9/20/2015gas laws unit1 parts of chapters 2, 10, &11 all of chapter 12 – gas laws chemistry...

04/21/2304/21/23 Gas Laws Unit Gas Laws Unit 11

Parts of chapters 2, 10, &11Parts of chapters 2, 10, &11All of Chapter 12 – Gas LawsAll of Chapter 12 – Gas Laws

CHEMISTRYCHEMISTRY

GREATER LATROBE HIGH SCHOOLGREATER LATROBE HIGH SCHOOL


TABLE OF CONTENTSTABLE OF CONTENTS

Introduction to Gases Boyles Law Charles Law Gay-Lusacc’s Law Combined Gas Law Ideal Gas Law Dalton’s Law Graham’s Law Phase Diagrams Surface Tension

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contents

All yellow text will take you to additional information


Introduction to Gases – Introduction to Gases – Causes of change (Ch. 11.1 and 11.2)Causes of change (Ch. 11.1 and 11.2)

Energy as Heat:Energy as Heat:– Heat and Temperature are not the same Heat and Temperature are not the same

thing.thing. Heat Temperature Enthalpy – Total energy content of the Enthalpy – Total energy content of the

sample. sample.



Molar Heat Capacity –(CMolar Heat Capacity –(Cpp)– this is the )– this is the amount of heat required to raise the amount of heat required to raise the temperature of 1 gram of a substance 1 temperature of 1 gram of a substance 1 degree Celsius (or 1 mole 1 Kelvin).degree Celsius (or 1 mole 1 Kelvin).– Practical example: snow fallPractical example: snow fall– Calculations using CCalculations using Cpp – These are done to – These are done to

determine the amount of energy change (determine the amount of energy change (H) in H) in the process of warming or cooling matter. This the process of warming or cooling matter. This is calculated according to the following is calculated according to the following equation:equation:



H = mCH = mCp p T (T (H can be changed for Q) H can be changed for Q) – m = mass, Cm = mass, Cpp = Specific Heat capacity (will be = Specific Heat capacity (will be

given, given, T = Change in temperature (Final T – T = Change in temperature (Final T – Initial T)Initial T) m can also be replaced with n for moles if necessary)m can also be replaced with n for moles if necessary)



Problem #1: Problem #1: (Click for answer)– Assuming the density of water to be 1.0 g/mL. Assuming the density of water to be 1.0 g/mL.

How much heat is lost by 4.0 L of water cooling How much heat is lost by 4.0 L of water cooling from 87from 87 C to 21 C to 21 C? (Heat capacity for water = C? (Heat capacity for water = 4.180 J/g* 4.180 J/g* CC

Problem #2: Problem #2: (Click for answer)– If 980 kJ of energy is added to 6.2L of water at If 980 kJ of energy is added to 6.2L of water at

1818 C, what is the final temperature of the C, what is the final temperature of the water?water?



When heat is added or removed from a substance When heat is added or removed from a substance it can go through phase changes. Recall that it can go through phase changes. Recall that there are three major states (phases) of matter there are three major states (phases) of matter (solid, liquid, and gas). A (solid, liquid, and gas). A Heating Curve (in your (in your book on page 44) shows the relationship between book on page 44) shows the relationship between these states as the temperature changes. these states as the temperature changes. – ∆∆HHfusfus – energy required to melt one mole – energy required to melt one mole– ∆∆HHvap vap - energy required to vaporize one mole- energy required to vaporize one mole– ∆∆HHvap vap > ∆H> ∆Hfusfus


State Changes and Phase DiagrmasState Changes and Phase Diagrmas

Review of what we know at this point about solids, Review of what we know at this point about solids, liquids, and gases:liquids, and gases:– Solids – Constant shape, constant volumeSolids – Constant shape, constant volume– Liquids – Variable shape, constant volumeLiquids – Variable shape, constant volume– Gases – Variable shape, variable volumeGases – Variable shape, variable volume

We now need a more detailed comparison of each We now need a more detailed comparison of each of these.of these.– Solids– Liquids– Gases


Phase Diagrams – Vapor Phase Diagrams – Vapor Pressure and BoilingPressure and Boiling

Vapor Pressure – the pressure of a gas on top of a liquid. Vapor Pressure – the pressure of a gas on top of a liquid. This is dependent only on the temperature.This is dependent only on the temperature.– This is key in understanding the process of boiling. For example:This is key in understanding the process of boiling. For example:

What would happen to a an open jar of water over time?What would happen to a an open jar of water over time? What would happen to a closed jar of water over time?What would happen to a closed jar of water over time?

