Unit 3: Phases of Matter-key Regents hemistry 14 Mr ... · Unit 3: Phases of Matter-key Regents...

Unit 3: Phases of Matter-key Regents Chemistry ’14-‘15 Mr. Murdoch

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Unit 3: Phases of Matter

Student Name: _______________Key_________________

Class Period: ________



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Unit 3 Vocabulary:

1. Avogadro’s Hypothesis: Equal volumes of two ideal gases under the same conditions of temperature and pressure will contain equal number of molecules.

2. Boiling: the transition of a liquid into a gas at the vaporization point. 3. Condensing: the transition of a gas into a liquid at the vaporization

point. 4. Equilibrium: the condition that exists when the rates of two

opposing phase changes are equal. 5. Evaporating: the transition of the surface molecules of a liquid into a

gas below the boiling point. 6. Freezing: the transition of a liquid into a solid at the solidification

point. 7. Gas: the phase of matter with complete dissociation of particles of

matter with a great average distance between the particles compared to the particle size. Negligible attractive forces between particles.

8. Heat of Fusion: energy required to liquefy a gram of solid at melting point.

9. Heat of vaporization: energy required to gasify a gram of liquid at its vaporization point.

10. Ideal Gas: a gas in which the molecules are infinitely small and far apart, the molecules travel with straight-line motion, all collisions are elastic (no net energy loss), no attractive forces between gas molecules and speed of molecules is directly proportional to Kelvin. Gasses act most ideal at high temperature and low pressures.

11. Liquid: a phase of matter characterized as loosely organized yet held together by intermolecular or ionic attractive forces.

12. Melting: the transition of a solid into a liquid at the liquefaction point.

13. Pressure: Force exerted over an area.



Unit 2 Vocabulary (Cont’d)

14. Solid: a phase with matter arranged in regular geometric patterns

(lattices) with only vibrational motion, and no relative motion. 15. Specific Heat: the energy required to heat one gram of a substance

one Kelvin. 16. Sublimation: the transition of a solid directly into a gas, skipping

liquefaction. 17. Vapor Pressure: the pressure exerted by vapor in a vapor-liquid

mixture closed system at equilibrium. 18. Vapor-Liquid Equilibrium: a system where the rate of evaporation

equals the rate of condensing.



Unit 3 Homework Assignments:

Assignment: Date: Due:



Positive and negative charges attract each other. Attractive forces

between molecules are called intermolecular attractive forces (IMAF).

The strength of these forces determines what phase of matter a

substance is in at a given temperature. Substances having weak

attractive forces (London dispersion) between their molecules tend to

be gasses at room temperature, and substances with strong attractive

forces (ionic) tend to be solids at room temperature. Phases of matter

are simply stages of attraction. Gases are composed of molecules with

little or no attractive forces, allowing the molecules to fly freely past

each other. Liquids are made of molecules with stronger attractive

forces, allowing molecules to flow past each other, but still stay

together. Solids are made of molecules or ions with strong attractive

forces, which lock the molecules into a crystal lattice where the

particles are free to vibrate, but they cannot move relative to each

other.

Topic: Phases and Phase Change

Objective: How does closeness of matter affect their properties?



Properties of Solids:

1. Molecules, atoms, or ions arranged into a regular, geometric pattern

called a crystal lattice.

2. Molecules, atoms, or ions vibrate in place. They do NOT move

relative to each other.

3. Solids have a definite shape and definite volume.

Topic: Properties of Solids

Objective: How does closeness of solids affect their properties?



Properties of Liquids:

1. Molecules, atoms, or ions may flow past each other.

2. Viscosity is a resistance to flow due to intermolecular attractive

forces (IMAF).

3. Viscosity increases as temperature decreases and IMAF strength

increases.

4. Liquid molecules near a gaseous surface may escape IMAF and enter

the vapor phase at temperatures below the boiling point in a process

called evaporation. (This is the source for vapor pressure)

5. Liquids take on the shape of the container they are in and have a

definite volume. (In the absence of gravity or a container, liquids

form a perfect sphere.)

Topic: Properties of Liquids

Objective: How does closeness of liquids affect their properties?



Properties of Gases:

1. Gas molecules are extremely far apart compared to the size of the

molecules.

2. Gas molecules travel in a straight line until they collide with

something.

3. Collisions are elastic, meaning they don’t lose kinetic energy in the

collision.

4. Gas molecules move faster when it is hotter (higher Kelvin).

5. The gas phase is the only phase that is affected by changes in

pressure.

6. Gases spread out to take the shape and entire volume of whatever

container they are placed in.

Topic: Properties of Gases

Objective: How does closeness of gases affect their properties?



Phase Change Diagram:

Phase Symbols:

Formula(s) = solid phase

Formula(l) = liquid phase

Formula(g) = gaseous phase

Topic: Phase Change Diagram

Objective: How do the changes of phase relate to each other?



Phase equilibrium: Boiling and condensing both occur at the boiling

point (100.0˚C for pure water), freezing and melting both occur at the

melting point (0.0 ˚C for pure water). During the phase change, both

phases exist at equilibrium.

