Chapter 9Chapter 9

GasesGases

Gases and Gas PressureGases and Gas Pressure GasesGases – constituent atoms and – constituent atoms and

molecules that have little molecules that have little attraction for one anotherattraction for one another

Free to move in available volumeFree to move in available volume Some properties of gasesSome properties of gases

Mixtures are always homogenousMixtures are always homogenous Very weak attraction between gas Very weak attraction between gas

moleculesmolecules Identity of neighbor is irrelevantIdentity of neighbor is irrelevant

CompressibleCompressible – volume – volume contracts when pressure is contracts when pressure is appliedapplied

0.10% of volume of gas is 0.10% of volume of gas is occupied by moleculesoccupied by molecules

Exert a measurable pressure on Exert a measurable pressure on the walls of their containerthe walls of their container

Gases and Gas PressureGases and Gas Pressure Pressure – force exerted per Pressure – force exerted per

unit areaunit area

SI unit equals Pascal (Pa)SI unit equals Pascal (Pa) 1 Pa = 1 N/m1 Pa = 1 N/m22 (1 N = 1 (1 N = 1

(kg•m)/s(kg•m)/s22)) Alternative unitsAlternative units

Millimeters of mercury (mmHg)Millimeters of mercury (mmHg) Atmosphere (atm)Atmosphere (atm)

1.0 atm = 760 mmHg = 101, 1.0 atm = 760 mmHg = 101, 325 Pa325 Pa

1.0 atm = 760 torr1.0 atm = 760 torr

Pressure:Unit area

Force

Atmospheric PressureAtmospheric Pressure

- pressure created - pressure created from the mass of from the mass of the atmosphere the atmosphere pressing down on pressing down on the earth’s the earth’s surfacesurface

Standard Standard atmospheric atmospheric pressure at sea pressure at sea level – 760 mmHglevel – 760 mmHg

Gases and Gas PressureGases and Gas Pressure

Barometer:Barometer: long thin mercury long thin mercury

filled tube sealed filled tube sealed at once end and at once end and inverted into a inverted into a dish of mercurydish of mercury

Downward Downward pressure of Hg in pressure of Hg in column equals column equals outside outside atmospheric atmospheric pressurepressure

Measuring PressureMeasuring Pressure Manometer:Manometer:

U-tube filled with U-tube filled with mercury, with one end mercury, with one end connected to the gas – connected to the gas – filled container and filled container and the other end open to the other end open to the atmosphere.the atmosphere.

Pgas < Patm; liquid Pgas < Patm; liquid level in the arm level in the arm connected to the gas-connected to the gas-filled cylinder will be filled cylinder will be higherhigher

PPgasgas + P + PHgHg = P = Patm atm

Measuring PressureMeasuring Pressure

Pgas > Patm; liquid Pgas > Patm; liquid level in the arm level in the arm connected to the gas connected to the gas – filled cylinder will – filled cylinder will be lowerbe lower

Pgas = Patm + PHg Pgas = Patm + PHg (P(PHgHg = the difference = the difference

in the heights of the in the heights of the two mercury two mercury columns)columns)

ExampleExample

What is the What is the pressure of the gas pressure of the gas inside the inside the following following apparatus (in mm apparatus (in mm Hg) if the outside Hg) if the outside pressure is 750 pressure is 750 mm Hg? (1 cm Hg mm Hg? (1 cm Hg = 10 mm Hg)= 10 mm Hg)

The Gas LawsThe Gas Laws Ideal Gas: A gas whose behavior follows the gas laws

exactly.

The physical properties of a gas can be defined by four variables:

P pressure

T temperature

V volume

n number of moles

Pressure and Volume: Pressure and Volume: Boyle’s LawBoyle’s Law

Showing the relationship between Showing the relationship between pressure and volumepressure and volume P x V = k (constant value @ specific P x V = k (constant value @ specific

temp and constant moles of gas)temp and constant moles of gas) k = 1.40 x 10k = 1.40 x 1033

V = 1/P (inverse relationship)V = 1/P (inverse relationship)

Pressure and Volume: Pressure and Volume: Boyle’s LawBoyle’s Law

Can predict a new volume of pressure is Can predict a new volume of pressure is changedchanged

PP11VV11 = = kk = P = P22VV22 P P11VV11 = = PP22VV22

The Gas LawsThe Gas LawsBoyle’s Law

VP

1

(constant n and T)PV = k

ExampleExample

A sample of helium gas has a pressure A sample of helium gas has a pressure of 3.54 atm in a container with a of 3.54 atm in a container with a volume of 23.1 L. This sample is volume of 23.1 L. This sample is transferred to a new container and transferred to a new container and the pressure is measured to be 1.87 the pressure is measured to be 1.87 atm. Assume constant temperature. atm. Assume constant temperature.