– Here water will evaporate until condensation = evaporation. Here vapor Here water will evaporate until condensation = evaporation. Here vapor pressure is a max. pressure is a max.

Boiling – this occurs when vapor pressure equals Boiling – this occurs when vapor pressure equals atmospheric pressure. atmospheric pressure. – Substances with high vapor pressure at low temperatures are Substances with high vapor pressure at low temperatures are

volatile (boil / vaporize easily) i.e. perfumevolatile (boil / vaporize easily) i.e. perfume


Phase DiagramsPhase Diagrams

Phase Diagrams – Graphically represent the relationship Phase Diagrams – Graphically represent the relationship between the state of a substance and its pressure and between the state of a substance and its pressure and temperature. temperature. – Processes shown phase diagram Processes shown phase diagram (click to go to diagram)

(1) – melting – solid to liquid(1) – melting – solid to liquid (2) – Freezing – liquid to solid(2) – Freezing – liquid to solid (3) – Boiling – liquid to gas(3) – Boiling – liquid to gas (4) – Condensing – gas to liquid(4) – Condensing – gas to liquid (5) – Sublimation – solid directly to gas (dry ice)(5) – Sublimation – solid directly to gas (dry ice) (6) – Deposition – gas directly to solid (frost)(6) – Deposition – gas directly to solid (frost) (7) – Triple point – Temperature and pressure at which a substance (7) – Triple point – Temperature and pressure at which a substance

exists as a solid, liquid, and gas at the same time.exists as a solid, liquid, and gas at the same time.


Characteristics of GasesCharacteristics of Gases

Kinetic Molecular Theory – Describes the Kinetic Molecular Theory – Describes the behavior of gases at the molecular level behavior of gases at the molecular level (Based on an ideal gas)(Based on an ideal gas)– Ideal Gas – Imaginary gas (model) that Ideal Gas – Imaginary gas (model) that

describes the behavior of real gases at describes the behavior of real gases at conditions close to STP (this will be explained conditions close to STP (this will be explained later)later)


CharacteristicsCharacteristics of Gases (con’t) of Gases (con’t)

5 Assumptions of the KMT:5 Assumptions of the KMT:– 1. Gases consist of large numbers of tiny particles1. Gases consist of large numbers of tiny particles– 2. Particles of a gas are in constant motion and 2. Particles of a gas are in constant motion and

therefore have KE therefore have KE– 3. Collisions between particles of a gas and 3. Collisions between particles of a gas and

container wall are completely elastic.container wall are completely elastic.– 4. There are no forces of attraction or repulsion 4. There are no forces of attraction or repulsion

between particle of a gas between particle of a gas– 5. The Average kinetic energy of the particles of a 5. The Average kinetic energy of the particles of a

gas is directly proportional to the Kelvin gas is directly proportional to the Kelvin Temperature of the gas. Temperature of the gas. KE = ½ MVKE = ½ MV22


Ideal versus Real GasesIdeal versus Real Gases

Ideal GasesIdeal Gases– The volume of the particles The volume of the particles

is negligibleis negligible– There are no attractive There are no attractive

forces between moleculesforces between molecules– Collisions between particles Collisions between particles

are elasticare elastic

Real GasesReal Gases– Particles do have volumeParticles do have volume– As temp. decreases the As temp. decreases the

particles slow down and particles slow down and attractive forces increaseattractive forces increase

– Collisions between particles Collisions between particles are inelasticare inelastic

Throughout this Unit we are going to focus on Ideal Gases, Ideal gases differ from real gases by the following


CharacteristicsCharacteristics of Gases (con’t) of Gases (con’t)

Variables that describe a gas:Variables that describe a gas:– Temperature – measure of the average KE of the Temperature – measure of the average KE of the

Particles of a gas (must be in Kelvin – T in K)Particles of a gas (must be in Kelvin – T in K) K = K = C + 273C + 273

