Gases

AP Chemistry Rapid Learning Series - 16

© Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com 1

Rapid Learning CenterChemistry :: Biology :: Physics :: Math

Rapid Learning Center Presents …p g

Teach Yourself AP Chemistry Visually in 24 Hours

1/78 http://www.RapidLearningCenter.com

The Gas Laws

AP Ch i t R id L i S i

Rapid Learning Centerwww.RapidLearningCenter.com/© Rapid Learning Inc. All rights reserved.

AP Chemistry Rapid Learning Series

Wayne Huang, PhDKelly Deters, PhDRussell Dahl, PhD

Elizabeth James, PhDDebbie Bilyen, M.A.



Learning Objectives

How gases cause pressure

Th Ki ti M l l Th

By completing this tutorial you will learn…

The Kinetic Molecular Theory

How properties of a gas are related

How to use several gas laws

The difference between ideal and real gases

3/78

Diffusion and Effusion

Concept MapChemistry

Studies

Previous content

New content

Matter

Gas

One state is

Volume

Pressure

D it

Rates of Effusion Rates of Effusion and Diffusion

4/78

TemperatureTemperature

Moles Molar Mass

Density

Gas Laws

Have properties

Related to each other with



KineticKinetic Molecular Theory

5/78

Kinetic Molecular Theory

Theory – An attempt to explain why or how behavior orwhy or how behavior or properties are as they are. It’s based on empirical evidence.

Kinetic Molecular Theory (KMT) –An attempt to explain gas

6/78

p p gbehavior based upon the motion of molecules.



Assumptions of the KMT

All gases are made of atoms or molecules.

Gas particles are in constant, rapid, random motion

1

2 random motion.

The temperature of a gas is proportional to the average kinetic energy of the particles.

Gas particles are not attracted nor repelled from one another.

3

4

7/78

All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms).

The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant.

5

6

Calculating Average Kinetic Energy

3

Temperature is proportional to average kinetic energy…how do you calculate it?

Avg. KE = Average Kinetic Energy (in J, Joules)RTKEAvg

23. =

g g gy ( , )R = Gas constant (use 8.31 J/K mol)T = Temperature (in Kelvin)

Find the average kinetic energy of a sample of O2 at 28°C.Example:

( )3

8/78

Avg. KE = ? JR = 8.31 J/K molT = 28°C + 273 = 301 K Avg. KE = 3752 J

( ) KmoleKJKEAvg 30131.8

23. ××=



Gas Behavior

9/78

KMT and Gas Behavior

The Kinetic Molecular Th d itTheory and its assumptions can be used to explain gas behavior.

10/78



Definition: Pressure

Pressure – Force of gas particles running into a surface.

11/78

Pressure and Number of Molecules

A th b f C lli i P

If pressure is molecular collisions with the container…

As the number of molecules increases, there are more molecules to collide with the wall

Collisions between molecules and the wall increase

Pressure increases

12/78

As # of molecules increases, pressure increases.

Pressure (P) and # of molecules (n) are directly proportional (∝).

nP ∝



Pressure and Volume

A l C lli i P

If pressure is molecular collisions with the container…

As volume increases, molecules can travel farther before hitting the wall

Collisions between molecules and the wall decrease

Pressure decreases

13/78

As volume increases, pressure decreases.

Pressure and volume are inversely proportional.

VP 1∝

Definition: Temperature

Temperature – Proportional to the average kinetic energy of the molecules.

Energy due to motion(Related to how fast the molecules are moving)

14/78

As temperature increases

Molecular motion increases



Pressure and TemperatureIf temperature is related to molecular motion…and pressure is molecular collisions with the container…

As temperature increases, molecular motion increases

Collisions between molecules and the wall increase

Pressure increases

15/78

As temperature increases, pressure increases.

Pressure and temperature are directly proportional.

TP ∝

Pressure Inside and Outside a Container

16/78



Definition: Atmospheric Pressure

Atmospheric Pressure Pressure due

Lower atmospheric

Atmospheric Pressure – Pressure due to the layers of air in the atmosphere.

Less layers of air

Climb in altitude

17/78

pressureairaltitude

As altitude increases, atmospheric pressure decreases.

Pressure In Versus OutA container will expand or contract until the pressure inside = atmospheric pressure outside.

Expansion will lower the internal pressure.

Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain.

Contraction will raise the internal pressure.(Volume and pressure are inversely related)

The internal pressure is from low altitude (high pressure)Higher

18/78

The internal pressure is higher than the external pressure.The bag will expand in order to reduce the internal pressure.

