I. Physical Properties

Gases

Real vs. Ideal Gases: Ideal gas = an imaginary gas that

conforms perfectly to all the assumptions of the kinetic-molecular theory• We will assume that the gases

used for the gas law problems are ideal gases.

Real Gases vs. Ideal Gases:Real gas = a gas that does not

behave completely according to the assumptions of the kinetic-molecular theory.

 All real gases deviate to some degree from ideal gas behavior. However, most real gases behave nearly ideally when their particles are sufficiently far apart and have sufficiently high kinetic energy.

Causes of non-ideal behavior:The kinetic-molecular theory is

more likely to hold true for gases whose particles have NO attraction for each other.

Causes of non-ideal behavior:

1. High pressure (low volume): • Space taken up by gas particles

becomes significant•  Intermolecular forces are more

significant between gas particles that are closer together

Causes of non-ideal behavior:

2. Low temperature: •Gas particles move slower so intermolecular forces become more important

Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences.

theory developed in the late 19th century to account for the behavior of the atoms and molecules that make up matter

KMT (Kinetic Molecular Theory)

based on the idea that particles in all forms of matter are always in motion and that this motion has consequences

KMT (Kinetic Molecular Theory)

SOLID LIQUID GAS

can be used to explain the properties of solids, liquids, and gases in terms of the energy of particles and the forces that act between them

KMT (Kinetic Molecular Theory)

B. Kinetic Molecular Theory – Ideal gas

KMT describing particles in an IDEAL gas:• have no volume.• have elastic collisions. • are in constant, random motion.• don’t attract or repel each other.• have an avg. KE directly related

to Kelvin temperature.

C. Kinetic Molecular Theory - Real Gases

KMT describing particles in a REAL gas:• have their own volume• attract each other

Gas behavior is most ideal…• at low pressures• at high temperatures• in nonpolar atoms/molecules

D. Characteristics of Gases- using KMT

Gases expand to fill any container.KMT - gas particles move rapidly

in all directions without significant attraction or repulsion between particles

D. Characteristics of Gases- using KMT

Gases are fluids (like liquids). KMT - No significant attraction or

repulsion between gas particles; glide past each other

D. Characteristics of Gases- using KMT

Gases have very low densities.• KMT - particles are so much

farther apart in the gas state

sodium in the solid state:

sodium in the liquid state:

sodium in the gas state:

Gases can be compressed.• KMT - gas particles are far apart

from one anther with room to be “squished” together

D. Characteristics of Gases- using KMT

Gases undergo diffusion & effusion.• KMT – gas particles move in

continuous, rapid, random motion

D. Characteristics of Gases- using KMT

Effusion

E. Temperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

Always use Kelvin temperature when working with gases.

F. Pressure

areaforcepressure

Which shoes create the most pressure?

F. PressureWhy do gases exert

pressure? • Gas particles exert a

pressure on any surface with which they collide!More collisions =

increase in pressure!

F. PressureBarometer = measures

atmospheric pressureThe height of the Hg in

the tube depends on the pressure• The pressure of the

atmosphere is proportional to the height of the Hg column, so the height of the Hg can be used to measure atmospheric pressure! Mercury Barometer

F. PressurePressure UNITS

101.3 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

G. STP

Standard Temperature & Pressure

0°C (exact) 1 atm (exact)

273 K 101.3 kPa

760 mm Hg (exact)

760 torr (exact)

STP

II. The Gas Laws

Gases

GAS LAW PROBLEMS- MUST USE KELVIN

K = ºC + 273

A. Boyle’s Law

P

V

PV = k

Volume (mL)

Pressure (torr)

P·V (mL·torr)

10.0 760.0 7.60 x 103 20.0 379.6 7.59 x 103 30.0 253.2 7.60 x 103 40.0 191.0 7.64 x 103

A. Boyle’s LawThe pressure and volume

of a gas are INVERSELY related• at constant mass & temp

P

V

P1V1 = P2V2

A. Boyle’s Law•Real life application: When you breathe, your diaphragm moves downward, increasing the volume of the lungs. This causes the pressure inside the lungs to be less than the outside pressure so air rushes in.

