Gas Laws

2

Importance of Gases Airbags fill with N2 gas in an accident. Gas is generated by the

decomposition of sodium azide, NaN3.2 NaN3 (s) → 2 Na (s) + 3 N2 (g)

3

Cellular respiration is going on throughoutyour body! It involves two gases: Oxygen Carbon Dioxide

You breathe in oxygenand breathe out carbon dioxide.

Importance of Gases

C6H12O6 + O2 → CO2 + H2OGlucose

4

Kinetic Molecular TheoryGas molecules are always moving.Moving molecules have kinetic energy (KE).

At the same temperature,all gases have the same average KE.

As temperature goes up,average KE also increases.

5

Velocity of Gas MoleculesMolecules of a given gas have a range

of speeds.

6

General Properties of GasesGeneral Properties of Gases

There is a lot of There is a lot of ““freefree”” space spacein a gas.in a gas.

Gases can be expandedGases can be expandedinfinitely.infinitely.

Gases occupy containersGases occupy containersuniformly and completely.uniformly and completely.

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

7

More Properties of GasesMore Properties of Gases

tiny particles that are fartiny particles that are farapart relative to their sizeapart relative to their size

constant, random motionconstant, random motion

elastic collisionselastic collisions(with(with no loss of kinetic energy)no loss of kinetic energy)

the average kinetic energythe average kinetic energydepends upon temperaturedepends upon temperature

8

Pressure of air ismeasured with aBAROMETER

PressurePressure

developed in 1643 by Evangelista Torricelli

9

PressurePressureHg rises in tube untilHg rises in tube until

force of Hg (down)force of Hg (down)balances the force ofbalances the force ofatmosphere (pushingatmosphere (pushingup).up).

P of Hg pushing downP of Hg pushing downis related tois related to

Hg densityHg density column heightcolumn height

10

PressurePressureColumn heightColumn heightmeasures pressuremeasures pressureof atmosphereof atmosphere 1 standard 1 standard atmatm

= 760 mm Hg= 760 mm Hg = 760 = 760 torr torr (named after Torricelli)(named after Torricelli)

SI unit is PASCAL, PaSI unit is PASCAL, Pa 1 1 atm atm = 101.3 = 101.3 kPakPa

11

BoyleBoyle’’s Laws LawFor a sample of gas at a

constant temperature:

PV = k As P goes up, V goes down. As V goes up, P goes down.

Comparing a sample of gasat two different pressures:

P1V1 = P2V2

Robert BoyleRobert Boyle(1627-1691)(1627-1691)1414thth child of child of

Earl of Cork, Ireland.Earl of Cork, Ireland.

12

Boyle’s Law: PV=kAt constant T and n,

Volume is inversely proportional to Pressure.

13

Boyle’s Law Example #1 A gas occupies 4.0 L at 760 torr;

what volume will it occupy at 380 torr?

!

V2

=P

1V

1

P2

=760torr( ) 4.0 L( )

380 torr

P1V1 = P2V2

!

V2

= 8.0 L

14

Boyle’s Law Example #2 A gas occupies 12 L at 1.2 atm;

what volume will it occupy at 2.4 atm?

!

V2

=P

1V

1

P2

=1.2 atm( ) 12 L( )

2.4 atm

!

V2

= 6.0 L

P1V1 = P2V2

15

Temperature For gas laws, we must convert

temperature to Kelvin:

K = °C + 273

0-273-460AbsoluteZero

273032H20 freezes

2932068Roomtemperature

373100212H20 boils

KKelvin

°CCelsius

°FFahrenheit

The symbolis K, not °K

16

Jacques CharlesJacques Charles(1746-1823)(1746-1823)

Isolated boron.Isolated boron.Studied gases.Studied gases.

Balloonist.Balloonist.

CharlesCharles’’s Laws LawFor a sample of gas under constant pressure:

V = kTV and T (in Kelvin) are directly proportional:

As V goes up, T goes up.As V goes down, T goes down.

!

V1

T1

=V2

T2

For a sample of gas at two different temperatures:

17

Charles’s Law: V=kTAt constant P and n,

Volume is directly proportional to Temperature.

18

Charles’ Law Example A gas occupies 117 mL at 100°C.

At what temperature will it havea volume of 234 mL?