– Volume – amount of space matter occupiesVolume – amount of space matter occupies Can use L, mL, mCan use L, mL, m33, or cm, or cm33

– Pressure – Force per unit area Pressure – Force per unit area Measure with Measure with barometer Describes the forces that gases exert on the walls of a Describes the forces that gases exert on the walls of a

container.container. Atmospheric Pressure = 1 atm = 760 mmHg = 760 torr = 101.3 Atmospheric Pressure = 1 atm = 760 mmHg = 760 torr = 101.3

Kpa = 29.9 inHg = 14.7 psiKpa = 29.9 inHg = 14.7 psi


Characteristics of Gases (con’t)Characteristics of Gases (con’t)

STP – Standard Temperature and PressureSTP – Standard Temperature and Pressure– T = 273 K T = 273 K – P = 1 atmP = 1 atm

At these conditions all gasses occupy 22.4 L of space (standard molar volume)


Boyles LawBoyles Law

This Gas law discovered by Robert Boyle involves This Gas law discovered by Robert Boyle involves the relationship between the relationship between pressurepressure and and volumevolume at at constant temperature. constant temperature.

Go to the following web site to see an illustration Go to the following web site to see an illustration of this relationship (freeze mass and of this relationship (freeze mass and Temperature):Temperature):– http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html

This relationship is know as an This relationship is know as an inverse relationship. (when one variable gets . (when one variable gets bigger the other variable gets smaller)bigger the other variable gets smaller)

http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html


Boyle’s Law (con’t)Boyle’s Law (con’t)

Formula:Formula:– PV = k (k=constant, P=pressure, V=volume)PV = k (k=constant, P=pressure, V=volume)– Since the pressure times the volume is constant this can be written Since the pressure times the volume is constant this can be written

as follows: as follows: (click on the formula to go to some practice problems)(click on the formula to go to some practice problems)

– P1V1 = P2V2 oror

– P1V1 = P2V2

Boyle’s Law has many practical applications. It is applied Boyle’s Law has many practical applications. It is applied in things such as in things such as breathing, scuba diving, submarines, and weather balloons. We will discuss several of these. . We will discuss several of these.

http://fsc.fernbank.edu/chemistry/boyles.htm








http://chemistry2.csudh.edu/lecture_help/boyleslaw.html








http://www.onr.navy.mil/focus/blowballast/sub/work4.htm


Charles LawCharles Law

This gas law was discovered by Joseph Louis Gay-Lussac This gas law was discovered by Joseph Louis Gay-Lussac but is usually named for Jacques Charles Law because of but is usually named for Jacques Charles Law because of the work he did. It involves the relationship between the work he did. It involves the relationship between VolumeVolume and and TemperatureTemperature at constant pressure. at constant pressure.

Go to the following web sites to see an illustration of this Go to the following web sites to see an illustration of this relationship (freeze mass and Pressure):relationship (freeze mass and Pressure):

– http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html– http://fsc.fernbank.edu/chemistry/charles.html

This relationship is know as an This relationship is know as an direct relationship. (when . (when one variable gets bigger the other variable gets bigger)one variable gets bigger the other variable gets bigger)


http://fsc.fernbank.edu/chemistry/charles.html


Charles LawCharles Law

Formula:Formula:– V/T = k (k=constant, T=Temperature, V=volume)V/T = k (k=constant, T=Temperature, V=volume)– Since the volume divided by the temperature is constant the Since the volume divided by the temperature is constant the

following formula can be written: following formula can be written: (click on the formula to go to some (click on the formula to go to some practice problems (choose solving problems or word problems)practice problems (choose solving problems or word problems)

– V1/T1 = V2/T2 oror

– V1/T1 = V2/T2

Charles’s Law also has many practical applications. It is Charles’s Law also has many practical applications. It is applied in things such as effects of temperature changes applied in things such as effects of temperature changes on balloons, tires, on balloons, tires, hot air balloons, etc. , etc.

http://fsc.fernbank.edu/chemistry/charles.htm








http://chemistry2.csudh.edu/lecture_help/charleslaw.html








http://travel.howstuffworks.com/hot-air-balloon.htm


Gay-Lussac’s LawGay-Lussac’s Law

This Gas law discovered by Joseph Louis This Gas law discovered by Joseph Louis Gay-LussacGay-Lussac involves the relationship involves the relationship between between temperaturetemperature and and PressurePressure at at constant volume. constant volume.