Equilibrium: a condition where the rates of opposing changes are

equal. At equilibrium temperature a substance at the

melting/freezing point is melting at the SAME rate that it is freezing.

Equilibrium is the reason why water melts ABOVE 0°C and freezes

BELOW 0°C. AT 0°C a mixture of water and ice will NOT change

either direction. If you have a sealed flask containing 20.0 g of water

ice and 20.0 g of liquid water and keep it at EXACTLY 0.0°C, there will

still 20.0 g of water ice and 20.0 g of liquid water for as long as the

temperature remains 0.0°C.

Topic: Phase Equilibrium

Objective: What processes are at work during phase equilibrium?



Sublimation: During sublimation the attractive forces between solid

molecules are so weak that heating a solid causes it to go directly into

the gas phase. There are two common substances that undergo this

change; CO2(s), known as dry ice, and I2(s).

Water also undergoes sublimation. Ice cubes ‘shrink’ if left alone in

a freezer. The opposite of sublimation is the process of

DEPOSITION. This can be witnessed by the intricate patterns of ice

found on window surfaces on those cold, dry upstate New York

mornings when water vapor in the air contacts the cold glass and the

gaseous water phase changes directly into solid water.

Topic: Sublimation

Objective: What occurs during the solid to gaseous phase change?



Notes page:



Topic: Heating Curve Diagrams

Objective: What steps occur during constant heating of a substance?



Phase Change Diagrams (Heating/Cooling Curve) Page 14:

When applying CONSTANT HEAT to a solid (point A), the temperature

of the solid increases (segment AB) until the melting point (point B) is

reached, for example of water ice at 0˚C or 273 K. The solid then

absorbs potential energy as it transitions from solid to liquid, or melts

(segment BC). During the melting process, the temperatures (under

CONSTANT HEAT) of both solid and liquid water BOTH remain constant

until the ENTIRE solid has become liquid (point C). Once the substance

is completely in the liquid phase, the temperature will increase

(segment CD). Once the boiling temperature (point D) has been

reached, for liquid water at 100˚C or 373 K, the temperatures (under

CONSTANT HEAT) of both liquid water and water vapor will remain

constant in a process called boiling, or vaporization (segment DE).

Once ALL the water is in the gas phase (point E), the temperature

(under CONSTANT HEAT) of the water vapor will increase (segment EF).

Both phase changes are ENDOTHERMIC here, because heat is a

CONSTANT application at the same RATE.

Topic: Heating Curve Diagrams

Objective: What steps occur during constant heating of a substance?



Heating Curve for Water:

The graph below shows a sample of water initially in the solid phase as it constantly heated from 200 K to 420 K

Topic: Heating Curve for Water

Objective: What steps occur during constant heating of water?



Heating Curve for Water Page 16:

Note: The melting point of water is 0˚C (273 K) and the

boiling point of water is 100˚C (373 K).

1. How many minutes pass from the first appearance of the liquid phase until the substance is entirely in the gas phases?

a. The liquid first appears (point B) at 4 minutes, and the liquid has entirely transitioned to the gas phase (point E) at 34 minutes. The time from first liquid to last liquid is therefore:

34 mins - 4 mins = 30 mins

2. How many minutes will it take for this substance to completely undergo melting?

a. The solid starts to melt (point B) at 4 minutes, and is entirely transitioned to the liquid phase (point C) at 8 minutes. The time from last solid to all liquid is therefore:


3. For how many minutes is the water completely in a solid crystal lattice phase?

a. H20 is a crystal lattice in the solid phase (point A) from 0 minutes until the melting temperature (point B) is reached at 4 minutes. The time is therefore:


Topic: Heating Curve for Water

Objective: What steps occur during constant heating of water?



4. Which line segment represents when H20 is both in the liquid AND gas phases?

a. H20 is in both the liquid and the gas phase during boiling at 20 minutes (point D) to 34 minutes (point E). The time is therefore:


5. For how many minutes is the water completely in a phase WITHOUT a definite shape or volume?

a. The gas phase has no definite shape or volume. The H2O is completely in the gas phase from 34 minutes (point E) until the end of the heating curve (point F). The time therefore:

36 mins - 34 mins = 2 min

6. How many minutes will it take for the water to completely boil away, once the boiling point has been reached?

a. H2O reaches the boiling point in 20 minutes (point D), and complete boiling takes until 34 minutes (point E). The time therefore:


Watch the Phase Change Diagram for water

https://www.youtube.com/watch?v=JcDDzOID960





Student name: _________________________ Class Period: _______

Please carefully remove this page from your packet to hand in.

Phase and Phase Change homework

Circle your answer for multiple choice questions.

