Chapter 5: GasesChapter 5: Gases

5.1 Pressure5.1 Pressure

Gaseous State of MatterGaseous State of Matter

has no distinct has no distinct or or __________ so fills any container__________ so fills any container

is easily compressedis easily compressed completely with any completely with any

other gasother gas exerts pressure on its exerts pressure on its

surroundingssurroundings

Measuring PressureMeasuring Pressure barometer:barometer:

measures atmospheric measures atmospheric pressurepressure

invented by Torricelli, Italian invented by Torricelli, Italian scientist in 1643scientist in 1643

glass tube filled with glass tube filled with mercury is inverted in a dishmercury is inverted in a dish

mercury flows out of the mercury flows out of the tube until pressure of the tube until pressure of the Hg inside the tube is equal Hg inside the tube is equal to the atmospheric pressure to the atmospheric pressure on the Hg in the dishon the Hg in the dish

Measuring PressureMeasuring Pressure

pressure:pressure: results from mass of air being pulled results from mass of air being pulled

toward the earth by gravitytoward the earth by gravity varies with altitude and weather varies with altitude and weather

conditionsconditions

Measuring PressureMeasuring Pressure

manometer:manometer: measures pressure of gas in a containermeasures pressure of gas in a container gas has less pressure than atmosphere gas has less pressure than atmosphere

if the Hg is closer to chamberif the Hg is closer to chamber gas has more pressure than atmosphere gas has more pressure than atmosphere

if the Hg is further from chamberif the Hg is further from chamber

gas pressure =gas pressure =

Units of PressureUnits of Pressure : most common since use Hg : most common since use Hg in manometers and barometersin manometers and barometers

: equal to mmHg: equal to mmHg (atm)(atm) (Pa): (Pa): SI unit; equal to SI unit; equal to N/mN/m22

1atm = 760mmHg = 760torr = 101,325Pa = 1atm = 760mmHg = 760torr = 101,325Pa = 101.325kPa101.325kPa

Chapter 5: GasesChapter 5: Gases

5.2 Gas Laws5.2 Gas Laws

Boyle’s LawBoyle’s Law Discovered by Irish chemist, Discovered by Irish chemist,

Robert BoyleRobert Boyle Used a J-shaped tube to Used a J-shaped tube to

experiment with varying experiment with varying pressures in multistory home pressures in multistory home and effects on volume of and effects on volume of enclosed gasenclosed gas

P and V are P and V are proportionalproportional

PV = kPV = k holds precisely at very low holds precisely at very low

pressurespressures

Charles’ LawCharles’ Law discovered by French discovered by French

physicist, Jacques Charles physicist, Jacques Charles in 1787in 1787

first person to fill balloon first person to fill balloon with hydrogen gas and with hydrogen gas and make solo balloon flightmake solo balloon flight

V and T are V and T are proportionalproportional

V = kTV = kT

Charles’ LawCharles’ Law

for any gas, at -273.2°C, the volume for any gas, at -273.2°C, the volume is zero is zero

since negative volumes cannot exist, there since negative volumes cannot exist, there cannot be a temperature lower than cannot be a temperature lower than absolute absolute zerozero (-273.2°C or 0 K) (-273.2°C or 0 K)

never actually been reached never actually been reached

(0.000001 K has been)(0.000001 K has been) Kelvin system has no negative valuesKelvin system has no negative values

Example

Avogadro’s LawAvogadro’s Law

Discovered by Italian Discovered by Italian chemistry, Avogadro in 1811chemistry, Avogadro in 1811

proposed that equal volumes proposed that equal volumes of gases at the same of gases at the same temperature and pressure temperature and pressure contain the same number of contain the same number of particlesparticles

V = knV = kn V and n are V and n are

proportionalproportional Example

Gay-Lussac’s LawGay-Lussac’s Law discovered in discovered in

1802 by Joseph 1802 by Joseph Gay-LussacGay-Lussac

P = kTP = kT P and T are P and T are

proportionalproportional

Example

Example: Pressure ConversionsExample: Pressure Conversions

The pressure of a gas is measured as 49 The pressure of a gas is measured as 49 torr. Represent this pressure in torr. Represent this pressure in atmospheres, Pascals, and mmHg.atmospheres, Pascals, and mmHg.

Try This!Try This!

The pressure of a gas is measured as The pressure of a gas is measured as 2.5 atm. Represent this pressure in 2.5 atm. Represent this pressure in Torr, Pascals, and mmHg.Torr, Pascals, and mmHg.

