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BEHAVIOR OF GASESBEHAVIOR OF GASESChapter 14Chapter 14

BEHAVIOR OF GASESBEHAVIOR OF GASESChapter 14Chapter 14

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Importance of Importance of GasesGases

Importance of Importance of GasesGases

Airbags fill with NAirbags fill with N22 gas in an accident. gas in an accident.

Gas is generated by the decomposition Gas is generated by the decomposition of sodium azide, NaNof sodium azide, NaN33..

2 NaN2 NaN3(s)3(s) ---> 2 Na ---> 2 Na(s)(s) + 3 N + 3 N2(g)2(g)

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THREETHREESTATES OF STATES OF

MATTERMATTER

THREETHREESTATES OF STATES OF

MATTERMATTER

Click picture above to view movie on states of matter

Click here for water production

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General Properties of GasesGeneral Properties of GasesThere is a lot of "free" There is a lot of "free"

space in a gas.space in a gas.

Gases can be expanded Gases can be expanded infinitely.infinitely.

Gases occupy containers Gases occupy containers uniformly and completely.uniformly and completely.

Gases diffuse and mix Gases diffuse and mix rapidly.rapidly.

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Kinetic Theory Revisited1. Gases consist of hard, spherical

particles (usually molecules or atoms)

2. Small- so the individual volume is considered to be insignificant

3. Large empty space between them

4. Easily compressed and expanded

5. No attractive or repulsive forces

6. Move rapidly in constant motion

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Kinetic Theory Revisited Recall: that the average kinetic

energy of a collection of gas particles is directly proportional to the Kelvin temperature of the gas.

Click for movie No Kinetic energy lost during

collisions. All particles have same energy at

same temperature.

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Properties of GasesProperties of GasesProperties of GasesProperties of GasesGas properties can be modeled Gas properties can be modeled

using math. Model depends using math. Model depends on-on-

VV = volume of the gas ( = volume of the gas (LL)) TT = temperature ( = temperature (KK)) nn = amount ( = amount (molesmoles)) PP = pressure ( = pressure (atm or atm or

kilopascalkilopascal))

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PressurePressurePressurePressurePressure of air is Pressure of air is

measured with a measured with a BAROMETERBAROMETER (developed by (developed by TorricelliTorricelli in 1643) in 1643)

Barometer calibrated for Barometer calibrated for column width and column width and pool width/depthpool width/depth

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PressurePressurePressurePressureHg rises in tube until Hg rises in tube until

gravitational force gravitational force of Hg (down) of Hg (down) balances the force balances the force of atmosphere of atmosphere (pushing up). (pushing up).

P of Hg pushing down P of Hg pushing down related to related to

Hg densityHg density column heightcolumn height

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PressurePressurePressurePressureColumn height measures Column height measures

PressurePressure of atmosphere of atmosphere

1 standard atm1 standard atm

= 760 mm Hg= 760 mm Hg

= 76 cm Hg= 76 cm Hg

= 760 torr= 760 torr

= 29.9 inches= 29.9 inches

= about 33 feet of water= about 33 feet of water

SI unit is SI unit is PASCALPASCAL, Pa, , Pa, where 1 atm = 101.325 kPawhere 1 atm = 101.325 kPa

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Gas -Volume, Temp, & Pressure

Click to view movie

1. Amount of Gas When we inflate a ball, we are

adding gas molecules. Increasing the number of gas

particles increases the number of collisions

• thus, the pressure increases If temp. is constant- doubling the

number of particles doubles pressure

Pressure and the Number of Molecules are Directly Related

Fewer molecules means fewer collisions.

Gases naturally move from areas of high pressure to low pressure because there is empty space to move in - spray can is example.

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Expanding Gas Uses?

The bombardier The bombardier beetle uses beetle uses decomposition of decomposition of hydrogen peroxide hydrogen peroxide to defend itself. The to defend itself. The gas acts as a gas acts as a propellant.propellant.

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The Shuttle Uses a Solid Booster and Uncontrolled Expanding

Gases Can Be Disastrous

1 atm

If you double the number of molecules

If you double the number of molecules

You double the pressure.