1. Which sample below has molecules that flow past each other, but are still attracted to each other? a) C6H12O6(s) b) C6H12O6(l) c) C6H12O6(g)

2. NaCl(l) is stated to be boiling when it undergoes a phase change to: a) Solid b) Liquid c) Gas d) Aqueous

3. What phase of matter has a definite volume, but not definite shape? a) Solid b) Liquid c) Gas

4. Motor oil viscosities are tested at different temperatures to see how effective the oil is at preventing friction. As temperature increases, what happens to the viscosity of the motor oil? a) Increases b) Decreases c) Stays the same

5. Explain your answer to question #4 in terms of molecular motion and average kinetic energy.

As molecular motion becomes more rapid, kinetic energy increases,

causing increased temperature

6. Which of the following phase changes are exothermic? a) LiF(s) LiF(l) b) LiF(l) LiF(s)

c) LiF(s) LiF(g) d) LiF(l) LiF(g)



7. Explain why your choice for question #6 IS exothermic.

Liquid has higher kinetic energy than solid, so energy is given off

from the liquid to solid stage

8. Explain why the other three options for question #6 are NOT exothermic.

Solid liquid absorbs (endo); liquid gas absorbs (endo)

9. Which phase of matter has the strongest attractive forces? a) Solid b) Liquid c) Gas

10. Explain why attractive forces between particles of a substance determine which phase a substance is in at a given temperature.

Greater attractive forces in solids; decreases to liquid and decrease

even more to gas phase

A beaker containing 100.0 g of substance ‘X’ in the solid phase is heated over a hot plate at the CONSTANT rate of 1000.0 J/minute. The results were logged in the table below.



Answer the following questions based on the chart on pg. 20.

11. Draw a graph (page 21) of the data on pg. 20, placing temperature on the Y-axis and time on the X-axis. Label ON THE GRAPH the three phases, and the positions where phase change occurs.

12. At what time does liquid substance ‘X’ first appear? _4 mins_

13. When is substance ‘X’ initially completely in the liquid phase? _8

mins_

14. At what time does gas substance ‘X’ first appear? _14 mins_

15. When is substance ‘X’ initially completely in the gas phase? 21 mins

16. How long does it take for substance ‘X’ to transition from the first

liquid present to the total gas phase? _17 mins_ (21 mins - 4 mins)

17. How long does substance ‘X’ exist completely in the liquid phase?

_6 mins_ (14 mins - 8 mins)

18. How long does substance ‘X’ exist completely as a crystal lattice?

_4 mins_ (4 mins - 0 mins)

19. How long does substance ‘X’ exist in a phase with no definite shape

or volume? _2 mins_ (23 mins - 21 mins)





Phase and Phase Change homework

20. Describe the changes, if any, of the VISCOSITY of substance ‘X’ during heating in the liquid phase.

As temperature increases, viscosity will decrease

21. Describe the changes, if any, in the KINETIC ENERGY of substance

‘X’ between 4 minutes and 8 minutes.

As there is NO increase in temperature there is no change in the

average KE of the water

22. Describe the changes, if any, in the POTENTIAL ENERGY of substance ‘X’ between 4 minutes and 8 minutes.

As there is NO increase in temperature (avg KE), then the added

energy increases the PE of the water

23. Water melts at 273 K and boils at 373 K. Circle and describe which

substance, water or substance ‘X’, has the stronger intermolecular forces.

‘X’ has stronger attractive forces as it has a higher boiling point than

does water



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Notes page:



Melting and boiling are both endothermic because energy has to be

constantly added in order for the change to be completed.

Condensing and freezing are exothermic because energy has to be

constantly removed in order for the change to be completed.

Heat of Fusion:

Heat of Fusion: the amount of heat energy needed to melt one

gram of a solid at the substance’s melting point. The heat of fusion

for water (H2O) is 334 J/g as found in Reference Table B. To calculate

the amount of heat energy (in joules) required to melt a sample, use

the equation (Reference Table T) of:

q = mHf (Hf = Heat of Fusion for a substance)

Heat of Fusion is the property used to defrost a car’s windshield.

Excess engine coolant heat is piped into the defroster to continually

ADD heat (q) to the air that flows onto the inside of the windshield,

adding heat to the ice on the windshield until all the ice has melted

(reached the Heat of Fusion) for the ice.

Topic: Energy for Phase Change

Objective: How do we quantify the energy required to phase change?



Heat of Fusion may be used to determine the energy (J) required to

freeze a liquid at the freezing point (same temperature as melting

point). With freezing you REMOVE heat (q) from liquid. Water

placed into ice cube trays in the freezer has the heat removed by the

freezer components. This removed heat is transferred to the coils at

the back of the refrigerator, which is why the back of a refrigerator is

warmer, as heat is being removed from the inside of the refrigerator.

i. How many joules of energy does take to completely melt 100.0 grams of water at water’s melting point?

a. q = mHf 100.0 g x 334 J/g = 33400 J or 3.34 x 104 J

ii. How many joules of energy does it take to freeze 50.0 g of water at water’s freezing point?

a. q = mHf 50.0 g x 334 J/g = 16700 J or 1.67 x 104 J

Watch Bozeman Science Endothermic and Exothermic video

https://www.youtube.com/watch?v=L-G7pLufXAo





Heat of vaporization:

Heat of Vaporization: the amount of heat energy needed to

completely boil one gram of liquid at the substance’s boiling point.

