Gases and Gas Gases and Gas LawsLaws

Chemistry– Unit 11: Chapter 14 Chemistry– Unit 11: Chapter 14

Kinetic Molecular Theory (THIS IS IMPORTANT!!)

• Particles in an ideal gas…– have no volume.– have elastic collisions. – are in constant, random, straight-line

motion.– don’t attract or repel each other.– have an avg. KE directly related to

Kelvin temperature.

Real Gases

• 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

Characteristics of Gases• Gases expand to fill any container.

– random motion, no attraction

• Gases are fluids (like liquids).– no attraction

• Gases have very low densities.– no volume = lots of empty space

Characteristics of Gases• Gases can be compressed.

– no volume = lots of empty space

• Gases undergo diffusion & effusion.– random motion

The Meaning of Temperature

• Kelvin temperature is an index of the random motions of gas particles (higherT means greater motion.)

(KE)32avg RT

Kinetic Energy of Gas ParticlesKinetic Energy of Gas ParticlesAt the same conditions of temperature, all gases have the same average kinetic energy.

m = mass

v = velocity

2

2

1mvKE

Temperature

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.

ºF =

ºC =

Absolute Zero: _____________________________________

Absolute Zero = ____K = _____ºCTemperature at which motion stops. 0 - 273

32 C 59 o +

32) - F( 95 o

Example Conversions:  (a) Convert 355 ºF to ºC and

K    

(b) Convert -40 ºC to K and ºF

oC = 179oC

K = 179 oC + 273

= 452 K

oF = -40

oFK = - 40 oC + 273

= 233 K

32 C40- 5

9

o+=

32) - (355 95=

Pressure

area

forcepressure

Which shoes create the most pressure?

Measuring Pressure

The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century.

The device was called a “barometer”.

Baro = weight Meter = measure

Pressure• Barometer

– measures atmospheric pressure

Mercury Barometer

Aneroid Barometer

Pressure

2m

NkPa

• KEY UNITS AT SEA LEVEL

101.3 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

14.7 psi** All of these amounts are equal to each other (conversion factors), just in different units!

Pressure ConversionsA. What is 475 mm Hg expressed in atm?

1 atm 760 mm Hg

B. The pressure of a tire is measured as 29.4 psi.What is this pressure in mm Hg?

760 mm Hg 14.7 psi = 1.52 x 103 mm Hg

= 0.625 atm475 mm Hg x

29.4 psi x

Pressure Conversions

A. What is 2 atm expressed in torr?

B. The pressure of a tire is measured as 32.0 psi.What is this pressure in kPa?

IQ 11) List the properties of an ideal gas.

2) What is this theory called?

3) How is an ideal gas different from a real gas?

IQ #31) What are 2 ways to increase the

pressure of a gas (Think about the formula for pressure)?

2) What effect does temperature have on kinetic energy?

3) What do you expect would happen to the volume of a balloon if the gas inside it made more collisions with it?

Boyle’s Law

P

V

PV = k

Robert Boyle (1627-Robert Boyle (1627-1691). Son of Earl of 1691). Son of Earl of Cork, Ireland.Cork, Ireland.

Boyle’s Law

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

P

V

PV = k

Boyle’s LawThis means Pressure This means Pressure and Volume are and Volume are INVERSELY INVERSELY PROPORTIONALPROPORTIONAL if if moles and temperature moles and temperature are constant (do not are constant (do not change). change). For example, P goes up For example, P goes up as V goes down.as V goes down.

PP11VV11 = P = P22 V V22

As the volume of As the volume of the air trapped in the air trapped in the bicycle pump is the bicycle pump is reduced, its reduced, its pressure goes up, pressure goes up, and air is forced and air is forced into the tire.into the tire.

GIVEN:V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:P1V1= P2V2 V2= P1V1

P2

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

kT

VV

T

Charles’ Law

kT

VV

T

Charles’ Law• The volume and absolute

temperature (K) of a gas are directly related – at constant mass & pressure

Charles’s Charles’s LawLaw

V and T are V and T are directly directly proportional.proportional.

VV11 V V22

==

TT11 T T22 • If one temperature If one temperature

goes up, the volume goes up, the volume goes up!goes up!