2 atm

As you remove molecules from a container

4 atm

As you remove molecules from a container the pressure decreases

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

Changing the Size of the Container

In a smaller container molecules have less room to move.

Hit the sides of the container more often.

As volume decreases pressure increases. Think air pump

1 atm

4 Liters

As the pressure on a gas increases

2 atm

2 Liters

As the pressure on a gas increases the volume decreases

Pressure and volume are inversely related

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What happens to the air in a diver’s lungs the deeper

they go?

Temperature Raising the temperature

of a gas increases the pressure if the volume is held constant.

The molecules hit the walls harder.

The only way to increase the temperature at constant pressure is to increase the volume.

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

and heat it to 600 K one of 2 things happens

300 K

Either the volume will increase to 2 liters at 1 atm

300 K600 K

300 K 600 K

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

The Gas Laws Describe HOW gases behave. Can be predicted by theory. Amount of change can be calculated

with mathematical equations.

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A. Boyle’s Law

P

VPV = k

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A. Boyle’s Law The pressure and volume of

a gas are inversely related

• at constant mass & temp

P

V

PV = k

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A. Boyle’s Law

Click to view movie

Boyle’s Law At a constant temperature pressure

and volume are inversely related. As one goes up the other goes down P x V = K (K is some

constant) Easier to use P1 x V1=P2 x V2

Click for movie

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GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

P1V1 = P2V2

Boyle’s Gas Law Problems A gas occupies 100. mL at 150. kPa.

Find its volume at 200. kPa.

BOYLE’S LAW

P V

(150.kPa)(100.mL)=(200.kPa)V2

V2 = 75.0 mL

A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is change to 1.5 atm what is the new volume?

A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change to volume to 43 L?

Examples

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kT

VV

T

B. Charles’ Law

45

kT

VV

T

B. Charles’ Law The volume and absolute

temperature (K) of a gas are directly related

• at constant mass & pressure

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Charles’ Law

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B. Charles’ Law

Click for movie

B. Charles’ Law The volume of a gas is directly

proportional to the Kelvin temperature if the pressure is held constant.

V = K x T(K is some constant) V/T= K V1/T1= V2/T2

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GIVEN:

V1 = 473 cm3

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

V1T2 = V2T1

Gas Law Problems A gas occupies 473 cm3 at 36°C. Find

its volume at 94°C.

CHARLES’ LAW

T V

(473 cm3)(367 K)=V2(309 K)

V2 = 562 cm3

Examples What is the temperature (ºC) of a gas

that is expanded from 2.5 L at 25ºC to 4.1L at constant pressure.

What is the final volume of a gas that starts at 8.3 L and 17ºC and is heated to 96ºC?

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kT

PP

T

C. Gay-Lussac’s Law

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kT

PP

T

C. Gay-Lussac’s Law The pressure and absolute

temperature (K) of a gas are directly related

• at constant mass & volume

C. Gay-Lussac’s Law The temperature and the pressure

of a gas are directly related at constant volume.

P = K x T(K is some constant) P/T= K P1/T1= P2/T2

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GIVEN:

P1 = 765 torr

T1 = 23°C = 296K

P2 = 560. torr

T2 = ?

WORK:

P1T2 = P2T1

E. Gas Law Problems A gas’ pressure is 765 torr at 23°C.

At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW

P T

(765 torr)T2 = (560. torr)(296K)

T2 = 217 K = -56°C

Examples What is the pressure inside a 0.250 L

can of deodorant that starts at 25ºC and 1.2 atm if the temperature is raised to 100ºC?

At what temperature will the can above have a pressure of 2.2 atm?

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Too Much Pressure and Not Enough

Volume!!!!

Putting the pieces together The Combined Gas Law Deals with

the situation where only the number of molecules stays constant.

(P1 x V1)/T1= (P2 x V2)/T2

Allows us to figure out one thing when two of the others change.

The combined gas law contains all the other gas laws!

If the temperature remains constant.

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’ 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

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= kPVPTVT

PVT

D. Combined Gas Law

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1

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GIVEN:

V1 = 7.84 cm3

P1 = 71.8 kPa

T1 = 25°C = 298 K

V2 = ?