The heat of vaporization for water (H2O) is 2260 J/g as found in

Reference Table B. To calculate the amount of heat energy (in

joules) required to melt a sample, use the equation (Reference Table

T) of:

q = mHv (Hv = Heat of Vaporization for a substance)

Heat of Vaporization is the property for ‘drying’ a substance that

was in solution. If you heat a solution to remove the water, you are

left with the ‘dry’ solute. Heat of Vaporization removes the water.

Heat of vaporization may also be used for determining how many

joules are released when a gas condenses back into a liquid. Use

the same calculation. Which causes greater danger, having boiling

water at 100°C splashed on you, or steam at 100°C? The water is

hot (334 J/g) and your skin absorbs that heat. The steam was also

hot (2260 J/g), and as the steam touches your skin, it releases more

(in J/g) heat of vaporization, causing greater injury.

Topic: Energy for Phase Change

Objective: How do we quantify the energy required to phase change?



Heat of vaporization is absorbed from your skin as your sweat

evaporates, removing heat from your skin (and blood), leaving your

body cooler.

i. How many joules of energy are required to completely boil away

100.0 g of water at water’s boiling point?

a. q = mHv 100.0 g x 2260 J/g = 226000 J or 2.26 x 105 J

ii. How many joules of energy does it take to completely condense

50.0 g of water at water’s boiling point?

a. q = mHv 50.0 g x 2260 J/g = 113000 J or 1.13 x 105 J



Notes page:





Energy Required for Phase Change homework

Circle your answer for the multiple choice questions.

1. Which of the following phase changes requires only Heat of Fusion?

a) H2O(s) H2O(g)

b) H2O(g) H2O(l)

c) H2O(l) H2O(g)

d) H2O(s) H2O(l)

2. Which of the following phase change is endothermic?

a) H2O(s) H2O(l)

b) H2O(g) H2O(l)

c) H2O(l) H2O(s)

d) H2O(g) H2O(s)

A mixture of 50.0 g of water ice (H2O(s)) and 30.0 g water (H2O(l)) is in a

sealed flask at 0.0°C.

3. What will happen to the amount of ice in the flask if the mixture is

left alone at a constant 0.0°C?

a) Increase b) Decrease c) Stay the same

4. Explain your answer for question #3.

At the EXACT melting point liquid and solid are in equilibrium



5. What will happen to the amount of ice in the flask if the temperature

of the flask is lowered to -10°C?

a) Increase b) Decrease c) Stay the same

6. Explain your answer for question #5.

Below the melting (freezing) point the equilibrium shifts towards

more solid than liquid

Calculate the energy (in joules) required for the following. Show your setup and calculations!

7. Melt 20.0 g of H2O(s) at 0°C.

8. Boil 30.0 g H2O(l) at 100°C

9. Freeze 200.0 g of H2O(l) at 0°C

10. Boil 50.0 g H2O(l) at 100°C



Kinetic-Molecular Theory (Ideal Gas Law):

1. Gases are made of molecules that are extremely small and very far

apart. This allows gases to be easily compressed.

2. Gas molecules move in a straight-line motion until they encounter

another molecule.

3. Any collision a gas molecule makes will be elastic with negligible loss

of energy. Each time a gas molecule collides against an obstacle, it

bounces off with the same energy it had before the collision.

4. There are negligible intermolecular attractive forces between ideal

gas molecules. All matter has attractive forces, but distances

between gas molecules are so great that attractive forces may be

ignored. Gases with London dispersion forces behave the most like

an ideal gas should, and gases with hydrogen bond attractions

behave the least like an ideal gas should.

5. The average velocity of gas molecules is directly proportional to the

Kelvin temperature. Velocity increases with increased temperature.

Topic: Gases and Pressure

Objective: What assumptions are needed to set property standards?



Ideal Gas Behavior:

Under conditions of HIGH temperature and LOW pressure,

molecules behave more like an “ideal” gas. The smaller the gas

molecule, the closer to ideal it will behave. Therefore hydrogen (H2)

and helium (He2) are the gases that are the ‘most’ ideal.

Avogadro’s Hypothesis:

Avogadro’s Hypothesis states if the conditions of temperature and

pressure are the same, then equal numbers of gas molecules will

occupy equal volumes. This is because gas molecules will spread out

to an equal degree at the same temperature and pressure. We can

apply these same math properties equally to all gases we treat as

ideal.

Topic: Ideal Gas Behavior

Objective: What conditions are assumed for gas molecule behavior?



Deviations from the Ideal Gas Law - Where and Why?

The Ideal Gas law works well under STANDARD CONDITIONS of

temperature and pressure (STP) as seen in Reference Table A.

o Standard temperature is 0˚C or 273 K. (K is more significant)

o Standard pressure is 1.000 atmosphere (atm) or 101.3 kPa or

760.0 mmHg (torr).