A. Boyle’s Law•Ex: Halving the volume leads to twice the rate of collisions and a doubling of the pressure.

A. Boyle’s Law

• As the volume increases, the pressure decreases!

kTV

V

T

B. Charles’ LawVolume

(mL) Temperature

(K) V/T

(mL/K) 40.0 273.2 0.146 44.0 298.2 0.148 47.7 323.2 0.148 51.3 348.2 0.147

V1_ = V2__

T1 T2

V

T

B. Charles’ LawThe volume and

temperature (K) of a gas are DIRECTLY related • at constant mass &

pressure

B. Charles’ LawReal life application: Bread dough

rises because yeast produces carbon dioxide. When placed in the oven, the heat causes the gas to expand, and the bread rises even further.

B. Charles’ LawAs temperature decreases, the

volume of the gas decreases

Liquid nitrogen’s temp. is about 63K or -210 ºC or -346 ºF!

B. Charles’ LawAs

temperature decreases, the volume of the gas decreases

B. Charles’ Law

NUMBER OF PARTICLES & PRESSURE ARE CONSTANT!!!

kTP

P

T

C. Guy-Lussac’s LawTemperature

(K) Pressure

(torr) P/T

(torr/K) 248 691.6 2.79 273 760.0 2.78 298 828.4 2.78 373 1,041.2 2.79

_P1_ = P2__

T1 T2

P

T

C. Guy-Lussac’s LawThe pressure and

temperature (K) of a gas are DIRECTLY RELATED • at constant mass &

volume

C. Guy-Lussac’s LawWhen the temp. of a gas increases

(KE increases) and gas particles move faster and hit container walls more frequently and collisions are more forceful

C. Guy-Lussac’s LawReal life application: The air

pressure inside a tire increases on a hot summer day.

= kBoyles PV PGuy Lussac’sT

VCharles’ T

PVT

D. Combined Gas Law

P1V1

T1=

P2V2

T2

E. EXAMPLE ProblemsYou need your calculators!

C. Johannesson

GIVEN:

V1 = 473 cm3

T1 = 36°C = 309KV2 = ?T2 = 94°C = 367K

WORK:

Gas Law Problems1.) A gas occupies 473 cm3 at 36°C. Find its volume at

94°C. CHARLES’ LAW V1_ = V2__

T1 T2

T V

V2= (473 cm3)(367 K)

(309 K)

V2 = 562 cm3

V2 = V1T2

T1

C. Johannesson

GIVEN:

V1 = 100. mLP1 = 150. kPaV2 = ?P2 = 200. kPa

WORK:V2 = P1V1

P2

Gas Law Problems2.) A gas occupies 100. mL at 150. kPa. Find its

volume at 200. kPa.

BOYLE’S LAW P1V1 = P2V2

P V

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

200.kPa

V2 = 75.0 mL

C. Johannesson

GIVEN:V1 = 7.84 cm3

P1 = 71.8 kPaT1 = 25°C = 298 KV2 = ?P2 = 101.3 kPaT2 = 273 K

WORK:V2 = P1V1T2

P2T1

V2 = (71.8 kPa)(7.84 cm3)(273 K)

(101.3 kPa) (298 K)

V2 = 5.09 cm3

Gas Law Problems3.) A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its

volume at STP.

P T V

COMBINED GAS LAW P1V1 = __P2V2

T1 T2

C. Johannesson

GIVEN:

P1 = 765 torrT1 = 23°C = 296KP2 = 560. torrT2 = ?

WORK:

T2 = P2T1

P1

Gas Law Problems4.) A gas’ pressure is 765 torr at 23°C. At what

temperature will the pressure be 560. torr?

P T

T2 = (560. torr)(296K)

765 torrT2 = 217 K

GUY LUSSAC’S LAW P1_ = P2__

T1 T2

III. Standard Molar Volume,

Gas Densities & Molar Mass

Gases

A. Standard Molar VolumeThe volume occupied by one mole

of a gas at STP • 22.41410 L/mol or about 22.4

L/mol In other words, one mole of any

ideal gas at STP will occupy 22.4 L

B. Example Problems-standard molar volume

1.) A chemical reaction is expected to produce 0.0680 mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?

0.068 mol O2 22.4 L O2

1 mol O2

=

1.52 L O2

B. Example Problems-standard molar volume

1.) A chemical reaction is expected to produce 0.0680 mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?

0.068 mol O2 22.4 L O2

1 mol O2

=

1.52 L O2

B. Example Problems-standard molar volume

2.) A chemical reaction produced 98.0 mL of SO2 at STP. What mass (in grams) of the gas was produced?

