Gas LawsGas Laws

• highly compressible.• occupy the full volume of their containers.• exert a uniform pressure on all inner surfaces of a

container• diffuse (mix) easily and quickly• have very low densities.

Characteristics of GasesCharacteristics of Gases

Kinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular Theory

– Gases consist of a large number of molecules in constant random motion.

– Volume of individual molecules negligible compared to volume of container.

– Intermolecular forces (forces between gas molecules) negligible.

– Collision of gas particles are elastic so no kinetic energy is lost

– As temperature increases the gas particles move faster, hence increased kinetic energy.

Four Physical Quantities Four Physical Quantities for Gasesfor GasesFour Physical Quantities Four Physical Quantities for Gasesfor Gases

Phys. Qty. Symbol SI unit Other common units

pressure PPascal (Pa)

atm, mm Hg, torr, psi

volume V m3 dm3, L, mL, cm3

temp. T K °C, °F

moles n mol

TemperatureTemperatureTemperatureTemperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

Always use absolute temperature (Kelvin) when working with gases.

Calculate the missing temperatures

0C = _______ K 100C = _______ K

100 K = _______ C – 30C = _______ K

300 K = _______ C 403 K = _______ C

25C = _______ K 0 K = _______ C

Practise

Absolute zero is – 273C or 0 K

What is the approximate temperature for absolute zero in degrees Celsius and kelvin?

Calculate the missing temperatures

0C = _______ K 100C = _______ K

100 K = _______ C – 30C = _______ K

300 K = _______ C 403 K = _______ C

25C = _______ K 0 K = _______ C

Kelvin Practice

273 373

– 173 243

27 130

298 – 273

Absolute zero is – 273C or 0 K

PressurePressurePressurePressure

area

forcepressure

Which shoes create the most pressure?

Pressure (P ) is defined as the force exerted per unit area. The atmospheric pressure is measured using a barometer.

PressurePressurePressurePressure

2m

NkPa

KEY UNITS AT SEA LEVEL

101.325 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

14.7 psi•1 atm = 760 mmHg = 760 torr = 101325 Pa.

PressurePressurePressurePressure

Barometer• measures atmospheric pressure

Mercury Barometer

Aneroid Barometer

STPSTPSTPSTP

Standard Temperature & PressureStandard Temperature & Pressure

273 K

101.325 kPa

STP

STPSTPSTPSTP

Standard Laboratory ConditionsStandard Laboratory Conditions

25°C or 298 K

101.325 kPa

SLC

-BOYLES-CHARLE

-GAY-LUSSAC

-BOYLES-CHARLE

-GAY-LUSSAC

The Gas LawsThe Gas Laws

Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law

The pressure and volume of a gas are inversely related at constant mass & temp.

P

V

PV = k

2211 VPVP

A. Boyle’s LawA. Boyle’s LawA. Boyle’s LawA. Boyle’s Law

PractisePractisePractisePractise

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

kT

V

V

T

Charles’ LawCharles’ LawCharles’ LawCharles’ Law

The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

2

2

1

1

T

V

T

V

Charles’ LawCharles’ LawCharles’ LawCharles’ Law

PractisePractisePractisePractise

2. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K?

3. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

2. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K?

3. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K3.5 L / 300 K = V2 / 200 K

V2 = (3.5 L/300 K) x (200 K) = 2.3 L

V1 = 1 L, T1 = 22°C = 295 KV2 = ?, T2 = 100 °C = 373 KV1/T1 = V2/T2, 1 L / 295 K = V2 / 373 KV2 = (1 L/295 K) x (373 K) = 1.26 L

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A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L

= = 192 K

V1/T1 = V2/T2

kT

PP

T

Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law

The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law

Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1

GIVEN:

V1 = 473 cm3

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

P1V1T2 = P2V2T1

E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. 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

GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

P1V1T2 = P2V2T1

E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. 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

PractisePractisePractisePractise

A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

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 ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. 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

GIVEN:

P1 = 765 torr

T1 = 23°C = 296K

P2 = 560. torr

T2 = ?

WORK:

P1V1T2 = P2V2T1

E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. 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)(309K)

T2 = 226 K = -47°C

kn

V

V

n

Avogadro’s PrincipleAvogadro’s PrincipleAvogadro’s PrincipleAvogadro’s Principle

Equal volumes of all gases contain equal numbers of moles at constant temp & pressure.

2

2

1

1

n

V

n

V

2

2

1

1

n

V

n

V

The Ideal Gas EquationThe Ideal Gas EquationThe Ideal Gas EquationThe Ideal Gas Equation

The gas laws can be combined into a general equation that describes the physical behavior of all gases.