Go to the following web site to see an Go to the following web site to see an illustration of this relationship (freeze mass illustration of this relationship (freeze mass and volume):and volume):– http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html

This relationship is also a direct relationship.This relationship is also a direct relationship.



Gay-Lussac’s LawGay-Lussac’s Law

Formula:Formula:– P/T = k (k=constant, T=Temperature, P=pressure)P/T = k (k=constant, T=Temperature, P=pressure)– Since the volume divided by the temperature is constant the Since the volume divided by the temperature is constant the

following formula can be written: following formula can be written: (click on the formula to go to some (click on the formula to go to some practice problems (choose solving problems or word problems)practice problems (choose solving problems or word problems)

– P1/T1 = P2/T2

Gay-Lussac’s law also has many practical applications. Gay-Lussac’s law also has many practical applications. For example: This is why on the side of spray cans it will For example: This is why on the side of spray cans it will tell you to keep the can stored at certain temperatures. tell you to keep the can stored at certain temperatures. This is why inflatable devices such as tires or balls go flat This is why inflatable devices such as tires or balls go flat in the winter time. in the winter time.

http://fsc.fernbank.edu/chemistry/gaylussac.htm









The Combined Gas LawThe Combined Gas Law

This gas law encorporates all three of the This gas law encorporates all three of the basic gas law (Boyle’s, Charles’s, and Gay-basic gas law (Boyle’s, Charles’s, and Gay-Lussac’s) Lussac’s) – Click here to see a graphical illustration of this.

It is often used for converting a temperature, It is often used for converting a temperature, pressure, or volume to standard conditions pressure, or volume to standard conditions (STP)(STP)


The Combined Gas LawThe Combined Gas Law

Formula:Formula:

Sample problem: Sample problem: – 2.00 L of a gas is collected at 25.0°C and 745.0 2.00 L of a gas is collected at 25.0°C and 745.0

mmHg. What is the volume at STP?mmHg. What is the volume at STP? Click for answer

P1V1 P2V2

T1 T2


The Combined Gas ContinuedThe Combined Gas Continued

A sample of gas occupies a volume of 6.80 A sample of gas occupies a volume of 6.80 L at 790.0 mmHg and 300.0 K. What is the L at 790.0 mmHg and 300.0 K. What is the temperature of the gas when the volume is temperature of the gas when the volume is 7.00 L and the Pressure is 2.00 atm?7.00 L and the Pressure is 2.00 atm?


Ideal Gas LawIdeal Gas Law

The Ideal gas law encorporates The Ideal gas law encorporates all three variablesall three variables that we that we have been using plus it also incorporates have been using plus it also incorporates molesmoles into the into the equation. This is the most common equation used in equation. This is the most common equation used in working with gases. working with gases.

First written by Emil Clapeyron, and is sometimes First written by Emil Clapeyron, and is sometimes (although rarely called the Clapeyron equation) (although rarely called the Clapeyron equation)

The formula is as follows:The formula is as follows:– PV = nRTPV = nRT

P = pressure, V = Volume, n = moles, R = Gas constant, T = temperatureP = pressure, V = Volume, n = moles, R = Gas constant, T = temperature R = 8.314 L * kPa or 0.0821 L * atm or 62.36 mmHg * LR = 8.314 L * kPa or 0.0821 L * atm or 62.36 mmHg * L

----------- ----------- --------------- ----------- ----------- ---------------

Mol * KMol * K mol * K mol * K mol * K mol * K


Ideal Gas LawIdeal Gas Law

Example ProblemsExample Problems– What is the pressure (in atm) exerted by a 0.500 mol sample of Nitrogen What is the pressure (in atm) exerted by a 0.500 mol sample of Nitrogen

gas in a 10.0 L container at 298 K?gas in a 10.0 L container at 298 K?

– What mass would a sample of chlorine gas have if the pressure in a 10.0 L What mass would a sample of chlorine gas have if the pressure in a 10.0 L tank at 27 degrees Celsius is 3.50 atm? tank at 27 degrees Celsius is 3.50 atm?

