GaseGasess

Chapter 3

Ideal GasesIdeal GasesIdeal gases are imaginary gases that Ideal gases are imaginary gases that perfectlyperfectlyfit all of the assumptions of the kinetic fit all of the assumptions of the kinetic molecular theory.molecular theory. Gases consist of tiny particles that are far Gases consist of tiny particles that are far apartapart relative to their size.relative to their size. Collisions between gas particles and between Collisions between gas particles and between

particles and the walls of the container are particles and the walls of the container are elasticelastic

No kinetic energy is lost in elastic No kinetic energy is lost in elastic collisionscollisions

REVIEW p.92

Ideal GasesIdeal Gases (continued)(continued)

Gas particles are in constant, rapid motion. Gas particles are in constant, rapid motion. They They therefore possess KE (energy of motion)therefore possess KE (energy of motion) There are no forces of attraction between There are no forces of attraction between

gasgas particlesparticles The average KE of gas particles The average KE of gas particles depends on Tdepends on T, ,

not not on the identity of the particleon the identity of the particle

Physical CharacteristicsPhysical Characteristics

of Gasesof Gases

Pressure (P)Pressure (P) p.104

Is caused by the collisions of molecules with the walls of a container

P is defined as the force per unit area on a surface

SI units = Newton/meter2 (N/m2) = 1 Pascal (Pa)

N is the force that will increase the speed of a one-kg mass by 1 m/s each second that the force is applied

1 standard atmosphere = 101,325 Pa = 101.3 kPa

1 atm = 760 mm Hg = 760 torr

Measuring Measuring PressurePressure

The first device for The first device for measuring atmospheric P measuring atmospheric P was developed by was developed by Evangelista TorricelliEvangelista Torricelli during the 17during the 17thth century. century.The device was called a The device was called a “barometer”“barometer”

BaroBaro = weight= weight MeterMeter = measure= measure

An Early An Early BarometerBarometer

The normal pressure due to The normal pressure due to the atmosphere at sea level the atmosphere at sea level can support a column of can support a column of mercury that is 760 mm high. mercury that is 760 mm high.

Converting Celsius to Kelvin

Gas law problems involving T (temperature) require that the T be in KELVINS!

Kelvins = C + 273

°C = Kelvins - 273

STP

P = 1 atm (atmosphere, 760 torr, 760 mm Hg, 101.3kPa)

T = 273 K (0°C)

Molar volume of an ideal gas is 22.4 L at STP

REVIEW: Nature of REVIEW: Nature of GasesGases

Gases expand to fill their containersGases are fluid – they flowGases have low density

1/1000 the density of the equivalent liquid or solid

Gases are compressibleGases effuse and diffuse

The Gas LawsThe Gas Laws – p. 118

• Describe HOW gases behave.

• Can be predicted by the theory.

• Amount of change can be calculated with mathematical equations.

Boyle’s LawBoyle’s Law

Pressure Volume = a constant P1V1 = P2V2 (T = constant)

Robert Boyle (1627-1691)

P is inversely proportional to V when T is held constant.

Graph of Boyle’s Law

Boyle’s Law says the pressure is inverse to the volume.

Note that when the volume goes up, the pressure goes down

P1V1 = P2V2 (T = constant)

Changing the Container SizeChanging the Container Size

• In a In a smaller container molecules have less room to move.

• Thus, hit the sides of the container more often.

• As V decreases, P increases.

1 atm

4 Liters

• As the As the pressure pressure on a gas on a gas increasesincreases

2 atm

2 Liters

• the volume the volume decreasesdecreases– Gas particles are closer

together, shorter distance to go before they impact the containers wall

– More molecule impacts per unit time increase in P

• P and V are inversely related

- Page 419

Example: A 250.0 mL sample of Cl2 is collected when the barometric pressure is 105.2 kPa. What is the volume of the sample after the barometer drops to 100.3 kPa?

Given: P1 = 105.2 kPa Find: V2 = ?

V1 = 250.0 mL

P2 = 100.3 kPa

Use Boyles Law: P1 V1 = P2 V2

Set up for Unknown: V2 = P1 V1

P2

Solve: V2 = 105.2 kPa x 250.0 mL

100.3 kPa

= 262.2 mL

Charles’s LawCharles’s LawThe V of a gas is directly proportional to T, and extrapolates to zero at zero Kelvin (-273.15°C).