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

Charles’s original balloonCharles’s original balloon

Modern long-distance balloonModern long-distance balloon

GIVEN:V1 = 473 cm3

T1 = 36°C = 309KV2 = ?

T2 = 94°C = 367K

WORK:V1 = V2

T1 T2

Gas Law Problems

• A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.

CHARLES’ LAW

T V

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

V2 = 562 cm3

V2 = V1T2

T1

kT

PP

T

Gay-Lussac’s Law

kT

PP

T

Gay-Lussac’s Law• The pressure and absolute

temperature (K) of a gas are directly related – at constant mass & volume

Gay-Lussac’s LawP and T are P and T are directlydirectly proportional.proportional.

PP11 P P22

==

TT11 T T22

• If one temperature goes If one temperature goes

up, the pressure goes up!up, the pressure goes up!

Joseph Louis Gay-Joseph Louis Gay-Lussac (1778-1850)Lussac (1778-1850)

GIVEN:P1 = 765 torr

T1 = 23°C = 296KP2 = 560.0 torr

T2 = ?

WORK:P1 = P2

T1 = T2

Gas Law Problems

• A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S

LAWP T

T2 = (560.0 torr)(296K)/765 torr

T2 = 217 K = -56°C

T2 = T1P2

P1

Combined Gas Law

• The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!

P1 V1 P2 V2

= T1 T2

Combined Gas Law

If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!

=

P1 V1

T1

P2 V2

T2

Boyle’s Law

Charles’ Law

Gay-Lussac’s Law

GIVEN:

V1 = 7.84 cm3

P1 = 71.8 kPa

T1 = 25°C = 298 KV2 = ?

P2 = 101.325 kPaT2 = 273 K

WORK:

P1V1T2 = P2V2T1

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

=(101.325 kPa) V2 (298 K)

V2 = 5.09 cm3

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

Combined Gas Law Problem #2

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Set up Data TableP1 = 0.800 atm V1 = 180 mL T1 = 302

KP2 = 3.20 atm V2= 90 mL T2 = ??

CalculationP1 = 0.800 atm V1 = 180 mL T1 = 302 KP2 = 3.20 atm V2= 90 mL T2 = ??

P1 V1 P2 V2

= P1 V1 T2 = P2 V2 T1

T1 T2

T2 = P2 V2 T1

P1 V1

T2 = 3.20 atm x 90.0 mL x 302 K

0.800 atm x 180.0 mL

T2 = 604 K - 273 = 331 °C

= 604 K

Learning Check (Group)

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

IQ 2

1. Covert -56 °C to Kelvin.

2. How many torr in 37.8 psi?

3. Covert 0 °C into °F.

One More Practice Problem (Group)

A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

Try This One on Your Own!

A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

STP

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.3 kPa-OR-

STP allows us to compare amounts of

gases between different pressures and temperatures

STP allows us to compare amounts of

gases between different pressures and temperatures

Avogadro’s Avogadro’s HypothesisHypothesis

Equal volumes of gases at the Equal volumes of gases at the same T and Psame T and P have the same have the same number of number of moleculesmolecules..

V = n (RT/P) = knV = n (RT/P) = kn

V and n are V and n are directlydirectly related. related.

twice as twice as many many moleculesmolecules

P proportional to n

The gases in this

experiment are all

measured at the same T

and V.

Avagadro’s Hypothesis and the Kinetic Molecular Theory

IDEAL GAS LAWIDEAL GAS LAW

Brings together gas Brings together gas properties.properties.

Can be derived from Can be derived from experiment and experiment and theory.theory.

BE SURE YOU KNOW BE SURE YOU KNOW THIS EQUATION!THIS EQUATION!

P V = n R TP V = n R T

Using PV = nRTUsing PV = nRTP = PressureV = VolumeT = Temperaturen = number of molesR is a constant, called the Ideal Gas Constant

R = 0.0821R = 0.0821L • atm

Mol • K

Sample Problem Sample Problem #1#1

How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC?

SolutionSolution

1. Get all data into proper units1. Get all data into proper units V = 27,000 LV = 27,000 L

T = 25 T = 25 ooC + 273 = 298 KC + 273 = 298 K

P = 745 mm Hg (1 atm/760 mm Hg) P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm = 0.98 atm

And we always know R, 0.0821 L atm / mol And we always know R, 0.0821 L atm / mol KK

Using PV = nRTUsing PV = nRTHow much N2 is req’d to fill a small room with a

volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?