– Answers to Ideal Gas Law Problems


Dalton’s LawDalton’s Law

This gas law was discovered by John Dalton. It This gas law was discovered by John Dalton. It deals with the partial pressures (The pressure of deals with the partial pressures (The pressure of an individual gas in a mixture of gases) of a gas. an individual gas in a mixture of gases) of a gas. It is primarily used for situations when a gas has It is primarily used for situations when a gas has been collected over water. been collected over water.

Go to the following web site for an explanation of Go to the following web site for an explanation of Dalton’s Law:Dalton’s Law:

http://www.fordhamprep.com/gcurran/sho/sho/lessons/lesson74.htm

http://www.fordhamprep.com/gcurran/sho/sho/lessons/lesson74.htm



FormulaFormula– PPtt = P = P11 + P + P22 + P + P33 + ….. + P + ….. + Pnn (P = Pressure) (P = Pressure)– This formula can also be written as follows:This formula can also be written as follows:

PPatm atm = P= Pgasgas + P + PHH22O O (P=pressure, atm=atmosphere) (P=pressure, atm=atmosphere)

– The second formula is the form which we will use the The second formula is the form which we will use the most often. This is the form of Dalton’s equation that most often. This is the form of Dalton’s equation that allows us to correct for allows us to correct for water vapor pressure (click to see water vapor chart) when a gas is collected when a gas is collected over water. over water.

http://www.cms.k12.nc.us/allschools/providence/duncan_files/h2o_vapor_chart.htm





Example Problem:Example Problem:– 2.50 L of gas is collected over water at 25 degrees C 2.50 L of gas is collected over water at 25 degrees C

and 795 mmHg. What is the volume of gas at STP?and 795 mmHg. What is the volume of gas at STP?

– Click for answer


Graham’s LawGraham’s Law

Graham’s Law is used to give the rates of effusion Graham’s Law is used to give the rates of effusion and diffusion.and diffusion.– EffusionEffusion – Gas escapes a container through a small pin – Gas escapes a container through a small pin

size hole.size hole.– DiffusionDiffusion – Gradual mixing of two gases because of – Gradual mixing of two gases because of

random motion of particlesrandom motion of particles Definition: The rates of effusion and diffusion are Definition: The rates of effusion and diffusion are

inversely proportional to the square roots of a inversely proportional to the square roots of a gases molar mass.gases molar mass.– Analogy – Big guy –vs- Small guyAnalogy – Big guy –vs- Small guy


Graham’s LawGraham’s Law

Formula:Formula:– Rate A Square root (MRate A Square root (MBB))

-------- = --------------------------- = -------------------

Rate B Square root (MRate B Square root (MAA))

Sample ProblemSample Problem


TemperatureTemperature

The temperature of a substance is a The temperature of a substance is a measure of the average kinetic energy of measure of the average kinetic energy of that substances particles. that substances particles. – Think of this as the intensity of energy.Think of this as the intensity of energy.– This is an intensive property – the measure of This is an intensive property – the measure of

temperature does not depend on the amount of temperature does not depend on the amount of the sample of material.the sample of material.

This is different than This is different than heat

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HeatHeat Heat is the measure of the total amount of Heat is the measure of the total amount of

energy transferred from an object of high energy transferred from an object of high temperature to one of low temperaturetemperature to one of low temperature– Think of this as the quantity of energy Think of this as the quantity of energy – This is an extensive property – the measure of This is an extensive property – the measure of

heat depends on the amount of the sample of heat depends on the amount of the sample of material.material.

– It is always High It is always High Low Low

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Heating CurveHeating Curve(Chapter 2 – Page 44)(Chapter 2 – Page 44)

The picture The picture to the right to the right represents represents a heating a heating curve for curve for water. water.

LecturePLUS Timberlake 2

Heating Curve for Water

120 °Csteam

100 °C water steam

50°C liquid water

0 °C ice liquid

-10 °C ice

Heat added

What is happening to the temperature when the substance is changing states?

