# Chapter 17 Solids Phase Changes Thermal Processes.

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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.

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? Boiling 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 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. 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 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. 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 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? 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: Follow-up: How can you cook the potatoes faster? a)Yes b)No c) Wait, I’m confused. Am I still in Denver?

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...

1 cal/(gK) phase change 540 cal/g 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. 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...

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? –L F = 80 cal/g –c water = 1 cal/g ° C a) 0°C b) between 0°C and 50°C c) 50°C d) greater than 50°C

Q = mL f = (1000 g)  (80 cal/g) = 80,000 cal How much heat is needed to melt the ice? Q = mL f = (1000 g)  (80 cal/g) = 80,000 cal °° Q = c water m  ΔT = (1 cal/g °C)  (1000 g)  (50°C) = 50,000 cal How much heat can the water deliver by cooling from 50°C to 0°C? Q = c water m  ΔT = (1 cal/g °C)  (1000 g)  (50°C) = 50,000 cal Thus, there is not enough heat available to melt all the 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? –L F = 80 cal/g –c water = 1 cal/g ° C a) 0°C b) between 0°C and 50°C c) 50°C d) greater than 50°C

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 5 th -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 5 th -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 0 o 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.)

Fusion curve for water Fusion curve for most stuff remember: water is weird: it melts under pressure, and freezes under vacuum, when near the fusion curve Again: explaining why putting the ice/water under vacuum won’t help the root beer chill faster

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 2 ΔP 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).

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. V P Quasi-static = slow enough that system is always effectively in equilibrium area under the curve W = W

zero a) zero -153 J b) -153 J c) -41 J d) -26 J e) 41 J 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: The change in internal energy for the process from C to D is:

Internal Energy zero a) zero -153 J b) -153 J c) -41 J d) -26 J e) 41 J An ideal gas is taken through the four processes shown. The changes in internal energy for three of these processes is as follows: 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 ΔU CD = -41 J

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 V 2 = 5 V 1 and P 2 = 3 P 1. P1P1 V1V1 V 2 =5V 1 P 2 = 3P 1 a) 4 P 1 V 1 b) 7 P 1 V 1 8 c) 8 P 1 V 1 d) 21 P 1 V 1 29 e) 29 P 1 V 1

a) 4 P 1 V 1 b) 7 P 1 V 1 8 c) 8 P 1 V 1 d) 21 P 1 V 1 29 e) 29 P 1 V 1 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 V 2 = 5 V 1 and P 2 = 3 P 1. P1P1 V1V1 V 2 =5V 1 P 2 = 3P 1 Area under the curve: (4 V 1 )(P 1 ) + 1/2 (4V 1 )(2P 1 ) = 8 V 1 P 1

How much internal energy is gained 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 V 2 = 5 V 1 and P 2 = 3 P 1. P1P1 V1V1 V 2 =5V 1 P 2 = 3P 1 a) 7 P 1 V 1 b) 8 P 1 V 1 15 c) 15 P 1 V 1 d) 21 P 1 V 1 e) 29 P 1 V 1

How much internal energy is gained 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 V 2 = 5 V 1 and P 2 = 3 P 1. P1P1 V1V1 V 2 =5V 1 P 2 = 3P 1 a) 7 P 1 V 1 b) 8 P 1 V 1 15 c) 15 P 1 V 1 d) 21 P 1 V 1 e) 29 P 1 V 1 Ideal monatomic gas: U = 3/2 nRT Ideal gas law: PV = nRT P 2 V 2 = 15 P 1 V 1 Δ(PV) = 14 P 1 V 1 U = 3/2 PV ΔU = 21 P 1 V 1

How much heat is gained 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 V 2 = 5 V 1 and P 2 = 3 P 1. P1P1 V1V1 V 2 =5V 1 P 2 = 3P 1 a) 7 P 1 V 1 b) 8 P 1 V 1 15 c) 15 P 1 V 1 d) 21 P 1 V 1 e) 29 P 1 V 1

How much heat is gained 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 V 2 = 5 V 1 and P 2 = 3 P 1. P1P1 V1V1 V 2 =5V 1 P 2 = 3P 1 a) 7 P 1 V 1 b) 8 P 1 V 1 15 c) 15 P 1 V 1 d) 21 P 1 V 1 e) 29 P 1 V 1 First Law of Thermodynamics W = 8 P 1 V 1

at constant pressure a) at constant pressure if the pressure increases in proportion to the volume 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 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?

at constant pressure a) at constant pressure if the pressure increases in proportion to the volume 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 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? 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. U final = 2U initial b) (PV) final = 4 (PV) initial U final = 4U initial 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,

For an ideal gas (from the kinetic theory) First Law of Thermodynamics Constant Volume

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 C p - C v works well for real gases. Both C V and C P 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

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