(P = constant). See P. 371, Fig. 8Charles’s experiments showed that all gases expand to the same extent when heated.

VT

VT

P1

1

2

2 ( constant)

Temperature MUST be in KELVINS!

Jacques Jacques Charles (1746-Charles (1746-1823). Isolated 1823). Isolated boron and boron and studied gases. studied gases. Balloonist.Balloonist.

TemperatureTemperature

• Raising the T of a gas increases Raising the T of a gas increases the P, if the V is held constant.the P, if the V is held constant.

• The molecules hit the walls The molecules hit the walls harder, more often.harder, more often.

• If you start with 1 liter of gas at 1 If you start with 1 liter of gas at 1 atm pressure and 300 Katm pressure and 300 K

• and heat it to 600 K one of 2 and heat it to 600 K one of 2 things happensthings happens

300 K

• Either the volume will Either the volume will increase to 2 liters at 1 increase to 2 liters at 1 atmatm

300 K600 K

Charles’s LawCharles’s Law

300 K 600 K

•Or the pressure will Or the pressure will increase increase to 2 atm.to 2 atm.•Or someplace in Or someplace in betweenbetween

Gay Gay Lussac’s Lussac’s

LawLaw

Example: A balloon is inflated with 6.22 L of helium at a temperature of 36 °C. What is the volume of the balloon when the temperature is 22°C?

Given: V1 = 6.22 L Find: V2 = ?

T1 = 36 °C T1 = ? K

T2 = 22 °C T2 = ? K

T1 = Kelvin = C + 273 = 36 °C + 273 = 309

T2 = Kelvin = C + 273 = 22 °C + 273 = 295

T1 = 309 K

T2 = 295 K

V1 V2

T1 T2

= V1

T1

=V2 T2

6.22L

309K

= 5.94 L

295K=

Gay-Lussac’s LawGay-Lussac’s LawThe pressure and temperature of a gas The pressure and temperature of a gas arearedirectly related, provided that the directly related, provided that the volume volume remains constant. remains constant.

2

2

1

1

T

P

T

P

Temperature MUST be in KELVINS! (1778-(1778-

1850)1850)

LAWLAW RELAT-RELAT-IONSHIPIONSHIP LAWLAW CONSTANTCONSTANT

SS

Boyle’sBoyle’s PP V V PP11VV1 1 = P= P22VV22 T, nT, n

Charles’Charles’ VV T T VV11/T/T11 = V = V22/T/T22 P, nP, n

Gay-Gay-Lussac’sLussac’s PP T T PP11/T/T11 = P = P22/T/T22 V, nV, n

Summary of the Named Gas-LawsSummary of the Named Gas-Laws:Summary of the Named Gas-LawsSummary of the Named Gas-Laws:

The Combined Gas LawThe Combined Gas LawThe combined gas law expresses the The combined gas law expresses the relationship relationship between P, V and T of a fixed amount of gas.between P, V and T of a fixed amount of gas.

2

22

1

11

T

VP

T

VP

Temperature MUST be in Temperature MUST be in KELVINS!KELVINS!

• The combined gas law contains all the other gas laws!

• If the temperature remains constant...

– Meaning T1 = T2

P1 V1

T1

x=

P2 V2

T2

x

Boyle’s Law

• The combined gas law contains all the other gas laws!

• If the pressure remains constant...

P1 V1

T1

x=

P2 V2

T2

x

Charles’s Law

The combined gas law contains all the other gas laws!

If the volume remains constant...

P1 V1

T1

x P2 V2

T2

x

Gay-Lussac’s Law

=

Example: A student collects 285 mL of O2 gas at a temperature of 15°C and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11°C. What is the new pressure of the gas?

Known: V1 = 285 mL

T1 = 15°C P1 = 99.3 kPa

V2 = 292 mL

T2 = 11°C

Unknown: P2 = ?T1 = ? KT2 = ? K

T1 = Kelvin = C + 273 = 15°C + 273 = 288 K

T2 = Kelvin = C + 273 = 11°C + 273 = 284 K

Example: A student collects 285 mL of O2 gas at a temperature of 15°C and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11°C. What is the new pressure of the gas?