Will the volume of the gas increase of Will the volume of the gas increase of decrease?decrease?

What will be the new volume of the gas?What will be the new volume of the gas?

Volume and Temperature: Volume and Temperature: Charles’s LawCharles’s Law

Relationship between Volume and Relationship between Volume and TemperatureTemperature V = bT (b is a constant)V = bT (b is a constant) V / T = bV / T = b Can predict the new volume or Can predict the new volume or

temperaturetemperature (V(V11/T/T11)= (V)= (V22/T/T22))

=Tfinal

Vfinal

Tinitial

Vinitial

V T

The Gas LawsThe Gas LawsCharles’ Law

(constant n and P)= kT

V

ExampleExample

A 2.0 L sample of air is collected at A 2.0 L sample of air is collected at 298K and then cooled to 278 K. The 298K and then cooled to 278 K. The pressure is held constant at 1.0 atm.pressure is held constant at 1.0 atm. Does the volume increase of decrease?Does the volume increase of decrease? Calculate the volume of the air at 278 Calculate the volume of the air at 278

K?K?

Volume and Moles: Volume and Moles: Avogadro’s LawAvogadro’s Law

Relationship between volume of gas Relationship between volume of gas and number moles of gasand number moles of gas V is directly proportional to nV is directly proportional to n V = an or V / n = a (a = constant)V = an or V / n = a (a = constant) Can predict the new volume or new Can predict the new volume or new

moles of gas at constant pressure and moles of gas at constant pressure and temperaturetemperature

(V(V11/n/n11) = (V) = (V22/n/n22))

=nfinal

Vfinal

ninitial

Vinitial

The Gas LawsThe Gas LawsAvogadro’s Law

V n (constant T and P)= k

n

V

ExampleExample

Consider two samples of nitrogen Consider two samples of nitrogen gas (composed Ngas (composed N22 molecules). molecules). Sample 1 contains 1.5 mol of NSample 1 contains 1.5 mol of N22 and and has a volume of 36.7 L at 25oC and 1 has a volume of 36.7 L at 25oC and 1 atm. Sample 2 has a volume of 16.0 atm. Sample 2 has a volume of 16.0 L at 25L at 25ooC and 1 atm. Calculate the C and 1 atm. Calculate the number moles of Nnumber moles of N22 in sample 2 in sample 2

SummarySummary

Boyle’s Law Charles’ Law Avogadro’s Law

constant T & n

constant P & constant P & nn

constant P & constant P & TT

V = 1/PV = 1/P V = TV = T V = nV = n

PinitialVinitial = PfinalVfinal

=Tfinal

Vfinal

Tinitial

Vinitial

=nfinal

Vfinal

ninitial

Vinitial

9.3 The Ideal Gas law9.3 The Ideal Gas law

Different gasses show similar Different gasses show similar physical behavior (unlike solid or physical behavior (unlike solid or liquid)liquid)

Relationship of variable – gas lawsRelationship of variable – gas laws

Ideal gas – behavior follows the gas laws Ideal gas – behavior follows the gas laws exactlyexactly

9.3 The Ideal Gas law9.3 The Ideal Gas law

Describes how the volume of a gas Describes how the volume of a gas is affected by changes in pressure, is affected by changes in pressure, temperature and moles.temperature and moles.

PV = nRT;PV = nRT;R = 0.082058

K mol

L atm

9.39.3 The Ideal Gas LawThe Ideal Gas LawPV = nRT

Standard Temperature and Pressure (STP) for Gases

Ideal Gas Law:

P = 1 atm

T = 0 °C (273.15 K)

R is the gas constant and is the same for all gases.