!

T2

=V

2T

1

V1

=234mL( ) 373 K( )

117 mL !

V1

T1

=V2

T2

!

T2 = 746K = (746 - 273)°C = 473 °C

19

Lussac’s Law For a sample of gas in a fixed volume:

P = kT Pressure and Temperature (in K)

are directly proportional:

Joseph Gay-Lussac(1788 - 1850)

!

P1

T1

=P2

T2

As P goes up, T goes up.As P goes down, T goes down.

20

Lussac’s Law: P=kTAt constant V and n,

Pressure and Temperature (in K) are directly proportional.

DoWs #1 & Ws #2

Robert Boyle says…

22

Combined Gas Law

PV

T

PV

T

1 1

1

2 2

2

=

Temperature must be in Kelvin.

23

Combined Law Example V1 = 105 L, V2 = ?

P1 = 985 torr, P2= 760 torr

T1 = 27°C, T2 = 273 K

!

V2

=P1V1T2

P2T1

=985torr "105L " 273K

760torr " 300K

!

V2

=124L

24

Standard Temperature andPressure

1.00 atm 273 K

STP(0°C)

25

STP Problem (Ws #3: Question #2)

A 700.0 mL gas sample at STP is compressed to avolume of 200.0 mL, and the temperature isincreased to 30.0°C.What is the new pressure of the gas in kPa?

!

T1

= TSTP

= 0°C = (0 + 273)K = 273K

!

P1

= PSTP

=1atm =101.3kPa

!

P1V1

T1

=P2V2

T2

!

P2

=P1V1T2

T1V2

=101.3kPa" 700.0mL " 303K

273K " 200.0mL!

T2

= 30°C = (273+ 30)K = 303K

!

P2

= 394kPa

Now do Ws #3Jacques Charles says…

27

IDEAL GAS LAW

Combines the Gas Laws.Combines the Gas Laws.Can be derived fromCan be derived from

experiment and theory.experiment and theory.

P V = n R T

28

P V = n R T The Ideal Gas Law holds for all gases

and mixtures of gases, as long as theyare ideal gases.

P = Pressure (atm, kPa, torr, mm Hg)V = Volume (L)n = Number of moles (mol)T = Temperature (K)R = Gas Law Constant

29

0.0821 (L atm)/(K mol)or

8.31 (L kPa)/(K mol)

R is the Gas Law Constant

P V = n R T

The units of pressure (atm or kPa) indicate which value of R to use.

30

STP Problem What is the volume of 1.00 mol He at

STP?

!

V =nRT

P=

1.00mol " 0.0821L # atm

mol #K" 273K

1atm

!

V = 22.4LRemember this number!

31

1 mole of any gas at STPhas the same volume:

1 atm, 273 K

STP 22.4 Liters1 mol H2 at STP = 22.4 L

1 mol N2 at STP = 22.4 L

1 mol Ar at STP = 22.4 L

1 mol He at STP = 22.4 L

1 mol Air at STP = 22.4 L

1 moleof any gas

at STP

22.4 L

32

Ideal–Gas Equation and Molecular Weight

density = g/V = PMW/RT

n=g/MWPV = nRT

PV=gRT/MW

33

Using PV = nRTHow much N2 would fill a small room with a

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

R = 0.0821 L•atm/K•molSolution1. Get all data into proper units V = 27,000 L T = 25 oC + 273 = 298 K

!

P = 745mmHg "1atm

760mmHg

!

P = 0.980atm

34

Using PV = nRTHow much N2 would fill a small room with a

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

R = 0.0821 L•atm/K•molSolution1. Get all data into proper units2. Now calculate n = PV / RT

n = 1.1 x 103 mol (or about 30 kg of gas)

!

n = 0.980 atm" 27000 L

0.0821L # atm

K #mol" 298 K

35

PV=nRT Example #2 What volume would 4.00 mol of He

occupy at a pressure of 748 torr and atemperature of 30.0°C?

!

V =nRT

P=

4.00 mol( ) 0.0821 L " atm

K "mol

#

$ %

&

' ( 303 K( )

0.984 atm ( )!

P = 748torr "1atm

760torr= 0.984atm

V = 101L

36

PV=(g/MW)RT Calculate the MW (Molecular Weight)

of a compound if 0.109 g occupied112 mL at 100.0°C and 750. torr?