Pressure:

Pressure is force exerted over an area. The gases in an atmosphere

exert pressure due to gravity’s pull. On Earth the atmospheric

pressure is about 14.7 pounds per square inch (psi). Therefore every

square inch of surface area on Earth (including you) has about 14.7

pounds of force pressing on it when at sea level. Organisms born on

Earth are born into the environment already pre-pressurized to

withstand the atmospheric force. As you ascend (climb) in the

atmosphere, there is less atmospheric mass above you, and

therefore the atmospheric pressure drops. This causes body

pressures to equalize some (ears pop, feet and hands swell slightly).

Topic: Ideal Gas Behavior

Objective: What conditions are assumed for gas molecule behavior?



Pressure is measured in atmospheres (atm), kilopascals (kPa), or

millimeters of mercury (mmHg).

Pressure Conversions:

14.7 psi = 1.00 atm = 101.3 kPa = 760.0 mmHg

Vapor Pressure:

Vapor Pressure is the pressure exerted by a liquid’s vapor in a sealed

container in a vapor-liquid equilibrium at a given temperature. The

vapor pressure of a liquid is NOT dependent on the mass or volume

of the liquid. Vapor pressures are found on Reference Table H. The

stronger the attractive intermolecular forces between liquid

molecules, the lower the vapor pressure will be.

Substances that have high vapor pressures evaporate quickly.

Gasoline is a mixture of high vapor pressure liquids; it may be seen

to evaporate even as it fills your tank. Alcohols and acetone have

high vapor pressures, and therefore evaporate quickly. High vapor

pressure substances are known as volatile, and many are also

flammable, or easily burn or explode.

Topic: Vapor Pressure

Objective: How does evaporation affect the liquid-gas environment?



Boiling Point:

Boiling Point is the temperature at which a liquid’s vapor pressure

equals the pressure exerted on the liquid by outside forces. Use

Reference Table H to determine a liquid’s boiling point. Boiling point

temperatures increase as system pressure increases.

Normal Boiling Point:

Normal boiling point is the boiling point of a liquid under a standard

pressure of 1.000 atmosphere (atm). Substances with higher boiling

points have stronger attractive intermolecular forces holding their

molecules together in the liquid phase, requiring more energy input

to overcome the attractions and permit boiling.

The normal boiling point of water at sea level is 100°C. At sea level,

the atmosphere is exerting 1.000 atm of pressure.

As you ascend into the atmosphere, the mass of air above decreases,

as well as pressure. Lower atmospheric pressure cannot keep water

in a liquid state at higher temperatures, and the boiling point will be

below 100°C. There are special cooking directions for high-altitude

cooking due to the lower heat in boiling water at high altitudes.

Topic: Boiling Point

Objective: How do temperature and pressure affect boiling?



How to use Table H:

1) What is the vapor pressure of __________ at __________ ˚C?

a. What is the vapor pressure of ethanol at 40˚C? Start at the 40˚C

point on the X axis, trace a line up to the ethanol curve, then

move a cross and read the vapor pressure off of the y axis. The

vapor pressure is 17 kPa.

2) What is the boiling point of __________ at a pressure of _____ kPa?

a. What is the boiling point of water at a pressure of 30 kPa? Start

at the 30 kPa point on the Y axis, trace a line to the water curve,

then go down and read the temperature off the X axis. The

boiling point is70˚C.

3) What is the normal boiling point of __________?

a. The normal boiling point is the temperature at which s liquid boils

under standard pressure, 101.3 kPa. There is a dashed line going

across the table at 101.3 kPa and is labeled “101.3 kPa”. Follow

the 101.3 kPa line to the curve you want, and go down to read the

boiling point temperature off the X axis.

Topic: Boiling Point

Objective: How do temperature and pressure affect boiling?



How to read Reference Table H; Vapor Pressure of Four Liquids

Notice that the scales for the X axis are very different than the scales

for the Y axis. Each interval on the Y axis is 10 kPa and each interval on

the X axis is 5˚C.



Notes page:





Gases and Pressure homework

Circle your answer for the multiple choice questions.

1. In which way do real gases deviate from ideal gas behavior? a) Real gas molecules are extremely far apart b) Real gas molecules have attractive forces c) Real gas molecules move faster at higher temperatures d) Real gas molecules travel in a straight line

2. Under which conditions will O2 behave MOST ideally? a) 100 K and 1 atm b) 200 K and 1 atm

c) 300 K and 1 atm d) 400 K and 1 atm

3. Which gas behaves most like an ideal gas at STP? a) O2 b) N2 c) H2 d) F2

4. Describe the quality of an ideal gas that your choice above best matches.

Smaller molecules behave most ideally

5. Water boils at 100°C at sea level. At the highest point in New York State, Mt. Marcy, water will boil: a) At 100°C b) Below 100°C c) Above 100°C

6. Explain your answer for question 5 in terms of vapor pressure and boiling point.

Higher altitudes have less atmospheric mass above, reducing the vapor pressure, lowering boiling point



7. Which substance listed on Reference Table H has the strongest attractive forces holding its molecules together? a) Water b) Ethanol

c) Propanone d) Ethanoic acid

8. Explain your answer for question 7 above in terms of normal boiling point.

Ethanoic acid reaches vapor pressure at 1 atm between 115C and 120C, which is a higher normal boiling point than the other three choices