R = 0.082058K mol

L atm

The Ideal Gas LawThe Ideal Gas Law

= 22.414 LV =P

nRT=

What is the volume of 1 mol of gas at STP?

(1 atm)

(1 mol) 0.082058K mol

L atm(273.15 K)

ExampleExample A helium gas cylinder of the sort used to fill A helium gas cylinder of the sort used to fill

balloons have a volume of 43.8 L and balloons have a volume of 43.8 L and pressure of 1.51 x 10pressure of 1.51 x 1044 kPa at 25.0oC. How kPa at 25.0oC. How many moles of helium are in the tank?many moles of helium are in the tank?

What volume is occupied by 0.250 mol of What volume is occupied by 0.250 mol of carbon dioxide gas at 25.0carbon dioxide gas at 25.0ooC and 371 torr?C and 371 torr?

A 0.250 mol sample of argon gas has a A 0.250 mol sample of argon gas has a volume of 9.00L at a pressure of 875 mmHg. volume of 9.00L at a pressure of 875 mmHg. What is the temperature (in What is the temperature (in ooC) of the gas?C) of the gas?

Stoichiometric Stoichiometric Relationships with GasesRelationships with Gases

The reaction used in the deployment The reaction used in the deployment of automobile airbags is the high of automobile airbags is the high temperature decomposition of temperature decomposition of sodium azide, NaNsodium azide, NaN33, to produce N, to produce N22 gas. How many liters of Ngas. How many liters of N22 at 1.15 at 1.15 atm and 30.0atm and 30.0ooC are produced by the C are produced by the decompostion of 45.0g NaNdecompostion of 45.0g NaN33??2Na(s) + 3N2(g)2NaN3(s)

ExamplesExamples

Consider the reaction represented Consider the reaction represented by the equation by the equation

PP44(s) + 6 H(s) + 6 H22(g) (g) 4H 4H33(g)(g)

What is the amount of PWhat is the amount of P44 is required to is required to react with 5.39 L of hydrogen gas at react with 5.39 L of hydrogen gas at 27.027.0ooC and 1.25 atm?C and 1.25 atm?

ExampleExample

Ammonia is commonly used as a Ammonia is commonly used as a fertilizer to provide a source of fertilizer to provide a source of nitrogen for plants. A sample of nitrogen for plants. A sample of NHNH33(g) occupies a volume of 5.00 L (g) occupies a volume of 5.00 L at STP. What moles will this sample at STP. What moles will this sample occupy?occupy?

Combine Gas LawCombine Gas Law

is an expression obtained by is an expression obtained by mathematically combining Boyle’s mathematically combining Boyle’s and Charles’ lawand Charles’ law

PP11VV11 = = PP22VV22 @ constant n@ constant n

TT11 T T22

can predict P, V or T when can predict P, V or T when condition is changedcondition is changed

ExamplesExamples

A sample of diborane gas B2H6, a A sample of diborane gas B2H6, a substance that bursts into flames substance that bursts into flames when exposed to air, has a pressure when exposed to air, has a pressure of 0.454 atm at a temperature of -of 0.454 atm at a temperature of -1515ooC and a volume of 3.48L. If C and a volume of 3.48L. If condition are changed so that the condition are changed so that the temperature is 36temperature is 36ooC and the C and the pressure is 0.616 atm, what is the pressure is 0.616 atm, what is the new volume of the sample?new volume of the sample?

ExamplesExamples

Consider a sample of hydrogen gas Consider a sample of hydrogen gas of 63of 63ooC with a volume of 3.65L at a C with a volume of 3.65L at a pressure of 4.55 atm. The pressure pressure of 4.55 atm. The pressure is changed to 2.75 atm and the gas is changed to 2.75 atm and the gas is cooled to -35is cooled to -35ooC. Calculate the C. Calculate the new volume of the gasnew volume of the gas

13.6 Dalton’s Law of 13.6 Dalton’s Law of Partial PressurePartial Pressure

A.A. Gas laws apply to mixtures of gasesGas laws apply to mixtures of gases B.B. Dalton's law of partial pressure – Dalton's law of partial pressure – PPtotatotall = P = P11 + P + P22 + P + P33 + ….. at + ….. at constant constant