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Answers to Heat Capacity Answers to Heat Capacity ProblemsProblems

Problem #1:Problem #1:– First find mass: m = (1000 g / L)(4.0 L) = 4000 gFirst find mass: m = (1000 g / L)(4.0 L) = 4000 g H=(4000g)(4.180 J/g* H=(4000g)(4.180 J/g* C)(21-87) = 1103520 JC)(21-87) = 1103520 J

Problem #2:Problem #2:– First find mass: m = (1000 g / L)(6.2 L) = 6200 gFirst find mass: m = (1000 g / L)(6.2 L) = 6200 g– 980 000 J = (6200g)(4.180 J/g* 980 000 J = (6200g)(4.180 J/g* C)(TC)(Tff - 18 - 18 C) = 55.8 C) = 55.8 C C

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BarometerBarometer The barometerThe barometer The mercury barometer The mercury barometer

measures atmospheric pressure. It works this measures atmospheric pressure. It works this way. Completely fill a long glass tube with way. Completely fill a long glass tube with mercury. Turn it upside down, and place the mercury. Turn it upside down, and place the top below the surface of more mercury in an top below the surface of more mercury in an open basin. Rather than pouring out again, open basin. Rather than pouring out again, the mercury in the tube will only fall until the the mercury in the tube will only fall until the height of the column is about a meter. This is height of the column is about a meter. This is because the pressure of the air on the because the pressure of the air on the mercury in the basin is equal to the pressure mercury in the basin is equal to the pressure of the mercury in the column. The gap at the of the mercury in the column. The gap at the top of the tube is not air but a vacuum. The top of the tube is not air but a vacuum. The difference in height between the top of the difference in height between the top of the column and the top surface of the mercury in column and the top surface of the mercury in the basin is a measure of the weight of the air, the basin is a measure of the weight of the air, which changeswhich changes as the weather changes. as the weather changes.

Weather can be predicted by this:Weather can be predicted by this:– Low barometric pressure generally indicates Low barometric pressure generally indicates

stormy weatherstormy weather– High barometric pressure generally indicates High barometric pressure generally indicates

good weathergood weather

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http://www.sciencemuseum.org.uk/collections/exhiblets/weather/baromdiagram2.asp

http://www.sciencemuseum.org.uk/collections/exhiblets/weather/baromdiagram2.asp


Inverse Relationships DefinedInverse Relationships Defined

An inverse relationship An inverse relationship is when one of the is when one of the variables is increased variables is increased the other variable the other variable decreases. See the decreases. See the graph to the right for graph to the right for an illustration of this.an illustration of this.

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Direct Relationship definedDirect Relationship defined

In a direct In a direct relationship when relationship when one of the variables one of the variables gets bigger the other gets bigger the other variable also gets variable also gets bigger. See the bigger. See the graph to the right for graph to the right for an illustration of this.an illustration of this.

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Water Vapor PressureWater Vapor Pressure

This is pressure in a container of gas that This is pressure in a container of gas that results from the water evaporating slightly results from the water evaporating slightly as the gas is being collected over water. as the gas is being collected over water. The amount of evaporation increases as the The amount of evaporation increases as the temperature increases.temperature increases.

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Answer to Dalton’s ProblemAnswer to Dalton’s Problem

– 2.50 L of gas is collected over water at 25 degrees C 2.50 L of gas is collected over water at 25 degrees C and 795 mmHg. What is the volume of gas at STP?and 795 mmHg. What is the volume of gas at STP? You must use the combined gas lawYou must use the combined gas law

– 11stst – Determine the pressure of the dry gas ( – Determine the pressure of the dry gas (vapor pressure chart)

795 mmHg = P795 mmHg = Pgg + 23.8 mmHg + 23.8 mmHg PPg g = 771.2 mmHg= 771.2 mmHg

– 22ndnd – Use the combined gas law to solve. – Use the combined gas law to solve. (771.2 mmHg)(2.50L)/298K = (760 mmHg)V(771.2 mmHg)(2.50L)/298K = (760 mmHg)V22/273K/273K V = 2.32 LV = 2.32 L

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Answer to Combined GasAnswer to Combined GasLaw ProblemLaw Problem

The data is organized to the left. The data is organized to the left. Insert them into: Insert them into:– PP11VV11 / T / T11 = P = P22VV22 / T / T22

(745.0 mmHg)(2.00L) / 298K(745.0 mmHg)(2.00L) / 298K

= ( 760.0 mmHg)(V2) / 273K= ( 760.0 mmHg)(V2) / 273K

V2 = 1.79 LV2 = 1.79 L

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Answer to Ideal GasAnswer to Ideal GasLaw ProblemLaw Problem