!

MW =gRT

PV =

0.109g( ) 0.0821 L " atm

K "mol

#

$ %

&

' ( 373K( )

750

760atm

#

$ %

&

' ( 0.112L( )

!

MW = 30.2g /mol

37

Low Low densitydensity

High High densitydensity

Gas DensityGas Density

!

d =m

V=PMW

RT

38

Density What is the density of NO2 if it

occupies 1.00L at 1.24 atm and 50°C?

!

d =g

V=MWP

RT

!

d =

46.0g

mol

"

# $

%

& ' 1.24atm( )

0.0821 L ( atm

K (mol

"

# $

%

& ' 323K( )

!

d = 2.15g

L

!

MWNO2=14.0 + (2 "16.0) = 46.0g /mol

!

T = 50°C = (273+ 50)K = 323K

Do Ws #4Joseph Gay-Lussac says…

40

Dalton’s Law

Component (%) (torr)Nitrogen (N2) 78.03 593.0Oxygen (O2) 20.99 159.5Argon (Ar) 0.94 7.14Carbon dioxide (CO2) 0.03 0.23Other gases Trace 0.13

Total Pressure 760.

The partial pressure of a gas in a mixtureis simply the fraction of the total pressurewhich that gas exerts.

41

Gas Mixtures and PartialPressures

Ptotal = P1 + P2 + …ntotal = nA + nB + ...

...++=

=

V

RTn

V

RTnP

V

RTnP

BA

total

total

total

42

Partial Pressures Calculate the pressure of a 10.0 L container

with 0.200 mole methane, 0.300 mole H2and 0.400 mole N2 at 298K.

!

P =nRT

V=

0.900mol( ) 0.0821L " atm

K "mol

#

$ %

&

' ( 298K( )

10.0L( )!

Total_moles_of _ gas = 0.200 + 0.300 + 0.400 = 0.900mol

!

P = 2.20atm

43

4 Tools for SolvingGas Stoichiometry Problems

1) Use the Molar Mass to convert: Mass (g) → Moles Moles → Mass (g)

2) Use the Balanced Chemical Equation toconvert:

Moles of X → Moles of Y3) At STP: 1 mole of gas = 22.4 L4) If not at STP, use PV=nRT

44

Gas Stoichiometry #1 How many liters of oxygen will be formed at STP

from the decomposition of 112g of KClO3?

2 KClO3 → 2 KCl + 3 O2

!

LO2= 112 g KClO3( )

mol KClO3

122.6gKClO3

"

# $

%

& '

3mol O2

2mol KClO3

"

# $

%

& '

22.4L O2

1mol O2

"

# $

%

& ' = 30.7L O2

!

molO2= 112 g KClO3( )

1mol KClO3

122.6gKClO3

"

# $

%

& '

3mol O2

2mol KClO3

"

# $

%

& ' =1.37molO2

Method #1: using 1mol gas at STP = 22.4 L

!

V =nRT

P=

1.37mol( ) 0.0821L " atm

K "mol

#

$ %

&

' ( 273K( )

1atm

!

V = 30.7L

Method #2: using PV=nRT

45

How many g of NaN3 are required tofill an air bag with a volume of 45.5 Lat 22.0ºC and a pressure of 828 torr?

2NaN3 → 2Na + 3N2

Gas Stoichiometry #2

!

n =PV

RT=

828torr1atm

760torr

"

# $

%

& ' 45.5L( )

0.0821L ( atm

K (mol

"

# $

%

& ' 295K( )

= 2.05molN2

!

gNaN3

= 2.05molN2

2molNaN3

3molN2

"

# $

%

& ' 65.02gNaN

3

molNaN3

"

# $

%

& '

!

gNaN3

= 88.9g

DoWs #5

Amedeo Avogadrosays…

47

twice the molestwice the molesmeansmeanstwice the volumetwice the volume

Avogadro’s Hypothesis Equal volumes of gases at the same T and P have

the same number of molecules.

V = kn V and n are directly proportional.

Amedeo Avogadro(1776 - 1856)

!

V1

n1

=V2

n2

48

Avogadro’s Hypothesis: V=knAt constant P and T,

Volume is directly proportional to n (moles).