GasesGases

Chapter 14Chapter 14

SectionsSections

1.1. The Gas LawsThe Gas Laws

2.2. The Combined Gas Laws and The Combined Gas Laws and Avogadro’s PrincipleAvogadro’s Principle

3.3. Ideal Gas LawIdeal Gas Law

4.4. Gas StoichiometryGas Stoichiometry

The Gas LawsThe Gas Laws

Chapter 14.1Chapter 14.1

The Gas LawsThe Gas Laws

Boyles LawBoyles Law

Charles LawCharles Law

Gay-Lussac’s LawGay-Lussac’s Law

Each law relates two variables to the Each law relates two variables to the behavior of gasses.behavior of gasses.– PressurePressure– TemperatureTemperature– VolumeVolume– amountamount

Kinetic Theory ReviewKinetic Theory Review

K-M theory suggests gas particles behave K-M theory suggests gas particles behave differently than liquids and solids.differently than liquids and solids.

K-M assumes the following are true about K-M assumes the following are true about gasses:gasses:– Gas particles do not attract or repel each otherGas particles do not attract or repel each other– Gas particles are much smaller than the distances Gas particles are much smaller than the distances

between thembetween them– Gas particles are in constant, random motionGas particles are in constant, random motion– No kinetic energy is lost when gas particles collide No kinetic energy is lost when gas particles collide

with each other or with the walls of the containerwith each other or with the walls of the container– All gasses have the same average kinetic energy at a All gasses have the same average kinetic energy at a

given temperaturegiven temperature

The Nature of GassesThe Nature of Gasses

Actual gasses don’t obey all the Actual gasses don’t obey all the assumptions made by K-M theoryassumptions made by K-M theory– But their behavior approximates the theoryBut their behavior approximates the theory

Notice that all assumptions of K-M theory Notice that all assumptions of K-M theory are based on the four factors mentioned are based on the four factors mentioned about the gas laws:about the gas laws:– Number of particles presentNumber of particles present– TemperatureTemperature– PressurePressure– Volume of gasVolume of gas

Boyle’s LawBoyle’s Law

Robert Boyle (1627-1691), Irish, studied Robert Boyle (1627-1691), Irish, studied relationship between pressure and relationship between pressure and volume.volume.– An inverse relationshipAn inverse relationship

Boyle’s Law:Boyle’s Law:– The volume of a given amount of gas at The volume of a given amount of gas at

constant temperature varies inversely with the constant temperature varies inversely with the pressurepressure

– So at any two different times, for the same So at any two different times, for the same gas, the product of pressure and volume will gas, the product of pressure and volume will be the samebe the same

– PP11VV11 = P = P22VV22

Boyle’s LawBoyle’s Law

http://www.chem.iastate.edu/group/Greenbowe/sections/prhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.htmlojectfolder/flashfiles/gaslaw/boyles_law_graph.html

Boyle’s Law ProblemBoyle’s Law Problem

The volume of a gas at 99.0 kPa is 300.0 mL. If The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what is the the pressure is increased to 188 kPa, what is the new volume?new volume?– PP11VV11 = P = P22VV22

– 99.0kPa x 300.0mL = 188kPa x ?99.0kPa x 300.0mL = 188kPa x ?

99.0kPa x 300.0mL99.0kPa x 300.0mL = ? = ?188kPa188kPa

158.0mL

Charles’s LawCharles’s Law

Jacque Charles (1746-1823) FrenchJacque Charles (1746-1823) French– Studied relationship between volume and Studied relationship between volume and

temperaturetemperature– As temperature increases , so does volume of gasAs temperature increases , so does volume of gas

MUST BE IN KELVINMUST BE IN KELVIN

TTkk = T = Tcc + 273 + 273

– K-M Theory says as temperature increases, K-M Theory says as temperature increases, particles move faster, so must be further away.particles move faster, so must be further away.