Known: V1 = 285 mL

T1 = 15°C P1 = 99.3 kPa

V2 = 292 mL

T2 = 11°C

Unknown: P2 = ?T1 = 288 KT2 = 284 K

2

22

1

11

T

VP

T

VP T2P1V1

T1V2

P2 =

Example: A student collects 285 mL of O2 gas at a temperature of 15°C and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11°C. What is the new pressure of the gas?Known: V1 = 285 mL

T1 = 15°C P1 = 99.3 kPa

V2 = 292 mL

T2 = 11°C

Unknown: P2 = ?T1 = 288 KT2 = 284 K

T2P1V1

T1V2

P2 = (284 K)(99.3 kPa)(285 mL)

(288 K)(292 mL) =

= 95.6 kPa

The effect of adding The effect of adding gas.gas.

• When we blow up a balloon we are adding gas molecules.

• Doubling the number of gas particles doubles the pressure.

(of the same V, at the same T)

1 atm

• If you double the number of If you double the number of moleculesmolecules

• If you double the number of If you double the number of moleculesmolecules

• You double the pressure.You double the pressure.2 atm

• As you remove As you remove molecules from a molecules from a containercontainer4 atm

• As you remove As you remove molecules from a molecules from a container the pressure container the pressure decreasesdecreases

2 atm

• As you remove molecules from a container the pressure decreases

• Until the pressure inside equals the pressure outside

• Molecules naturally move from high to low pressure

1 atm

Pressure and the number of Pressure and the number of molecules are directly related.molecules are directly related.

• More molecules means more collisions.– Fewer molecules means fewer collisions.

• Gases naturally move from areas of high P to low P (because of empty space to move in).

Think of it terms of pressure.

• The same P at the same T should require that there be the same number of particles.

• The smaller particles must have a greater average speed to have the same KE.

• Measuring & Comparing the Volumes of Reacting Gases

Gay-Lussac’s law of combining volumes of gases states that at constant T and P, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

– Examples: • 1 liter of nitrogen gas reacts with 3 liters of hydrogen

gas to produce 2 liters of ammonia gas

N2(g) + 3H2(g) -----> 2NH3(g)• Since ALL the reactants and products are gases, the

mole ratio of N2(g):H2(g):NH3(g) of 1:3:2 is also the ratio of the volumes of gases

• Therefore, 10mL of N2 gas would react with 10 x 3 = 30mL of H2 gas to produce 10 x 2 = 20mL NH3 gas

Avogadro’s LawAvogadro’s LawFor a gas at constant T and P, the For a gas at constant T and P, the volume is directly proportional to volume is directly proportional to the the number of moles of gasnumber of moles of gas

or…at the same T and P equal V have equal # of molecules, regardless of what type of gas they contain.

VV == aannaa = proportionality constant = proportionality constant

VV = volume of the gas = volume of the gas

nn = number of moles of gas = number of moles of gas

Avogadro’s HypothesisAvogadro’s Hypothesis

2 Liters of

Helium

2 Liters of

Oxygen

• Has the same Has the same number of number of particles as ..particles as ..

• Only at STPOnly at STP

– 273K or 0ºC273K or 0ºC

– 1 atm 1 atm

• This way we compare gases at This way we compare gases at the same T and P.the same T and P.

This is where we getThis is where we get22.4 L =1 mole22.4 L =1 mole

Standard Molar Volume – p. 141

Equal volumes of all Equal volumes of all gases at the same gases at the same temperature and temperature and pressure contain the pressure contain the same number of same number of moleculesmolecules. .

- Amadeo Avogadro

Standard Molar Volume – the volume occupied by one mole of a gas at STP

Standard Molar VolumeStandard Molar Volume

One mole of any gas will occupy the same volume as one mole of any other gas (at the same T and P), despite mass differences!

Practice ProblemPractice Problem

• At STP, what is the volume of 7.08 mol of N2 gas?

7.08 mol N2 22.4 L N2

1 mol N2

= 159 L N2

Dalton’s Law of Partial Pressures - p. 144

For a mixture of gases in a container,

PPTotalTotal = = PP11 + + PP22 + + PP33 + . . . + . . .

This is particularly useful in calculating the pressure of gases collected over water.

• We can find out the pressure in the We can find out the pressure in the fourth container.fourth container.

• By adding up the pressure in the first 3.By adding up the pressure in the first 3.

2 atm

1 atm

3 atm

6 atm

Gas Stoichiometry – p. 150Gas Stoichiometry – p. 150• Using stoichiometry for gas reactions.Using stoichiometry for gas reactions.• As before, need to consider mole ratios when examining As before, need to consider mole ratios when examining

reactions quantitatively.reactions quantitatively.