9. Which sample of gas contains the same number of molecules as a 2.0 L sample of O2(g) at 300 K and 100 kPa? a) 1.0 L sample of H2(g) at 300 K and 100 kPa b) 2.0 L sample of N2(g) at 300 K and 150 kPa c) 2.0 L sample of F2(g) at 350 K and 150 kPa d) 2.0 L sample of Cl2(g) at 300 K and 100 kPa

10. Explain your answer for question 9 above in terms of Avogadro’s Hypothesis.

d) has the same P and T and V as the question; therefore the ame number of moles as stated by Avogadro’s hypothesis

Perform the following conversions, showing your work.

11. 2.0 atm = 2.0 atm x 101.3 𝑘𝑃𝑎

1 𝑎𝑡𝑚 = 202.6 kPa = 2Ō0 kPa

12. 1950. kPa = 1950. kPa x 1 𝑎𝑡𝑚

101.3 𝑘𝑃𝑎 = 19.24975 atm = 19.25 atm





Gases and Pressure homework

13. Why is the atmospheric pressure lower at the top of Mount Everest (highest point on Earth’s surface) than it is at the surface of the Dead Sea (lowest point on Earth’s surface)? Explain using the quantity of air above each location. Higher altitudes have less atmospheric mass above, reducing the vapor pressure, lowering boiling point

14. In which location in question 13 above will water boil at a higher temperature? Explain using your answer above.

Water will boil at a higher temperature at the Dead Sea surface as the atmospheric (and vapor pressure) will be higher there due to more atmospheric mass above the Dead Sea than on Mt. Everest

Based on Reference Table H, what is the vapor pressure of:

15. Ethanol @ 80°C? __102-105__ kPa

16. Propanone @ 20°C? __23-26__ kPa

17. Ethanoic acid @ 110°C? ___78-80___ kPa



Based on Reference Table H, what is the boiling point of:

18. Ethanol under 70 kPa pressure? _67.5-70.5_°C

19. Water under 10 kPa pressure? _43.5-47.5_°C

20. Ethanoic acid under 120 kPa pressure? _122.5 124.5_°C

21. Based on Reference Table H, what is the normal boiling point of propanone?

_54.5-57.5_ °C

22. Why do cooking directions require you to boil your food longer at higher elevations than the time required at lower elevations?

Higher altitudes have less atmospheric mass above, reducing the vapor pressure, lowering boiling point

23. A flask containing only air is filled partway with water then sealed. What will happen to the pressure inside the flask over time? Explain in terms of vapor pressure.

If the flask is sealed, vapor pressure will reach equilibrium as equal #s of molecules will be entering the gas phase as are returning to the liquid phase



The Gas Laws are relationships between temperature, pressure, and

volume of a gas. Gas law equations are used to determine what effect

changing one of those variables will have on any of the others.

What are the units for pressure, volume, and temperature?

Pressure: atm or kPa Volume: mL or L Temperature: K

The Gas Laws are based on one fundamental truth known as

Avogadro’s Hypothesis which states:

i. Equal volumes of two samples of ideal gases contain equal

numbers of particles under the same conditions of

temperature and pressure.

Watch Bozeman Science Gas Laws video

https://www.youtube.com/watch?v=gmN2fRlQFp4

Topic: The Gas Laws

Objective: How can models of particle behavior be used to predict?

https://www.youtube.com/watch?v=BxUS1K7xu30

https://www.youtube.com/watch?v=gmN2fRlQFp4



Consider two 4.00 L containers, both at 298 K and 1.00 atm.

Container ‘A’ holds nitrogen (N2) gas, while container ‘B’ holds

carbon dioxide (CO2) gas. If container ‘A’ holds 2.00 moles of N2 gas,

how many moles of CO2 gas must be present in container ‘B’?

i. As both containers hold equal volumes of gases under the same

conditions of temperature and pressure, Avogadro’s Hypothesis

states that both containers will hold equal numbers of molecules,

and then hold equal numbers of moles of molecules. Therefore, if

container ‘A’ holds 2.00 moles of gas, so container ‘B’ MUST also

hold 2.00 moles of gas.

This means that all gases behave more or less equally to changes in

temperature, pressure, and volume, so one equation may be used

to describe these changes and also equally applied to all gases, if the

assumption is made that they exhibit IDEAL gas behavior.

Do equal volumes of gases under the same conditions of

temperature and pressure have the same MASS?

NO! Each element has its own unique atomic mass.

Topic: The Gas Laws

Objective: How are models of particle behavior used for predictions?



Solving Gas Law problems:

1. Get rid of the words. Create a data table to organize the numbers.

a) If the units are volume (mL or L), pick out the V1 and V2.

b) If the units are pressure (atm or kPa), pick out P1 and P2.

c) If the units are temperature (K), pick out T1 and T2.