98.0 mL SO2 1 L SO2 1 mol O2

=

0.280 g SO2

22.4 L SO21000 mL SO2 1 mol SO2

64.064 g SO2

C. DensityThe ratio of an object’s mass to its

volume.D = M V The volume (and density) of a gas

will change when pressure and temperature change. • Densities are usually given in g/L at

STP

C. DensityFor an ideal gas:

Density (at STP) = molar mass standard molar volume

D. Example Problems- density

1.) What is the density of CO2 in g/L at STP?

44.009 g CO2 1 mol CO2

1 mol CO2 22.4 L CO2

=

1.96 g/L CO2

D. Example Problems- density2.) What is the molar mass of a gas

whose density at STP is 2.08 g/L.

2.08 g 22.4 L1 L 1 mol =

46.6 g/mol

IV. More Gas Laws: Ideal Gas

Law & Avogadro's Law

Gases

knV

V

n

A. Avogadro’s LawEqual volumes of gases

contain equal numbers of moles• at constant temp &

pressure

PVTVn

PVnT

B. Ideal Gas Law

= kUNIVERSAL GAS

CONSTANTR=0.0821

Latm/molKR=8.315

dm3kPa/molK

= R

You don’t need to memorize these values!

• Boyle’s, Charles, and Avogadro’s laws are contained within the ideal gas law!

• Merge the Combined Gas Law with Avogadro’s Law:

B. Ideal Gas Law

GAS CONSTANTR=0.0821

Latm/molKR=8.315

dm3kPa/molK

PV=nRT

You don’t need to memorize these values!

P= pressureV = volume n = molesT = temperature (in K)R = the ideal gas constant

C. Johannesson

GIVEN:P = ? atmn = 0.412 molT = 16°C = 289 KV = 3.25 LR = 0.0821Latm/molK

WORK:P = nRT

VP =(0.412 mol)(0.0821Latm/molK) (289K)

3.25L

P = 3.01 atm

C. Ideal Gas Law Problems1.) Calculate the pressure in atmospheres of

0.412 mol of He at 16°C & occupying 3.25 L.

PV = nRT

C. Johannesson

GIVEN:

V = ?n = 85 gT = 25°C = 298 KP = 104.5 kPaR = 8.315 dm3kPa/molK

C. Ideal Gas Law Problems2.) Find the volume of 85 g of O2 at 25°C

and 104.5 kPa. PV = nRT

= 2.7 mol

WORK:

85 g O2 1 mol O2 = 2.7 mol 31.998 g O2 O2

V = nRT PV =(2.7 mol)(8.315 dm3kPa/molK) (298K)

104.5 kPa

V = 64 dm3

C. Johannesson

GIVEN:

V = 1.00 Ln = ?T = 28°C = 301 KP = 98.7 kPaR = 8.314 LkPa/molK

C. Ideal Gas Law Problems3.) At 28C and 98.7 kPa, 1.00 L of an unidentified gas has

a mass of 5.16 g. Calculate the number of moles of gas present and the molar mass of the gas. PV = nRT

WORK:

n = PV RT

(98.7 kPa) (1.00L)(8.314 LkPa/molK) (301 K)

n = 0.0394 mol

m = 5.16g 0.0394 mol

=

= 131 g/mol

Gases

IV. Two More Laws: Dalton’s Law & Graham’s Law

A. 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 + P3 ...

A. Dalton’s Law

Ptotal = P1 + P2 + P3 ...

A. Dalton’s Law

When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 AND water vapor.

C. Johannesson

GIVEN:PH2 = ?Ptotal = 94.4 kPaPH2O = 2.72 kPa

WORK:Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

A. Dalton’s Law1.) 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 on gas formula sheet 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.

C. Johannesson

GIVEN:Pgas = ?Ptotal = 742.0 torrPH2O = 42.2 torr

WORK:Ptotal = Pgas + PH2O

742.0 torr = PH2 + 42.2 torr

Pgas = 699.8 torr

2.) 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 on gas formula sheet for

35.0°C.

Sig Figs: Round to least number of decimal places.

A. 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.

B. Graham’s LawDiffusion =

• Spreading of gas molecules throughout a container until evenly distributed.