– Relationship between temperature and volume is a Relationship between temperature and volume is a straight line with positive slopestraight line with positive slope

V1

T1

V2

T2

=

Charles’s LawCharles’s Law

Law: The volume of a gas is directly proportional to the Law: The volume of a gas is directly proportional to the temperature, expressed in Kelvin, at a constant pressuretemperature, expressed in Kelvin, at a constant pressure

Charles’s Law ProblemCharles’s Law Problem

A gas at 89A gas at 89°C occupies a volume of 0.67L. C occupies a volume of 0.67L. At what Celsius temperature will the At what Celsius temperature will the volume increase to 1.12L?volume increase to 1.12L?

V1

T1

V2

T2

=

0.67L89°C + 273

1.12L?

= ? x 0.67L362 °K

1.12L=

? 1.12L x 362°K0.67L

= = 605.1 K

605.1°K – 273K = 332°C

Gay-Lussac’s LawGay-Lussac’s Law

Joseph Gay-Lussac studied relationship Joseph Gay-Lussac studied relationship between temperature and pressure of a between temperature and pressure of a constained gas at fixed volume:constained gas at fixed volume:– Found direct proportion exists between Found direct proportion exists between

temperature (KELVIN) and pressuretemperature (KELVIN) and pressure– Mathematically:Mathematically:

P1

T1

P2

T2

=

Gay-Lussac’s Law ProblemGay-Lussac’s Law Problem

A gas in a sealed container has a pressure A gas in a sealed container has a pressure of 125kPa at a temperature of 30of 125kPa at a temperature of 30°°C. If the C. If the pressure of the container is increased to pressure of the container is increased to 201kPa, what is the new temperature?201kPa, what is the new temperature?

Convert temp to K: 30+273 = 303KConvert temp to K: 30+273 = 303K

P1

T1

P2

T2

= 125kPa303K

201kPaT2

=

T2201kPa x 303K125kPa

= = 487K

487K – 273 = 214°C°C

Combined Gas Law and Combined Gas Law and Avogadro’s PrincipleAvogadro’s Principle

Chapter 14, Section 2Chapter 14, Section 2

Combined Gas LawCombined Gas Law

Combine Boyle’s Law, Charles’s Law, and Combine Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law into one:Gay-Lussac’s Law into one:– PP11VV11/T/T11 = P = P22VV22/T/T22

– Let’s you work out problems involving more Let’s you work out problems involving more variablesvariables

Use known variables under one set of conditions to Use known variables under one set of conditions to find a value of a missing variable.find a value of a missing variable.

Avogadro’s PrincipleAvogadro’s Principle

1811, Avogadro suggested that equal 1811, Avogadro suggested that equal volumes of gasses at the same volumes of gasses at the same temperature and pressure contain the temperature and pressure contain the same number of particles:same number of particles:– K-M theory says the same thingK-M theory says the same thing

Particles are so far apart that size doesn’t matterParticles are so far apart that size doesn’t matter

– 1 mole of particles is 6.02 x 101 mole of particles is 6.02 x 102323 particles particles– 1 mole of gas particles is 22.4 L at 0.001 mole of gas particles is 22.4 L at 0.00°° C C

and 1.00 atm pressureand 1.00 atm pressureStandard Temperature and Pressure (STP)Standard Temperature and Pressure (STP)

Example ProblemsExample ProblemsDetermine the volume of a container that holds Determine the volume of a container that holds 2.4 mole of gas at STP:2.4 mole of gas at STP:– 2.4 mol x 22.4L/1mol = 53.76 L = 54L2.4 mol x 22.4L/1mol = 53.76 L = 54L

What volume will 1.02 mol of carbon monoxide What volume will 1.02 mol of carbon monoxide gas occupy at STP?gas occupy at STP?– 1.02 mol x 22.4L/1 mol = 22.8 L1.02 mol x 22.4L/1 mol = 22.8 L

A balloon will rise off the ground when it contains A balloon will rise off the ground when it contains 0.0226 mol of He in a volume of 0.460L. How 0.0226 mol of He in a volume of 0.460L. How many moles of He are needed to make the many moles of He are needed to make the balloon rise when volume is 0.885L, assuming balloon rise when volume is 0.885L, assuming temperature and pressure are constant?temperature and pressure are constant?– 0.0226 mol x 0.865L/0.460L = 0.0425 mol0.0226 mol x 0.865L/0.460L = 0.0425 mol