• At times you will be able to use 22.4 L/mol at STP

grams (x) grams (x) moles (x) moles (x) moles (y) moles (y) grams (y) grams (y)

molar mass of y molar mass of y mole ratio from balanced mole ratio from balanced equation equation molar mass of xmolar mass of x

P, V, T (x)P, V, T (x)

P, V, T (y)P, V, T (y)

PV = nRTPV = nRT

Ideal Gas LawIdeal Gas Law p. 131

PPVV = = nnRRTTP = pressureV = volume (L)n = # of molesR = ideal gas constant (proportionality constant) @ STP

= 0.08206 L atm/ mol·value depends on units chosen)

T = temperature (K)

Holds closely at Holds closely at PP < 1 atm < 1 atm

Real Gases vs. IdealRecall….Ideal Gas: Although no gas is truly

ideal, many gases follow the ideal gas law very closely at sufficiently low pressures.

Real gas laws try to predict the true behavior of a gas better than the ideal gas law by putting in terms to describe attractions and repulsions between molecules.

Hard Sphere Gas - accounts for very small V of particles

P (V-nb) = n R T - no intermolecular forces

van der Waals Gas [ P + a (n / V)2 ] [ V - nb ] =n R T

ExampleCH4 burns in O2, producing CO2 and H2O vapor. A 1.22 L CH4 cylinder, at 15°C, registers a pressure of 328 kPa.

a) What volume of O2 at STP will be required to react completely with all of the CH4?

First: CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

PV = nRT

(8.31 kPa•L/K•mol)(288 K)(8.31 kPa•L/K•mol)(288 K)

(328 kPa)(1.22 L)(328 kPa)(1.22 L)= 0.167 mol CH= 0.167 mol CH44

P = 328 kPa, V = 1.22 L, T = 288 K

0.167 mol CH0.167 mol CH44 2 mol O2 mol O22

1 mol CH1 mol CH44 xx = 7.48 L O2

n = n =

xx 1 mol O1 mol O22

22.4 L O22.4 L O22

Example - continuedExample - continued

CHCH44(g) + 2O(g) + 2O22(g) (g) CO CO22(g) + 2H(g) + 2H22O(g)O(g)

b) What volume of CO2 (at STP) is produced if only 2.15 g of the CH4 was burned?

2.15 g CH4 2.15 g CH4 1 mol CH1 mol CH44

16.05 g CH16.05 g CH44 xx

1 mol CO1 mol CO22

1 mol CH1 mol CH44 xx

= 3.00 L CO= 3.00 L CO22

1 mol CO1 mol CO22

22.4 L CO22.4 L CO22xx

Gas DensityGas Densitymolar mass

molar volume

massDensity

volume

… … so at STP…so at STP…

molar mass

22.4 LDensity

Density and the Ideal Gas LawCombining the formula for density with the Ideal

Gas law, substituting and rearranging algebraically:

MPD

RT

D = DensityD = DensityM = Molar MassM = Molar MassP = PressureP = PressureR = Gas ConstantR = Gas ConstantT = Temperature in T = Temperature in KelvinsKelvins

PV = nRT and if: n = m/M and ρ = D = m/V   

This equation can be modified to:

= mRTMPV

VMP =

mRT M

P =

DRT

Density and the Ideal Gas LawCombining the formula for density with the Ideal

Gas law, substituting and rearranging algebraically:

M = Molar MassM = Molar Mass

P = PressureP = Pressure

R = Gas ConstantR = Gas Constant

T = Temperature in T = Temperature in KelvinsKelvins

PV = nRT and if: n = m/M and D = m/V   

This equation can be modified to:

= mRTMPV

VMP =

mRT

MP =

DRT

PM =

DRT

Or solving for M (Molar Mass):

• Graham’s Law of Effusion p. 145– States that the rates of effusion of gases

at the same T and P are inversely proportional to the square roots of their molar masses

rate of effusion of A √MB

rate of effusion of B √MA

where M = mass

=

Diffusion:Diffusion: describes the describes the mixing of gases.mixing of gases. The The rate rate of of diffusion is the diffusion is the rate of gas mixingrate of gas mixing.

(Graham’s law)

DiffusionDiffusion

EffusionEffusionEffusion:Effusion: describes describes the the passage passage of gas of gas into an into an evacuateevacuatedd chamber.chamber.