Effusion =• Passing of gas molecules

through a tiny opening in a container

B. Graham’s Law

KE = ½mv2

Speed of diffusion/effusion• Kinetic energy is determined by

the temperature of the gas.• At the same temp & KE, heavier

molecules move more slowly.Larger m smaller v

B. Graham’s LawKE = kinetic energym= molar massv = velocity

KE = ½mv2

C. Johannesson

B. Graham’s LawGraham’s Law = Rate of effusion 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

mm

vv

1.) Determine the relative rate of effusion for krypton and bromine.

1.381

Kr effuses 1.381 times faster than Br2.

Kr

Br

Br

Kr

mm

vv 2

2

A

B

B

A

mm

vv

g/mol83.80 g/mol159.80

C. Graham’s Law Problems

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

C. Johannesson

2.) 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

mm

vv

2

2

2

2

H

O

O

H

mm

vv

g/mol 2.02g/mol32.00

m/s 12.3vH 2

C. Graham’s Law Problems

3.980m/s 12.3

vH 2

m/s49.0 vH 2

Put the gas with the unknown

speed as “Gas A”.

C. Johannesson

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

Amg/mol32.00 16

A

B

B

A

mm

vv

A

O

O

A

mm

vv 2

2

Amg/mol32.00 4.0

16g/mol32.00 mA

2

Amg/mol32.00 4.0

g/mol2.0

C. Graham’s Law Problems

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.

Gases VI. Gas Stoichiometry at Non-STP Conditions

A. Gas Stoichiometry Moles Liters of a Gas:

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

Non-STP• Given liters of gas?

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

start with stoichiometry conv.

B. Volume Mass Stoich.Thinking:Gas volume A moles A moles B mass B

If gas A is at STP:

? L “A” 1 mol “A” ? mol “B” molar mass(g) “B” 22.4 L “A” ? mol “A” 1 mol “B”

Mole Ratio

B. Volume Mass Stoich.If gas “A” is NOT at STP:

Use PV = nRT to find moles of “A”

? mol “A” ? mol “B” molar mass (g) “B”

? mol “A” 1 mol “B”

C. Johannesson

WORK:n = PV RTn= (97.3 kPa) (15.0 L) (8.315dm3kPa/molK) (294K)

n = 0.597 mol O2

B. Volume Mass1.) 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 Ln = ?T = 21°C = 294 KR = 8.315 dm3kPa/molK

4 Al + 3 O2 2 Al2O3 15.0 L

non-STP ? gGiven liters: Start with

Ideal Gas Law and calculate moles of O2.

NEXT

2 mol Al2O3

3 mol O2

B. Volume Mass1.) 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.

1 mol SO2

22.4L SO2

B. Volume Mass2.) What mass of sulfur is required to

produce 12.6 L of sulfur dioxide at STP according to the equation:

S8 (s) + 8 O2 (g) 8 SO2 (g)

12.6 L SO2 = 18.0 g

S8

256.582 g S8

1 molS8

12.6L? g

1 mol S8

8 molSO2

C. Mass VolumeThinking:Mass A moles A moles B gas volume B

If gas “B” is at STP:

? g “A” 1 mol “A” ? mol “B” 22.4 L “B” molar mass (g) “A” ? mol “A” 1 mol “B”

C. Mass Volume If gas “B” is NOT at STP:

? g “A” 1 mol “A” ? mol “B” molar mass (g) “A” ? mol “A”

• Then use PV = nRT to find volume of “B”

C. Mass Volume1.) What volume of chlorine gas at STP is needed to react

completely with 10.4 g of sodium to form NaCl?

2 Na + Cl2 2 NaCl 10.4 g ? L

10.4 g Na

22.990 g Na

22.4 L Cl21 mol Cl2

2 mol Na 1 mol Cl2

1 mol Na

= 5. 07 L Cl2

1 molCaCO3

100.09g CaCO3

C. Mass Volume2.) What volume of CO2 forms from 5.25 g

of CaCO3 at 103 kPa & 25ºC?

5.25 gCaCO3 = 1.26 mol CO2

CaCO3 CaO + CO2

1 molCO2

1 molCaCO3

5.25 g ? Lnon-STPLooking for liters: Start with stoich

and calculate moles of CO2.

Plug this into the Ideal Gas Law to find liters.

C. Johannesson

WORK:V = nRT PV = (1.26mol)(8.315dm3kPa/molK) (298K)

(103 kPa)

V = 30.3 dm3 CO2

C. Mass Volume

2.) What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?

GIVEN:P = 103 kPaV = ?n = 1.26 molT = 25°C = 298 KR = 8.315 dm3kPa/molK