1900 Torr1900 Torr

253312.5 Pa253312.5 Pa

1900 mmHg1900 mmHg

Example: Boyle’s LawExample: Boyle’s Law

Consider a 1.53-L sample of gaseous SOConsider a 1.53-L sample of gaseous SO22 at a pressure of 5.6 x 10at a pressure of 5.6 x 1033 Pa. If the Pa. If the pressure is changed to 1.5 x 10pressure is changed to 1.5 x 1044 Pa at Pa at constant temperature, what will be the constant temperature, what will be the new volume of the gas?new volume of the gas?

Example: Charles’ Law & Example: Charles’ Law & Temp.Temp.

A sample of gas at 15°C and 1 atm has a A sample of gas at 15°C and 1 atm has a volume of 2.58 L. What volume will this volume of 2.58 L. What volume will this gas occupy at 38°C and 1 atm?gas occupy at 38°C and 1 atm?

Example: Avogadro’s LawExample: Avogadro’s LawSuppose we have a 12.2-L sample of gas Suppose we have a 12.2-L sample of gas containing 0.50 mol Ocontaining 0.50 mol O22 at a pressure of 1 atm at a pressure of 1 atm and temperature of 25°C. If all of this Oand temperature of 25°C. If all of this O22 were were converted to Oconverted to O33 (ozone) at the same temperature (ozone) at the same temperature and pressure, what would be the volume of Oand pressure, what would be the volume of O33??

Example: Gay-Lussac’s LawExample: Gay-Lussac’s Law

The gas in an aerosol can is at a pressure The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a warn the user not to keep the can in a place where temperature exceeds 52°C. place where temperature exceeds 52°C. What would the gas pressure be in the What would the gas pressure be in the can at 52°C?can at 52°C?

Ch. 5 GasesCh. 5 Gases

5.3 Ideal Gas law5.3 Ideal Gas law

Combinations of Gas LawsCombinations of Gas Laws

Combined Gas LawCombined Gas Law::

Ideal Gas LawIdeal Gas Law

Universal Gas Constant (R)Universal Gas Constant (R) combined proportionality constantcombined proportionality constant

equals 0.08206 L*atm/mol*Kequals 0.08206 L*atm/mol*K equals 8.314 J/mol*Kequals 8.314 J/mol*K

Real Gases vs. Ideal GasesReal Gases vs. Ideal Gases

an ideal gas is a hypothetical an ideal gas is a hypothetical substance that obeys the ideal gas substance that obeys the ideal gas lawlaw

real gases approach this behavior real gases approach this behavior closest at low pressure and high closest at low pressure and high temperaturetemperature

Ch. 5 GasesCh. 5 Gases

5.4 Gas Stoichiometry5.4 Gas Stoichiometry

Molar Volume Molar Volume

L = 1 L = 1 mol at STPmol at STP

Standard Standard Temperature Temperature and Pressureand Pressure (STP):(STP):

Molar Mass and DensityMolar Mass and Density

Example 1Example 1

CaO is made from decomposition of CaO is made from decomposition of CaCOCaCO33. Find the volume of CO. Find the volume of CO22 at STP at STP made from decomposition of 152 g made from decomposition of 152 g CaCOCaCO33..

Example 2Example 2A sample of CHA sample of CH44 gas having a volume of gas having a volume of 2.80 L at 25°C and 1.65 atm was mixed 2.80 L at 25°C and 1.65 atm was mixed with sample of Owith sample of O22 gas having volume of gas having volume of 35.0 L at 31°C and 1.25 atm. The mixture 35.0 L at 31°C and 1.25 atm. The mixture formed COformed CO22 and H and H22O. Find the volume of O. Find the volume of COCO22 at 2.50 atm and 125°C. at 2.50 atm and 125°C.

Example 2Example 2

Find the limiting reactant:Find the limiting reactant:

Example 2Example 2

Find the amount of product produced:Find the amount of product produced:

Example 2Example 2

Find the volume of product produced:Find the volume of product produced:

Example 3Example 3

If the density of a sample of gas is 1.95 If the density of a sample of gas is 1.95 g/L at 1.50 atm and 27°C, find the molar g/L at 1.50 atm and 27°C, find the molar mass of gas. What could the identity of mass of gas. What could the identity of the gas be?the gas be?