Ideal Gas LawIdeal Gas Law

Chapter 14, Section 3Chapter 14, Section 3

Ideal Gas LawIdeal Gas Law

PP11VV11/T/T11 = P = P22VV22/T/T22 can be interpreted as can be interpreted as saying that PV/T is a constant, saying that PV/T is a constant, kk– kk is based on the amount of gas present, is based on the amount of gas present, nn– Experiments have shown that Experiments have shown that kk = = nnR, where n R, where n

is number of moles and R is a constant.is number of moles and R is a constant.

Ideal Gas Law is therefore:Ideal Gas Law is therefore:– PV = nRTPV = nRT

Ideal Gas Constant (R), is:Ideal Gas Constant (R), is:– 0.0821 L*atm/mol-K (most often used)0.0821 L*atm/mol-K (most often used)– 8.314 L*kPa/mol K8.314 L*kPa/mol K– 62.4 L*mm Hg/mol K62.4 L*mm Hg/mol K

Real vs. IdealReal vs. Ideal

Ideal Gas AssumptionsIdeal Gas Assumptions– Particles take up no spaceParticles take up no space– No intermolecular forcesNo intermolecular forces– Follows ideal gas law under all temperatures and Follows ideal gas law under all temperatures and

pressurespressures

Real GasReal Gas– All gas particles take up some spaceAll gas particles take up some space– All gas particles are subject to intermolecular forcesAll gas particles are subject to intermolecular forces– Most gasses act like ideal gasses at many/most Most gasses act like ideal gasses at many/most

temperatures and pressurestemperatures and pressures– Deviations occur at extremely high pressures and Deviations occur at extremely high pressures and

extremely low temperatures because intermolecular extremely low temperatures because intermolecular forces start to show effects.forces start to show effects.

Applying Ideal Gas LawApplying Ideal Gas Law

If you know any three variables, you can If you know any three variables, you can solve for the fourth:solve for the fourth:– You can solve for You can solve for nn (number of moles) (number of moles)– Combined gas law, you cannot.Combined gas law, you cannot.

You can use ideal gas law allows you to You can use ideal gas law allows you to solve for molar mass and density, if mass solve for molar mass and density, if mass is knownis known– n n (number of moles) = (number of moles) = mm (mass) / (mass) / MM (molar (molar

mass)mass)– Sustitute Sustitute m/Mm/M for for nn

PV = PV = mmRT/RT/MM

Solving for DensitySolving for Density

Density is Density is mm (mass) / (mass) / VV (volume) (volume)– Substitute Substitute DD (Density) for (Density) for m/Vm/V– M = M = mmRT/PV = DRT/PRT/PV = DRT/P– Or D = MP/RTOr D = MP/RT

Example ProblemExample Problem

Calculate the number of moles of gas Calculate the number of moles of gas contained in a 3.0L vessel at 3.00x10contained in a 3.0L vessel at 3.00x1022K K with a pressure of 1.50 atmwith a pressure of 1.50 atm– Find: molesFind: moles– Known: V = 3.0LKnown: V = 3.0L– Known: T = 3.00x10Known: T = 3.00x1022

– Known: P = 1.50 atmKnown: P = 1.50 atm– Known: R = 0.0821 L*Atm/mol*KKnown: R = 0.0821 L*Atm/mol*K– PV=nRTPV=nRT– Solve for nSolve for n– N = PV/RTN = PV/RT

Example ProblemExample Problem

nn = PV/RT – substitute in known = PV/RT – substitute in known information:information:

nn = = (1.50atm)(3.0L)(1.50atm)(3.0L)

(0.0821 L*atm/mol*K)(3.00 x 10(0.0821 L*atm/mol*K)(3.00 x 1022 K) K)

n = 0.18 mole

Evaluation: 1 mole of gas occupies 22.4L at STP. The volume is much less than 1 mole. Temperature and pressures are not dramatically different than STP. Slightly higher pressure means more gas. Slightly higher temperature means a little less gas. Units of the answer is moles – every unit cancels except 1/ 1/mol.