NOTE: If temperature is given in °C, convert it to K by adding

°C + 273 to get K, as Kelvin MUST be used in gas laws.

2. Write the gas law equation down. Write it on your homework page.

3. Circle the variable you are solving for. Then use basic algebra to

rearrange the equation.

4. Substitute the numbers into the newly rearranged equation.

Ensure you don’t ‘lose’ numbers and units in the process. Enter the

numbers into your calculator.

5. Round your answer. Gas laws are multiplication and division, so

round to the fewest number of sig figs in the original values. Make

sure you have cancelled all units you can and place the proper units

on your answer.

Topic: Solving Gas Law Problems

Objective: How can we use Gas Laws to solve gas reaction problems?



Relationships between variables of Pressure (P), Volume (V), and

Temperature (T):

1. Pressure vs. Volume (Constant Temperature)-as gas pressure is

increased, gas volume is decreased (indirect relationship).

Boyle’s Law (Pressure vs Volume at constant temperature)

P x V = k: P1V1 = k, P2V2 = k, therefore P1V1 = P2V2

Note: “k” here means a constant relationship, NOT Kelvin

A 10.0 L (V1) sample of gas (at constant temperature) under a

pressure of 1.0 atm (P1) is trapped in a cylinder with a moveable

piston. The cylinder is moved until the pressure doubles to 2.0

atm (P2). The new volume of the gas in the cylinder is now 5.0 L

(V2). If pressure is doubled, volume is halved.

Topic: Gas Property Relationships

Objective: What relationships exist between P, V, & T in gases?



Constant Temperature Example:

A sample of gas occupies a volume of 2.00 L at Standard

Temperature and Pressure (STP). If the pressure is increased to 2.00

atm at constant temperature, what will be the new gas volume?

a) Set up a table of givens: (P1 is given as STP, look at Ref. Table A =

1.00 atm)

P1 = 1.00 atm; V1 = 2.00 L; P2 = 2.00 atm; V2 = X

b) Rearrange the Combined Gas Law Equation (Ref. Table T) to solve

for the desired variable, omitting the variable that is held constant

(temperature in this case):

has temperature (T) omitted to become:

P1V1 = P2V2

c) Algebraically solve for V2 to give the equation of: P1V1 / P2 = V2

d) Substitute the given numbers and units into the new equation and

solve:

1.00 𝑎𝑡𝑚 ∗ 2.00 𝐿

2.00 𝑎𝑡𝑚= 1.00 𝐿




Temperature (T):

2. Volume vs. Temperature (Constant Pressure)-as gas temperature is

increased, gas volume is increased (direct relationship).

Charles’ Law (Volume vs. Temperature at constant pressure)

V / T = k: V1 / T1 = k, V2 / T2 = k therefore V1 / T1 = V2 / T2


A 1.0 L (V1) sample of gas (at constant pressure) at a temperature

of 200.0 K (T1) is trapped in a cylinder with a moveable piston.

The cylinder is heated until the temperature doubles to 400.0 K

(T2). The new gas volume will be 2.0 L (V2). If the temperature is

doubled, then volume doubles as well.





Constant Pressure example:

A sample of gas at 300.0 K occupies a volume of 5.00 L. If the

temperature is doubled under constant pressure, what will be the new

volume?

a) Set up a table of givens:

V1 = 5.00 L; T1 = 300.0 K; V2 = X; T2 = 600.0 K (T1 doubled)

b) Rearrange the Combined Gas Law Equation (Ref. Table T) to solve for

the desired variable, omitting the variable that is held constant

(pressure in this case):

has pressure (P) omitted to become:

V1 / T1 = V2 / T2

c) Algebraically solve for V2 to give the equation of: V1T2 / T1 = V2


solve:

5.00 𝐿 ∗ 600.0 𝐾

300.0 𝐾= 10.0 𝐿




Temperature (T):

1. Temperature vs. Pressure (Constant Volume)-as gas temperature is

increased, gas pressure is increased (direct relationship).

Gay-Lussac’s Law (Pressure vs. temperature at constant pressure)

P/ T = k: P1 / T1 = k, P2 / T2 = k, therefore P1/T1 = P2/T2


A sample of gas at 300.0 K (T1) and 2.0 atm (P1) is trapped in a

rigid cylinder (constant volume). When the cylinder is heated

until the temperature doubles to 600.0 K (T2), the new pressure of

the gas will be 4.0 atm (P2). If temperature is doubled, then

pressure doubles as well.





Constant Volume example:

A 10.0 L sample of gas in a rigid container at 1.00 atm and 200.0 K is

heated to 800.0 K. Under constant volume, what will the new pressure

of the gas be?


P1 = 1.00 atm T1 = 200.0 K P2 = X T2 = 800.0 K

b) Rearrange the Combined Gas Law Equation (Ref. Table T) to solve for

the desired variable, omitting the variable that is held constant

(volume in this case):

has volume (V) removed to become:

P1 / T1 = P2 / T2

c) Algebraically solve for P2 to give the equation of: P1T2 / T1 = P2


solve:

1.00 𝑎𝑡𝑚 ∗ 800.0 𝐾

200.0 𝐾= 4.00 𝑎𝑡𝑚



The three given Gas Laws, Boyle’s Law, Charles’ Law, and Gay-Lussac’s

Law, are combined to give the Combined Gas Law found on Reference

Table T as shown.