Ch. 5 GasesCh. 5 Gases

5.5 Dalton’s Law of Partial 5.5 Dalton’s Law of Partial PressuresPressures

Partial PressurePartial Pressure

for a mixture of gases, the total for a mixture of gases, the total pressure is the sum of the pressure is the sum of the pressures each gas would exert if pressures each gas would exert if it were aloneit were alone

using the ideal gas law, can using the ideal gas law, can change to:change to:

Partial PressuresPartial Pressures

Example 1Example 1

47 L He and 12 L O47 L He and 12 L O22 at 25°C and 1.0 at 25°C and 1.0 atm were pumped into a tank with a atm were pumped into a tank with a volume of 5.0 L. Calculate the partial volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure of each gas and the total pressure in the tank.pressure in the tank.

Example 1Example 1

Find moles of each gas:Find moles of each gas:

Example 1Example 1

Find the new P of each gas:Find the new P of each gas:

Find the total pressure of the gases:Find the total pressure of the gases:

Partial PressurePartial Pressure

shows that the identities of the shows that the identities of the gases do not matter, just the gases do not matter, just the number of molesnumber of moles

so, for ideal gases:so, for ideal gases:1.1. size of gas molecule is not importantsize of gas molecule is not important2.2. forces between molecules is not forces between molecules is not

importantimportant these are the things that would these are the things that would

change with the identity of the gaschange with the identity of the gas

Mole FractionMole Fraction

Mole FractionMole Fraction: : ratio of ratio of number of number of moles of a moles of a certain certain component of component of a mixture to a mixture to number of number of moles total in moles total in mixturemixture

Water DisplacementWater Displacement when gas is collected using water displacement, when gas is collected using water displacement,

there is always a mixtures of gasesthere is always a mixtures of gases the pressure of water vapor varies with the pressure of water vapor varies with

temperature and will be given in a problemtemperature and will be given in a problem

Ch. 5 GasesCh. 5 Gases

5.6 Kinetic Molecular Theory5.6 Kinetic Molecular Theory

The Kinetic Molecular The Kinetic Molecular TheoryTheory

model of gas behavior so only an model of gas behavior so only an approximationapproximation

1.1. volume of particles is assumed to be zerovolume of particles is assumed to be zero

2.2. particles are in constant motionparticles are in constant motion

3.3. particles exert no forces on each other particles exert no forces on each other (no attraction or repulsion)(no attraction or repulsion)

4.4. kinetic energy kinetic energy is proportional tois proportional toKelvin Kelvin temperaturetemperature

Boyle’s Law: P and VBoyle’s Law: P and V

decrease in volume means that decrease in volume means that particles will hit wall more often and particles will hit wall more often and that will cause P increasethat will cause P increase

Gay-Lussac’s Law: P and TGay-Lussac’s Law: P and T

the speed of particles increases as T the speed of particles increases as T increases so they hit the wall more increases so they hit the wall more often and with greater force and P often and with greater force and P increasesincreases

Charles’ Law: V and TCharles’ Law: V and T

increase in T causes and increase in increase in T causes and increase in particle speed so they hit the wall more particle speed so they hit the wall more oftenoften

to keep P constant, the V must increaseto keep P constant, the V must increase

Avogadro’s’ Law: V and nAvogadro’s’ Law: V and n

increase in number of gas molecules increase in number of gas molecules would cause increase in P if V were would cause increase in P if V were held constantheld constant

to keep P constant, V must increaseto keep P constant, V must increase

Dalton’s LawDalton’s Law

Kinetic Molecular Theory Kinetic Molecular Theory assumes that all particles are assumes that all particles are independent of each otherindependent of each other

TemperatureTemperature

Kelvin temperature is a sign of the Kelvin temperature is a sign of the random motions of gas particles random motions of gas particles

higher higher TT means greater motion means greater motion

Ch. 5 GasesCh. 5 Gases

5.7 Effusion and Diffusion5.7 Effusion and Diffusion

EffusionEffusion passing of gas through a small hole into an passing of gas through a small hole into an

evacuated chamberevacuated chamber

DiffusionDiffusion mixing of gasesmixing of gases

Graham’s Law of EffusionGraham’s Law of Effusion

effusion rates depend directly on the effusion rates depend directly on the average velocity of the particlesaverage velocity of the particles

the faster they are moving, the more the faster they are moving, the more likely they are are to go through the likely they are are to go through the holehole

ExampleExample

Compare the effusion rates of hydrogen and Compare the effusion rates of hydrogen and oxygen gas.oxygen gas.

So HSo H22 effuses 4 times faster than O effuses 4 times faster than O22

CompareCompare

Ch. 5 GasesCh. 5 Gases

5.8 Real Gases5.8 Real Gases

Ideal GasesIdeal Gases

Real Real GasesGases