixture will ride the fusion curve from point 1 to point 2.This implies that temperature goes up.

Fusion curve for water

1

2ΔP

ΔT

The larger ΔT, the more heat transfers per unit time. Thus, the colder the ice bath, the faster the root beer will chill, and the warmer the bath, the slower the root beer will chill

When two states exist in the same system (like, ice and water), the system MUST be on the equilibrium curve (in the case, the fusion curve).

Thermal Processes

The Zeroth Law of ThermodynamicsIf object A is in thermal equilibrium with object B, and object C is also in thermal equilibrium with object B, then objects A and C will be in thermal equilibrium if brought into thermal contact.

That is, temperature is the only factor that determines whether two objects in thermal contact are in thermal equilibrium or not.

Object B can then be a thermometer, providing a scale to compare objects:

Temperature

Kinetic Energy and Temperature

Comparing pressure in the kinetic theory (monatomic ideal gas) with the ideal gas law allows us to relate

average kinetic energy and temperature

Internal Energy

In the case where each molecule consists of a single atom, this is all linear kinetic energy of atoms:

The internal energy of an ideal monatomic gas is the sum of the kinetic energies of all its

molecules.

Conservation of EnergyIf a system does work on the external world, and no heat is added, its internal energy decreases.

Internal energy changes with heat input

If heat is added to a system, this is an increase in internal energy.

Assuming constant volume (so W = 0):

The First Law of Thermodynamics

The change in a system’s internal energy is related to the heat Q and the work W by conservation of

energy:

It is vital to keep track of the signs of Q and W.

Combining these gives the first law of thermodynamics.

The First Law of ThermodynamicsJogger is warm: heat transfer to the environment

She is doing work on the environment (force*distance)

Internal energy is decreasing

The First Law of Thermodynamics

State function of a system depend only on the state of the system (temperature, pressure, etc), not on how a system arrived in that state.

The internal energy of a system depends only on its temperature. It is a state function.

The work done and heat added are specific to a process.

There is no “work” or “heat” in a system... those are just terms to describe the change in internal energy.

Boiling water:When 1 g of water boils at 100o C under 1 Atm. The volume of the steam at 100o C is 1671 cm3. Find the work done in the expansion and calculate the change in internal energy of the system

Lv = 22.6 x 105 J/kg

Boiling water:When 1 g of water boils at 100o C under 1 Atm. The volume of the steam at 100o C is 1671 cm3. Find the work done in the expansion and calculate the change in internal energy of the system

Lv = 22.6 x 105 J/kg

Q = 0.001 kg x 22.6 x 105 J/kg = 2260 J

W = (101 x 103 N/m2) x (1671 cm3 -1 cm3)x(10-6 m3 /cm3) = 169 J

ΔU = Q - W = 2091 J

Reversible Thermal Processes

We also assume they are reversible (frictionless pistons, etc.):

For a process to be reversible, it must be possible to return both the system and its surroundings to the same states

they were in before the process began.

We will discuss 4 idealized processes with Ideal Gases:•Constant Pressure•Constant Volume•Constant Temperature•Q= 0 (adiabatic)

We will assume that all processes we discuss are “quasi-static” – they are slow enough that the system is always “in equilibrium” (fluid volumes have the same temperature throughout, etc.)

Constant pressureWork done by an expanding gas, constant pressure:

Work is area under the PV graph

Isobaric process

so changing volume implies changing temperature

Examples: piston against atmosphere, or vertical piston with constant weight on top

Work is area under the PV graph

imagining any general process as approximated by a number of constant pressure processes:

Constant VolumeIf the volume stays constant, nothing moves and no work is done.

so changing pressure implies changing temperature

Change in internal energy is related only to the net heat input

Isovolumetric process

Constant Temperature

If the temperature is constant, the pressure varies inversely with the volume.

Isothermal processes

Constant TemperatureA system connected to a large heat reservoir is usually thought to be held at constant temperature. Volume can change, pressure can change, but the temperature remains that of the reservoir.

if W < 0 (work done on the system)than Q<0 (heat flows out of the system)

if W > 0 (work done by the system)than Q>0 (heat flows out into the system)

T = constant W = Q

Work in an Constant-Temperature Process

The work done is the area under the curve:

For you calculus junkies:

Adiabatic ProcessAn adiabatic process is one in which no heat flows into or out of the system.

One way to ensure that a process is adiabatic is to insulate the system.

The adiabatic P-V curve is similar to the

isothermal one, but is steeper.

Q = 0

Another way to ensure that a process is effectively adiabatic is to have the volume change occur very quickly.