P2 = 101.325 kPa

T2 = 273 K

WORK:

P1V1T2 = P2V2T1

(71.8 kPa)(7.84 cm3)(273 K)

=(101.325 kPa) V2 (298 K)

V2 = 5.09 cm3

E. Gas Law Problems A gas occupies 7.84 cm3 at 71.8 kPa &

25°C. Find its volume at STP.

P T VCOMBINED GAS LAW

Examples A 15 L cylinder of gas at 4.8 atm

pressure at 25ºC is heated to 75ºC and compressed to 17 atm. What is the new volume?

If 6.2 L of gas at 723 mm Hg at 21ºC is compressed to 2.2 L at 4117 mm Hg, what is the temperature of the gas?

The Fourth Part Avagadro’s Hypothesis click for movie V is proportional to number of

molecules at constant T and P V is proportional to moles One mole = 6.02 x 1023 particles One mole = 22.4 L at STP movie V = K n ( n ) is the number of moles Gets put into the Ideal Gas Law

Ideal Gases Remember in this chapter we assume

the gases behave ideally. Ideal gases don’t really exist, but

assuming it makes the math easier. So we get a close approximation.

Particles have no volume. No attractive forces. However, real gases do behave like

ideal gases at high temperature and low pressure.

The Ideal Gas Law P x V = n x R x T Pressure times Volume equals the

number of moles times the Ideal Gas Constant (R) times the temperature in Kelvin.

This time R does not depend on anything, it is really constant

R = 0.0821 (L atm)/(mol K) or R = 62.4 (L mm Hg)/(K mol) We now have a new way to count

moles. By measuring T, P, and V. We aren’t restricted to STP.

n = PV/RT

The Ideal Gas Law

Examples How many moles of air are there in a

2.0 L bottle at 19ºC and 747 mm Hg? What is the pressure exerted by 1.8 g

of H2 gas exert in a 4.3 L balloon at 27ºC?

Density The Molar mass of a gas can be

determined by the density of the gas. D= mass = m

Volume V Molar mass = mass = m

Moles n n = PV

RT

Molar Mass Formulas Molar Mass = m

(PV/RT) Molar mass = m RT

V P Molar mass = DRT

P

Dalton’s Law of Partial Pressures The total pressure inside a container

is equal to the partial pressure due to each gas.

The partial pressure is the contribution by that gas.

PTotal = P1 + P2 + P3

For example

We can find out the pressure in the fourth container.

By adding up the pressure in the first 3.2 atm

1 atm

3 atm

6 atm

Examples What is the total pressure in a balloon

filled with air if the pressure of the oxygen is 170 mm Hg and the pressure of nitrogen is 620 mm Hg?

In a second balloon the total pressure is 1.3 atm. What is the pressure of oxygen if the pressure of nitrogen is 720 mm Hg?

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B. Dalton’s Law The total pressure of a mixture

of gases equals the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + ...When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

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GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

B. Dalton’s Law Hydrogen gas is collected over water at

22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Look up water-vapor pressure for 22.5°C.

Sig Figs: Round to least number of decimal places.

The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.

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GIVEN:

Pgas = ?

Ptotal = 742.0 torr

PH2O = 42.2 torr

WORK:

Ptotal = Pgas + PH2O

742.0 torr = PH2 + 42.2 torr

Pgas = 699.8 torr

A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas?

Look up water-vapor pressure for 35.0°C.

Sig Figs: Round to least number of decimal places.

B. Dalton’s Law

The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

Diffusion

Effusion Gas escaping through a tiny hole in a container.

Depends on the speed of the molecule.

Molecules moving from areas of high concentration to low concentration.

Perfume molecules spreading across the room.

Graham’s Law The rate of effusion and diffusion is

inversely proportional to the square root of the molar mass of the molecules.

Kinetic energy = 1/2 mv2

m is the mass v is the velocity.

Chem Express

Bigger molecules move slower at the same temp. (by Square root)

Bigger molecules effuse and diffuse slower

Helium effuses and diffuses faster than air - escapes from balloon.

Graham’s Law

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C. Graham’s Law Graham’s LawGraham’s Law

• Rate of diffusion of a gas is inversely related to the square root of its molar mass.