NOTE: Ref Table T does NOT state

temperature in K. You MUST

convert °C to K by adding 273 to °C!

If a variable is given as a constant, it may be ignored using this

equation. Set up a table of the given variables and solve for the

unknown. There MAY NOT be any CONSTANT. If no constant is stated,

use all the given variables.

Topic: Combined Gas Law

Objective: How do we combine the three laws into a combined law?

You MUST use

KELVIN temps!



Combined Gas Law examples:

1. A 2.00 L sample of a gas at STP is heated to 500.0 K and compressed

to 200.0 kPa. What will the new volume of the gas be?


P1 = 101.3 kPa; V1 = 2.00 L; T1 = 273 K; P2 = 200.0 kPa; V2 = X; T2 = 500.0 K

b) Rearrange the Combined Gas Law equation to solve for the

desired new volume:

solved for V2 becomes the equation of:

(P1V1T2) / (P2T1) = V2

c) Substitute the given numbers and units into the new equation and

solve:

101.3 𝑘𝑃𝑎 ∗ 2.00 𝐿 ∗ 500.0 𝐾

200.0 𝑘𝑃𝑎 ∗ 273 𝐾= 1.86 𝐿




2. A 2.00 L sample of a gas at 1.00 atm and 300.0 K is heated to 500.0 K

and compressed to a volume of 1.00 L. What will the new pressure

of the gas be?


P1 = 1.00 atm; V1 = 2.00 L; T1 = 300.0 K; P2 = X; V2 = 1.00 L; T2 = 500.0 K


desired new volume:

solved for P2 becomes the equation of:

(P1V1T2)/ (V2T1) = P2


solve:

1.00 𝑎𝑡𝑚 ∗ 2.00 𝐿 ∗ 500.0 𝐾

1.00 𝐿 ∗ 300.0 𝐾= 3.33 𝑎𝑡𝑚




3. A 2.00 L sample of a gas at 300.0 K and a pressure of 80.0 kPa is

placed into a 1.00 L container and pressurized to 240.0 kPa. What

will the new temperature of the gas be?


P1 = 80.0 kPa; V1 = 2.00 L; T1 = 300.0 K; P2 = 240.0 kPa; V2 = 1.00 L; T2 = X


desired new volume:

solved for T2 becomes the equation of:

(P2V2T1)/ (P1V1) = T2


solve:

(240.0 𝑘𝑃𝑎 ∗ 1.00 𝐿 ∗ 300.0 𝐾)

80.0 𝑘𝑃𝑎 ∗ 2.00 𝐿= 450. 𝐾



Notes page:





The Gas Laws homework

Answer the following questions, showing ALL work, including conversions, units, and proper significant figures. Circle your answer.

1. A sample of hydrogen gas has a volume of 1.00 L at a pressure of 100.0 kPa. If the temperature is kept constant and the pressure is raised to 140.0 kPa, what is the new volume of the hydrogen gas sample?

2. A gas sample occupies 10.0 mL at 1.00 atm of pressure. If the gas sample is placed into a sealed 20.0 mL container and the temperature remains constant, what is the new pressure of the gas sample?



3. When 500.0 mL of hydrogen gas is heated from 30.0°C to 60.0°C at a constant pressure, what will be the volume of the gas at 60.0°C?

4. Temperatures in New York State can normally range from near 100°F (38°C) to -20°F (-29°C) across the state from season to season. Assuming no air is lost in a tire (constant volume of 10.00 L) that started with a pressure of 35 psi (241 kPa); find the change in tire pressure from the hottest temperature to coldest temperature given above.

241 kPa - 189 kPa = 52 kPa (∆P) which is ≈ 7.5 psi loss



5. A sample of gas at 200.0 K occupies a volume of 10.0 L at a pressure of 2.00 atm. To what temperature must the gas be raised to double both the gas volume and the gas pressure?

6. A 2.00 L sample of gaseous oxygen exerts a pressure of 2.00 atm at 273 K. If the temperature of the oxygen is raised to 546 K and the volume decreased to 1.00 L, what will be the final pressure of the gas?

7. A gas sample occupies 10.0 L of volume at STP. How much volume will it occupy at a pressure of 2.00 atm and temperature of 300.0 K?



A CO2 fire extinguisher contains 15.5 L of CO2(g) under a pressure of 500.0 atm at 22.0°C. When activated, the pressure of the CO2 drops to 95.0 kPa and the volume of CO2 released is 458 L.

8. What is the temperature of the released CO2?

9. CO2 undergoes sublimation at a temperature of -78.5°C. What phase does the CO2 in the scenario above exit the fire extinguisher?

Solid CO2 - the temperature that the CO2 exits at (16.3 K) is -256.7˚C, which is FAR below the sublimation point of CO2



Notes page:

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