V, TV, T

where Pwhere P11, P, P22, ….refer to the pressure of the , ….refer to the pressure of the individual gases in the mixtureindividual gases in the mixture

Mole Fraction (X) =Xi =

Ptotal

PiXi = ntotal

ni or

Total moles in mixture

Moles of component

Partial PresssurePartial Presssure

C.C. Partial pressures refer to the Partial pressures refer to the pressure each individual gas would pressure each individual gas would exert if it were alone in the exert if it were alone in the container (Pcontainer (P11, P, P22, …), …)

1.1. Total pressure depends on the Total pressure depends on the total molar amount of gas presenttotal molar amount of gas present

PPtotaltotal = n = ntotaltotal (RT/V) (RT/V)

ExamplesExamples

A 2.0 L flask contains a mixture of A 2.0 L flask contains a mixture of nitrogen gas and oxgyen gas at nitrogen gas and oxgyen gas at 25.025.0ooC. The total pressure of the gas C. The total pressure of the gas mixture is 0.91 atm, and the mixture mixture is 0.91 atm, and the mixture is known to contain 0.050 mol of Nis known to contain 0.050 mol of N22. . Calculate the partial pressure of Calculate the partial pressure of oxygen and the moles of oxygen oxygen and the moles of oxygen presentpresent

ExamplesExamples

Mixture of helium and oxygen are use Mixture of helium and oxygen are use in the “air” tanks of underwater in the “air” tanks of underwater divers for deep dives. For a particular divers for deep dives. For a particular dive, 12.0L of Odive, 12.0L of O22 at 25.0 at 25.0ooC and 1.0 C and 1.0 atm and 46.0 L of He at 25atm and 46.0 L of He at 25ooC and 1.0 C and 1.0 atm were both pumped into a 5.0 L atm were both pumped into a 5.0 L tank. Calculate the partial pressure, tank. Calculate the partial pressure, moles fraction of each gas and the moles fraction of each gas and the total pressure in the tank at 25.0total pressure in the tank at 25.0ooCC

The Kinetic Molecular The Kinetic Molecular Theory of GasTheory of Gas

A.A. Model that can explain the behavior of Model that can explain the behavior of gasesgases..AssumptionsAssumptions

1.1. A gas consists of particles in A gas consists of particles in constant random motionconstant random motion

2.2. Most of the volume of a gas is empty Most of the volume of a gas is empty spacesspaces

3.3. The attractive and repulsive forces The attractive and repulsive forces between molecules of gases are negligiblebetween molecules of gases are negligible

4.4. The total kinetic energy of the gas The total kinetic energy of the gas particles is constant at constant Tparticles is constant at constant T

5.5. Average Ek α TAverage Ek α T

The Kinetic-Molecular The Kinetic-Molecular Theory of GasesTheory of Gases

averagespeed

molarmass

The Kinetic-Molecular The Kinetic-Molecular Theory of GasesTheory of Gases

Graham’s Law: Diffusion Graham’s Law: Diffusion and Effusion of Gasesand Effusion of Gases

Diffusion: The mixing of different gases by molecular motion with frequent molecular collisions

Graham’s Law: Diffusion Graham’s Law: Diffusion and Effusion of Gasesand Effusion of Gases

Effusion: The escape of a gas through a pinhole into a vacuum without molecular collisions.

Graham’s Law: Rate1

m

The Behavior of Real The Behavior of Real GasesGases

The volume of a real gas is larger than predicted by the ideal gas law.

The Behavior of Real The Behavior of Real GasesGases

Attractive forces between particles become more important at higher pressures.

The Behavior of Real The Behavior of Real GasesGases

PV2

n2

= nRT+ V - n

van der Waals equation

ba

Correction for intermolecular

attractions.

Correction for molecular volume.

examplesexamples

Assume that you have 0.500 mol of Assume that you have 0.500 mol of NN22 in a volume of 0.600L at 300K. in a volume of 0.600L at 300K. Calculate the pressure in the Calculate the pressure in the atmosphere using both the ideal gas atmosphere using both the ideal gas law and the van der Waals equation. law and the van der Waals equation. For N2, a = 1.35 (LFor N2, a = 1.35 (L22·atm)mol·atm)mol22, and b , and b = 0.0387 L/mol= 0.0387 L/mol