• The equation shows the ratio of Gas A’s speed to Gas B’s speed.

A

B

B

A

m

m

v

v

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Determine the relative rate of diffusion for krypton and bromine.

1.381

Kr diffuses 1.381 times faster than Br2.

Kr

Br

Br

Kr

m

m

v

v2

2

A

B

B

A

m

m

v

v

g/mol83.80

g/mol159.80

C. Graham’s Law

The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.

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An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.

Am

g/mol32.00 16

A

B

B

A

m

m

v

v

A

O

O

A

m

m

v

v2

2

Am

g/mol32.00 4.0

16

g/mol32.00 mA

2

Am

g/mol32.00 4.0

g/mol2.0

C. Graham’s Law

The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.

Square both sides to get rid of the square

root sign.

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A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?

A

B

B

A

m

m

v

v

2

2

2

2

H

O

O

H

m

m

v

v

g/mol 2.02

g/mol32.00

m/s 12.3

vH 2

C. Graham’s Law

3.980m/s 12.3

vH 2

m/s49.0 vH 2

Put the gas with the unknown

speed as “Gas A”.

You may need to review Chapter 12 click for movie

At STP At STP determining the amount of

gas required or produced is easy. 22.4 L = 1 mole For example

How many liters of O2 at STP are required to produce 20.3 g of H2O?

Not At STP Chemical reactions happen in

MOLES. If you know how much gas - change

it to moles Use the Ideal Gas Law n = PV/RT If you want to find how much gas -

use moles to figure out volume V = nRT/P

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A. Gas Stoichiometry Moles Moles Liters of a Gas: Liters of a Gas:

• STP - use 22.4 L/mol • Non-STP - use ideal gas law

Non-Non-STPSTP• Given liters of gas?

–start with ideal gas law• Looking for liters of gas?

–start with stoichiometry conversions.

98

1 molCaCO3

100.09g CaCO3

B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of

CaCO3 at 103 kPa & 25ºC?

5.25 gCaCO3

= 0.0525 mol CO2

CaCO3 CaO + CO2

1 molCO2

1 molCaCO3

5.25 g ? Lnon-STP

Looking for liters: Start with stoich and calculate moles of CO2.

Plug this into the Ideal Gas Law to find liters.

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WORK:

PV = nRT

(103 kPa)V=(.0525mol)(8.314 LkPa/molK) (298K)

V = 1.26 L CO2

B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g

of CaCO3 at 103 kPa & 25ºC?

GIVEN:

P = 103 kPaV = ?

n = 0.0525 molT = 25°C = 298 KR = 8.314 LkPa/molK

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WORK:

PV = nRT

(97.3 kPa) (15.0 L)= n (8.314 LkPa/molK) (294K)

n = 0.597 mol O2

B. Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0

L of O2 at 97.3 kPa & 21°C?

GIVEN:

P = 97.3 kPaV = 15.0 L

n = ?T = 21°C = 294 KR = 8.314 LkPa/molK

4 Al + 3 O2 2 Al2O3 15.0 L

non-STP ? g

Given liters: Start with Ideal Gas Law and

calculate moles of O2.

NEXT

101

2 mol Al2O3

3 mol O2

B. Gas Stoichiometry Problem How many grams of Al2O3 are formed

from 15.0 L of O2 at 97.3 kPa & 21°C?

0.597mol O2

= 40.6 g Al2O3

4 Al + 3 O2 2 Al2O3

101.96 g Al2O3

1 molAl2O3

15.0Lnon-STP

? gUse stoich to convert moles of O2 to grams Al2O3.

Example #1 HCl(g) can be formed by the

following reaction 2NaCl(aq) + H2SO4 (aq)

2HCl(g) + Na2SO4(aq) What mass of NaCl is needed to

produce 340 mL of HCl at 1.51 atm at 20ºC?

Example #2 2NaCl(aq) + H2SO4 (aq)

2HCl(g) + Na2SO4 (aq) What volume of HCl gas at 25ºC and

715 mm Hg will be generated if 10.2 g of NaCl react?

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Too Little Pressure and Too Little Volume!!!!

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What is a Turbocharger?