PV = nRTPV = nRT– P(10.0L) = (0.500)(0.0821)(298K)P(10.0L) = (0.500)(0.0821)(298K)– P = 1.22 atmP = 1.22 atm

PV = nRTPV = nRT– (3.50atm)(10.0L) = n(0.0821)(300K)(3.50atm)(10.0L) = n(0.0821)(300K)– n=1.42 moles * 70.90 g Cln=1.42 moles * 70.90 g Cl22

------------------------------------- = ------------------------------------- = 101 grams101 grams

1 mole of Cl1 mole of Cl22

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Phase DiagramsPhase Diagrams

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Combined Gas Law GraphCombined Gas Law Graph

Return

Trace across the white area to see what happens as all three change.

Trace across a colored area to see what happens when one variable is held constant.


Water Vapor Pressure Vapor ChartWater Vapor Pressure Vapor ChartTemp (C.) Press. Temp (C.) Press. Temp (C.) Press.

(mmHg) (mmHg) (mmHg)

1 4.9 21 18.6 41 58.3

2 5.3 22 19.8 42 61.5

3 5.7 23 21.1 43 64.8

4 6.1 24 22.4 44 68.3

5 6.5 25 23.8 45 71.9

6 7.0 26 25.2 46 75.6

7 7.5 27 26.7 47 79.6

8 8.0 28 28.3 48 83.7

9 8.0 29 30.0 49 88.0

10 9.2 30 31.8 50 92.5

11 9.8 31 33.7 55 118.0

12 10.5 32 35.7 60 149.4

13 11.2 33 37.7 65 187.5

14 12.0 34 39.9 70 233.7

15 12.8 35 42.2 75 289.1

16 13.6 36 44.6 80 355.1

17 14.5 37 47.1 85 433.6

18 15.5 38 49.7 90 525.8

19 16.5 39 52.4 95 633.9

20 17.5 40 55.3 100 760.0


Solids Solids

In a solid:In a solid:– The particles are packed very closely together.The particles are packed very closely together.– The particles are very orderedThe particles are very ordered– The particles vibrate about fixed postions.The particles vibrate about fixed postions.– The particles in a solid exhibit very strong The particles in a solid exhibit very strong

Intermolecular Forces Intermolecular Forces

Back to State Changes


LiquidsLiquids

Particles in a liquid:Particles in a liquid:– Can slide/move past one anotherCan slide/move past one another– Exhibit weaker Intermolecular ForcesExhibit weaker Intermolecular Forces– Are arranged more randomlyAre arranged more randomly– Exhibit:Exhibit:

Surface TensionSurface Tension Capillary ActionCapillary Action

Back To State Changes


GasesGases

Particles in a gas:Particles in a gas:– Are very spread out (lots of empty space Are very spread out (lots of empty space

between particle)between particle)– Move rapidly in random, straight pathsMove rapidly in random, straight paths– Don’t exhibit intermolecular forcesDon’t exhibit intermolecular forces– Are very compressibleAre very compressible

Back To State Changes


Surface TensionSurface Tension

Definition – The measure of a liquids Definition – The measure of a liquids tendency to decrease its surface area to a tendency to decrease its surface area to a minimum. minimum. – What does this really mean?What does this really mean?

Water spiders, capillary action, and water skiing are Water spiders, capillary action, and water skiing are all illustrations of surface tension. all illustrations of surface tension.

Water Spider PictureWater Spider Picture

Back To Liquids


Capillary ActionCapillary Action

Capillary action occurs when the adhessive Capillary action occurs when the adhessive forces exceed the cohessive forces.forces exceed the cohessive forces.– Adhesion – liquid particles attract to a solid Adhesion – liquid particles attract to a solid

surface.surface.– Cohesion – liquid particles attract to each otherCohesion – liquid particles attract to each other

This explains the meniscus seen in glass This explains the meniscus seen in glass ware and how trees get water to the top of ware and how trees get water to the top of the tree.the tree.

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9/20/2015gas laws unit1 parts of chapters 2, 10, &11 all of chapter 12 – gas laws chemistry...

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