Example Problem – Using Molar Example Problem – Using Molar MassMass

What is the molar mass of a pure gas that What is the molar mass of a pure gas that has a density of 1.40g/L at STP?has a density of 1.40g/L at STP?– Find: molar mass (g/mole)Find: molar mass (g/mole)– Known: D = 1.40g/LKnown: D = 1.40g/L– Known: T=0.00Known: T=0.00°C°C– Known: P = 1.00atmKnown: P = 1.00atm– Known: R = 0.0821 L*atm/mol*KKnown: R = 0.0821 L*atm/mol*K– Step1: Convert T to KelvinStep1: Convert T to Kelvin

TTkk = T = Tcc+273+273

TTkk = 273 = 273

Molar Mass ProblemMolar Mass Problem

Use density form of ideal gas lawUse density form of ideal gas law– M = DRT/PM = DRT/P– Substitute known values:Substitute known values:

M = M = (1.40 g/L)(0.0821 L*atm/mol*K)(273K)(1.40 g/L)(0.0821 L*atm/mol*K)(273K)1 atm1 atm

M = 31.4 g/mol

Gas StoichiometryGas Stoichiometry

Chapter 14, Section 4Chapter 14, Section 4

Gas Stoichiometry – Volume onlyGas Stoichiometry – Volume only

Example:Example:– 2C2C44HH10(g)10(g) + 13 O + 13 O2(g)2(g) →→ 8 CO 8 CO2(g)2(g) + 10 H + 10 H22OO(g)(g)

– Remember: Avogadro’s principle states that 1 Remember: Avogadro’s principle states that 1 mole of gas is 22.4 Lmole of gas is 22.4 L

– For volume to volume calculations, you only For volume to volume calculations, you only need to know mole ratiosneed to know mole ratios

– 2 L of butane reacts, it involves2 L of butane reacts, it involves13 L of O13 L of O22 (2L x 13 O (2L x 13 O22/2 C/2 C44HH1010))

8 L of CO8 L of CO2 2 (2L x 8CO(2L x 8CO22/2/2 CC44HH1010))

10 L of H10 L of H22OO

Gas Stoichiometry – Volume & MassGas Stoichiometry – Volume & Mass

Ammonia is synthesized from hydrogen Ammonia is synthesized from hydrogen and nitrogenand nitrogen– NN2(g)2(g) + 3 H + 3 H2(g)2(g) →→ 2 NH 2 NH3(g)3(g)

– If 5.00 L of nitrogen reacts completely by this If 5.00 L of nitrogen reacts completely by this reaction at 3.0 atm and 298K, how many reaction at 3.0 atm and 298K, how many grams of ammonia are produced?grams of ammonia are produced?

– Analyze: you are given L, pressure, Analyze: you are given L, pressure, temperature. You are asked to find grams of temperature. You are asked to find grams of NHNH33..

– Known:Known:

– VVNN = 5.00L, P = 3.00 atm, T = 298K = 5.00L, P = 3.00 atm, T = 298K

Gas Stoichiometry (Cont)Gas Stoichiometry (Cont)

1.1. Determine the mole ratio neededDetermine the mole ratio needed2 volumes NH2 volumes NH33 /1 volume N /1 volume N22

2.2. Determine volume of ammonia Determine volume of ammonia produced:produced:

5.00L N5.00L N22 x x 2 volumes NH3 2 volumes NH3

1 volume N2

= 10L NH3

3. Rearrange Gas Law to solve for n (moles):

PV = nRT n = PVRT

4. Substitute values and solve for n

Gas Stoichiometry (Cont).Gas Stoichiometry (Cont).

4. Substitute values and solve for n

n = (3.00 atm)(10.00L)(0.0821 L*atm/mol*k)(298K)

= 1.23 mol NH3

5. Find the mass (M) of NH3 by finding the molecular mass and converting the moles in grams

Molecular mass = 1N(14.01amu) + 3 H(1.01amu) = 17.04amu

1.23 mol NH3 x 17.04 g/mol = 21.0 g NH3