In this case, heat has no time to flow in or out of the system.

Rapid Adiabatic Process

Thermal Processes

The different types of ideal thermal processes

Specific Heat for an Ideal Gas at Constant Volume

Specific heats for ideal gases must be quoted either at constant pressure or at constant volume. For a constant-volume process,

for an ideal gas (from the kinetic theory)First Law of Thermodynamics

Specific Heat for an Ideal Gas at Constant Pressure

At constant pressure, (some work is done)

Some of the heat energy goes into the mechanical work, so more heat input is required to produce the same ΔT

for an ideal gas (from the kinetic theory)First Law of Thermodynamics

Specific Heats for an Ideal Gas

Although this calculation was done for an ideal, monatomic gas, the difference Cp - Cv works well for real gases.

Both CV and CP can be calculated for a monatomic ideal gas using the first law of thermodynamics.

Specific Heats and Adiabats In Ideal Gas

The P-V curve for an adiabat is given by

for monotonic gases

Work of a Thermal CycleWork of a Thermal Cycle

In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is:

a) positive

b) zero

c) negative

V

P

In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is:

a) positive

b) zero

c) negative

The gas expands at a higher pressure

and compresses at a lower pressure.

In general, clockwise = positive work;

counterclockwise = negative work.

V

P

Work of a Thermal CycleWork of a Thermal Cycle

How much work is done by the gas in this process, in terms of the initial pressure and volume?

One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1.

P1

V1 V2 =5V1

P2 = 3P1

a) 4 P1V1

b) 7 P1V1

c) 8 8 P1V1

d) 21 P1V1

e) 29 29 P1V1

a) 4 P1V1

b) 7 P1V1

c) 8 8 P1V1

d) 21 P1V1

e) 29 29 P1V1

How much work is done by the gas in this process, in terms of the initial pressure and volume?


P1

V1 V2 =5V1

P2 = 3P1

Area under the curve:(4 V1)(P1) + 1/2 (4V1)(2P1) = 8 V1P1

How much internal energy is gained by the gas in this process, in terms of the initial pressure and volume?


P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

How much internal energy is gained by the gas in this process, in terms of the initial pressure and volume?


P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

Ideal monatomic gas: U = 3/2 nRT

Ideal gas law: PV = nRT

P2V2 = 15 P1V1

Δ(PV) = 14 P1V1

U = 3/2 PV

ΔU = 21 P1V1

How much heat is gained by the gas in this process, in terms of the initial pressure and volume?


P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

How much heat is gained by the gas in this process, in terms of the initial pressure and volume?


P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

First Law of Thermodynamics

W = 8 P1V1

The Second Law of Thermodynamics

We observe that heat always flows spontaneously from a warmer object to a cooler one, although the opposite would not violate the conservation of energy.

This direction of heat flow is one of the ways of expressing the second law of thermodynamics:When objects of different temperatures are

brought into thermal contact, the spontaneous flow of heat that results is

always from the high temperature object to the low temperature object. Spontaneous heat flow never proceeds in the reverse

direction.

Heat EnginesA heat engine is a device that converts heat into work. A classic example is the steam engine. Fuel heats the water; the vapor expands and does work against the piston; the vapor condenses back into water again and the cycle repeats.

All heat engines have: a working substance a high-temperature reservoir a low-temperature reservoir a cyclical engine

PHYS2010 practice for Chap14-18, Fall 2009

12) An ideal monatomic gas undergoes a reverrsible expansion to 1.5 times its original

volume. In which of these processes does the gas perform the least amount of work?

A) at constant temperature

B) at constant pressure

C) if the pressure decreases in proportion to the volume (i.e. PV=constant)

D) adiabatically

E) if the pressure increases in proportion to the volume (i.e., P/V=constant)

suggested time: 1-2 minutes

Provided in lecture

notes on: 12/1


13) An expandable container holds 1.50 mole of He gas with an initial pressure of 650

kPa and an initial volume of 2.10 L. The gas expands isothermally to a final pressure

of 350 kPa. How much heat is gained by the gas in this process?

A) 1370 J

B) 685 J

C) 792 J

D) 1280 J

E) 1700 J

suggested time: 4 minutes

Provided in lecture

notes on: 12/1


14) 1.50 moles of an ideal monatomic gas are initially at a temperature of 317 K. If the

gas gains 2670 J of heat and performs 770 J of work, what is its final temperature?

A) 419 K

B) 756 K

C) 687 K

D) 359 K

E) 526 K

suggested time: 3-4 minutes

Provided in lecture

notes on: 12/1

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Lecture 25 Phases & Phase Changes Thermal Processes