Solution2. Now plug in those values and solve for the

unknown.

PV = nRTn = (0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)n =

(0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)

n = 1.1 x 10n = 1.1 x 1033 mol (or about 30 kg of gas) mol (or about 30 kg of gas)

RT RTRT RT

Learning Check

Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

Learning Check #2

A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

IQ #31) What are “STP” conditions?

2) What is the volume of an ideal gas?

3) A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

273 K/ 1 atm273 K/ 1 atm

An ideal gas has no An ideal gas has no volume!!volume!!

6.4 g6.4 g

Deviations from Ideal Gas Deviations from Ideal Gas LawLaw

• Real molecules have volume.• The ideal gas consumes the

entire amount of available volume. It does not account for the volume of the molecules themselves.

• There are intermolecular forces.• An ideal gas assumes there are

no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.– Otherwise a gas could not

condense to become a liquid.

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-STPNon-STP– Given liters of gas?

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

•start with stoichiometry conv.

1 molCaCO3

100.09g

CaCO3

Gas Stoichiometry Problem #1

• What volume of CO2 forms from 5.25 g of CaCO3 at 1.02 atm & 25ºC?

5.25 gCaCO3

= 0.0524 mol CO2

CaCO3 CaO + CO2

1 molCO2

1 molCaCO

3

5.25 g ? L non-STPLooking for liters: Start with

stoich and calculate moles of CO2.

Plug this into the Ideal Gas Law to find

liters.

WORK:

PV = nRT(1.02 atm)V=(0.0524 mol)(0.0821 Latm/molK) (298K)

V = 1.26 L

Gas Stoichiometry Problem #1• What volume of CO2 forms from 5.25

g of CaCO3 at 1.02 atm & 25ºC?

GIVEN:

P = 1.02 atmV = ?

n = 0.0524 molT = 25°C = 298 KR = 0.0821 Latm/molK

WORK:

PV = nRT(.961 atm) (15.0 L)= n (0.0821 Latm/molK) (294K)

n = 0.597 mol O2

Gas Stoichiometry Problem #2• How many grams of Al2O3 are formed from

15.0 L of O2 at 97.3 kPa & 21°C?

GIVEN:P = 97.3 kPa= .961atm

V = 15.0 L

n = ?T = 21°C = 294 KR = 0.0821

Latm/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

Gas Stoichiometry Problem #2• How many grams of Al2O3 are formed

from 15.0 L of O2 at 97.3 kPa & 21°C?

0.597 mol 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.

Gas Stoichiometry Problem Gas Stoichiometry Problem #3#3

2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)

Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with in a flask with a volume of 2.50 L. What is the volume a volume of 2.50 L. What is the volume of Oof O22 at STP? at STP?

Bombardier Bombardier beetle uses beetle uses decomposition of decomposition of hydrogen hydrogen peroxide to peroxide to defend itself.defend itself.

Gas Stoichiometry Problem Gas Stoichiometry Problem #3#3

2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)

Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a in a flask with a volume of 2.50 L. What is the volume of Ovolume of 2.50 L. What is the volume of O22 at at STP? STP?

SolutionSolution 1.1 g1.1 g HH22OO22 1 mol H 1 mol H22OO22 1 mol O 1 mol O22 22.4 L 22.4 L

OO22 34 g H34 g H22OO22 2 mol H 2 mol H22OO22 1 mol 1 mol OO22 = 0.36 L O2 at

STP!!

Gas Stoichiometry: Practice!A. What is the volume at STP of

4.00g of CH4?

•How many grams of He are present in 8.0 L of gas at STP?

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.

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

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.

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.

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.

Graham’s Law• DiffusionDiffusion

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

• EffusionEffusion– Passing of gas molecules through a

tiny opening in a container

Graham’s Law

KE = ½mv2

• Speed of diffusion/effusionSpeed 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

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

• 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

Graham’s Law

The first gas is “Gas A” and the second gas is “Gas B”.

Relative rate mean find the ratio “vA/vB”.

• 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

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”.

• 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

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.