Chapter 17
Solids
Phase Changes
Thermal Processes
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Vapor Pressure
The 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.
Boiling 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?
Boiling 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?
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.
There are similar
curves describing the
pressure/temperature
of transition from
solid to liquid,
and solid to gas
When the liquid reaches the critical point,
there is no longer a distinction between liquid
and gas; there is only a “fluid” phase.
Fusion Curve
The 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.
Curve 1
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 2
Fusion Curve
The 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.
Curve 1
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 2
Ice melts under pressure!
This is how an ice skate works
Fusion curve for water
Phase Equilibrium
The 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 Change
When 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 Heat
The 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 Heat
The latent heat of fusion is the heat needed to go
from solid to liquid;
the latent heat of vaporization from liquid to gas.
Boiling Potatoes
Will potatoes cook faster if the
water is boiling faster?
a) Yes
b) No
c) Wait, I’m confused. Am
I still in Denver?
Boiling Potatoes
Will potatoes cook faster if the
water is boiling faster?
a)
Yes
b)
No
c)
Wait, I’m confused.
Am I still in Denver?
The water boils at 100°C and remains at that temperature until all
of 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, the
potatoes will not cook any faster.
Follow-up: How can you cook the potatoes faster?
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...
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/(gK) of heat
as it cools. The steam, however, first has to undergo a phase
change into water and that process releases 540 cal/g, which is a
very large amount of heat. That immense release of heat is what
makes steam burns so dangerous.
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
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/g
– cwater = 1 cal/g °C
a) 0°C
b) between 0°C and 50°C
c) 50°C
d) greater than 50°C
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/g
a) 0°C
b) between 0°C and 50°C
c) 50°C
d) greater than 50°C
– cwater = 1 cal/g °C
How much heat is needed to melt the ice?
Q = mLf = (1000 g)  (80 cal/g) = 80,000 cal
How much heat can the water deliver by cooling from 50°C to 0°C?
Q = cwater m T = (1 cal/g °C)  (1000 g)  (50°C) = 50,000 cal
Thus, there is not enough heat available to melt all the ice!!
Ice 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 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
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. (This is why you salt the sidewalk in winter.)
Again: explaining why putting the ice/water under
vacuum won’t help the root beer chill faster
Fusion curve for
most stuff
remember: water is weird: it melts
under pressure, and freezes under
vacuum, when near the fusion curve
Fusion curve for water
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
1
ΔP
2
ΔT
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).
Fusion curve for 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.
Chapter 18
The Laws of
Thermodynamics
Reversible (frictionless pistons, etc.) and quasi-static processes
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.
Quasi-static = slow enough that system
is always effectively in equilibrium
P
area under
W=
the curve
V
W
Internal Energy
An ideal gas is taken through
the four processes shown. The
changes in internal energy for
three of these processes is as
follows:
a) zero
b) -153 J
c) -41 J
d) -26 J
e) 41 J
The change in internal energy
for the process from C to D is:
Internal Energy
An ideal gas is taken through
the four processes shown. The
changes in internal energy for
three of these processes is as
follows:
a) zero
b) -153 J
c) -41 J
d) -26 J
e) 41 J
The change in internal energy
for the process from C to D is:
PV = nRT
so in a PV cycle, ΔT = 0
ΔT = 0 means that ΔU = 0
ΔUCD = -41 J
a) 4 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much work is done by the gas in this
process, in terms of the initial pressure and
volume?
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 4 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much work is done by the gas in this
process, in terms of the initial pressure and
volume?
Area under the curve:
(4 V1)(P1) + 1/2 (4V1)(2P1)
= 8 V1P1
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much internal energy is gained by the
gas in this process, in terms of the initial
pressure and volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much internal energy is gained by the
gas in this process, in terms of the initial
pressure and volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
Ideal monatomic gas: U = 3/2 nRT
Ideal gas law: PV = nRT
U = 3/2 PV
P2V2 = 15 P1V1
Δ(PV) = 14 P1V1
P2 = 3P1
P1
ΔU = 21 P1V1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much heat is gained by the gas in this
process, in terms of the initial pressure and
volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much heat is gained by the gas in this
process, in terms of the initial pressure and
volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
First Law of Thermodynamics
W = 8 P1V1
P2 = 3P1
P1
V1
V2 =5V1
Internal Energy
a) at constant pressure
An ideal gas undergoes a
reversible expansion to 2 times
its original volume. In which of
these processes does the gas
have the largest loss of internal
energy?
b) if the pressure increases in
proportion to the volume
c) if the pressure decreases in
proportion to the volume
d) at constant temperature
e) adiabatically
Internal Energy
a) at constant pressure
An ideal gas undergoes a
reversible expansion to 2 times
its original volume. In which of
these processes does the gas
have the largest loss of internal
energy?
b) if the pressure increases in
proportion to the volume
c) if the pressure decreases in
proportion to the volume
d) at constant temperature
e) adiabatically
Since U = 3/2 nRT, and PV=nRT, the largest loss in internal energy
corresponds to the largest drop in temperature, and so the largest
drop in the product PV.
a) PV doubles. Ufinal = 2Uinitial
b) (PV)final = 4 (PV)initial Ufinal = 4Uinitial
c) PV is constant, so U is constant
d) U is constant
e) Adiabatic, so ΔU = -W. This is the only process which reduces U!
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,
First Law of Thermodynamics
Constant Volume
For an ideal gas (from
the kinetic theory)
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
First Law of Thermodynamics
For an ideal gas (from
the kinetic theory)
Specific Heats for an Ideal Gas
Both CV and CP can
be calculated for a
monatomic ideal gas
using the first law of
thermodynamics.
Although this calculation was
done for an ideal, monatomic gas,
the difference Cp - Cv works well
for real gases.
Specific Heats and Adiabats In Ideal Gas
The P-V curve for an adiabat is
given by
for
monotonic
gases
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• Regular office hours Thursday
• Office hours from 9:00-3:30 Friday