11.5

nTV

P

VP

1

Boyle’s law

V n

Avogadro’s law

V T

Charles’s law

nTV R

P PV = nRTrearrangement

R is the proportionality constant, called the gas constant.

B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law

UNIVERSAL GAS CONSTANT

R = 8.3145 J/mol·KR=0.0821 Latm/molK

PV=nRT

R = 0.0821 liter·atm/mol·K R = 8.3145 J/mol·KR = 8.2057 m3·atm/mol·KR = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0821Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law Calculate the pressure in atmospheres of

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

IDEAL GAS LAW

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/molK

B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law

Find the volume of 85 g of O2 at 25°C and 104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol = 2.7 mol

32.00 g

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

IDEAL GAS LAW

SummaryDalton found that the total pressure of mixed gases is equal to the sum of their individual pressures (provided the gases do not react).

+

1 L oxygen

50 kPa

1 L nitrogen

100 kPa

1 L mixed gas

150 kPa

=

Note: all of these volumes are the same

Gas MixturesGas Mixtures

Each component of a gas mixture exerts a pressure independent of the other components. The total pressure is the sum of the partial pressures.

Practise

• Calculating partial pressures

Vapour Pressure Defined• Vapour pressure is the pressure exerted by a

vapour. E.g. the H2O(g) in a sealed container.

Eventually the air above the water is filled with vapour pushing down. As temperature , more molecules fill the air, and vapour pressure .

Collecting gases over water• Many times gases are collected over H2O• Often we want to know the volume of dry

gas at STP (useful for stoichiometry).For this we must make 3 corrections:1. The level of water inside and outside the

tube must be level (so pressure inside is equal to the pressure outside).

2. The water vapour pressure must be subtracted from the total pressure (to get the pressure of the dry gas).

3. Finally, values are converted to STP using the combined gas law.

Sample calculationA gas was collected over 21°C H2O. After equal-izing water levels, the volume was 325 mL. Give the volume of dry gas at STP (Patm=102.9 kPa).Step 1: Determine vapour pressure (pg. 464)

At 21°C vapour pressure is 2.49 kPaStep 2: Calculate the pressure of dry gas

Pgas = Patm - PH2O = 102.9 - 2.49 = 100.41 kPaStep 3: List all of the data

T1 = 294 K, V1 = 325 mL, P1 = 100.41 kPaStep 4: Convert to STP

(P1)(V1)(T2)(P2)(T1)

V2=(100.4 kPa)(325 mL)(273 K)

(101.325 kPa)(294 K)=

= 299 mL

Assignment1. 37.8 mL of O2 is collected by the downward

displacement of water at 24°C and an atmospheric pressure of 102.4 kPa. What is the volume of dry oxygen measured at STP?

2. Try questions 8 – 10 on page 465.3. 236 mL of H2 is collected over water at 22°C

and at an atmospheric pressure of 99.8 kPa. What is the volume of dry H2 at STP?

4. If H2 is collected over water at 22°C and an atmospheric pressure of 100.8 kPa, what is the partial pressure of the H2 when the water level inside the gas bottle is equal to the water level outside the bottle?

1)

V1 = 37.8 mL, P1 = 99.42 kPa, T1 = 297 K

V2 = ?, P2 = 101.3 kPa, T2 = 273 K

P1V1

T1

=P2V2

T2

(99.42 kPa)(37.8 mL)(297 K)

=(101.3 kPa)(V2)

(273 K)

(99.42 kPa)(37.8 mL)(273 K)(297 K)(101.3 kPa)

=(V2) = 34.1 mL

Vapor pressure at 24C = 2.98 kPaPgas = Patm - Pvapor

= 102.4 kPa - 2.98 kPa= 99.42 kPa = P1

3)

V1 = 236 mL, P1 = 97.16 kPa, T1 = 295 K

V2 = ?, P2 = 101.3 kPa, T2 = 273 K

P1V1

T1

=P2V2

T2

(97.16 kPa)(236 mL)(295 K)

=(101.3 kPa)(V2)

(273 K)(97.16 kPa)(236 mL)(273 K)

(295 K)(101.3 kPa)=(V2) = 209 mL

Vapor pressure at 22C = 2.64 kPaPgas = Patm - Pvapor

= 99.8 kPa - 2.64 kPa= 97.16 kPa = P1

Answers4 - Total pressure = PH2 + PH2O

100.8 kPa = PH2 + 2.64 kPa

100.8 kPa - 2.64 kPa = PH2 = 98.16 kPa

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Ideal and real gas

• Ideal gas:

• Ideal gas molecule have

- zero volume

-zero attraction between molecules

• Real gases behave ideally at

- high temperature

- low pressure