( g p )The external pressure is high altitude (low pressure).

Higher pressureLower

pressure Lower pressure



When Expansion isn’t Possible

Example: An aerosol can is left in a car trunk in the summer. What happens?

Rigid containers cannot expand.

CanExplodes!

happens?

The temperature inside the can begins to rise.As temperature increases, pressure increases.

Higher pressure

Lowerpressure

19/78

The internal pressure is higher than the external pressure.The can is rigid—it cannot expand, it explodes!

Soft containers or “movable pistons” can expand and contract.Rigid containers cannot.

Attacking StrategyAttacking Strategy for Gas Law Problems

20/78



General Strategy for Gas Law Problems

Id tif titi b th i it1

The following steps are a general way to approach these problems.

Identify quantities by their units.

Make a list of known and unknown quantities in symbolic form.

Look at the list and choose the gas law that relates all the quantities together.

Pl titi i d l

1

2

3

4

21/78

Plug quantities in and solve.4

Pressure UnitsSeveral units are used when describing pressure

Unit Symbol

atmospheres atm

Pascals, kiloPascals

millimeters of mercury

pounds per square inch

Pa, kPa

mm Hg

psi

22/78

1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi



Definition: Kelvin Scale

Kelvin (K) – Temperature scale with b l tan absolute zero.

Temperatures cannot fall below an absolute zero.A temperature scale with absolute zero is needed in Gas Law calculations because you can’t have negative pressures or volumes.

23/78

g p

KC =+ 273

Standard Temperature & Pressure

Standard Temperature andStandard Temperature and Pressure (STP) – 1 atm (or the equivalent in another unit) and 0°C (273 K).

24/78

Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!



Gas Laws

25/78

KMT and Gas Laws

The Gas Laws are the experimental observations of the gas behavior that the Kinetic Molecular Theory explains.

26/78



“Before” and “After” in Gas Laws

This section has 4 gas laws which have “before” and “after” conditions.

For example:

2

2

1

1

nP

nP=

Where P1 and n1 are pressure and # of moles “before”

27/78

and P2 and n2 are pressure and # of moles “after”.

Both sides of the equation are talking about the same sample of gas—with the “1” variables before a change, and the “2” variables after the change.

Avogadro’s LawAvogadro’s Law relates # of particles (moles) and Volume.

Where Temperature & Pressure are held constant.p

V = Volumen = # of moles of gas

2

2

1

1

nV

nV

=

Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?

The two volume units must match!

28/78

moles?

n1 = 0.15 molesV1 = 2.5 Ln2 = 0.55 molesV2 = ? L

moleV

moleL

55.015.05.2 2=

215.05.255.0 V

moleLmole=

× V2 = 9.2 L



Boyles’ Law - 1Boyles’ Law relates pressure and volume.

Where temperature and # of molecules are held constant.P = pressureV = volume2211 VPVP =The two pressure units must match and the two volume units must match!

Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 745 mm Hg?

P it d t t h t

29/78

P1 = 1.05 atmV1 = 2.5 LP2 = 745 mm HgV2 = ? L

Pressure units need to match—convert one:

=0.980 atm

745 mm Hg= ______ atm

mm Hg

atm1

7600.980

Boyles’ Law - 2Boyles’ Law relates pressure and volume.

Where temperature and # of molecules are held constant.P = pressureV = volume2211 VPVP =The two pressure units must match and the two volume units must match!

Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 745 mm Hg?

30/78

2980.05.205.1 VatmLatm ×=×P1 = 1.05 atmV1 = 2.5 LP2 = 745 mm HgV2 = ? L

V2 = 2.7 L=0.980 atm 2980.05.205.1 V

atmLatm=

×



Charles’ Law - 1Charles’ Law relates volume and temperature.

Where pressure and # of molecules are held constant.

21 VVV = VolumeT = Temperature

2

2

1

1

TV

TV

=

The two volume units must match and temperature must be in Kelvin!

Example: What is the final volume if a 10.5 L sample of gas is changed from 25°C to 50°C?

T t d t b i K l i !

31/78

V1 = 10.5 LT1 = 25°CV2 = ? LT2 = 50°C

Temperature needs to be in Kelvin!

= 298 K

= 323 K

25°C + 273 = 298 K

50°C + 273 = 323 K

Charles’ Law - 2Charles’ Law relates temperature and pressure.

Where pressure and # of molecules are held constant.

21 VVV = VolumeT = Temperature

2

2

1

1

TV

TV

=

The two volume units must match and temperature must be in Kelvin!

Example: What is the final volume if a 10.5 L sample of gas is changed from 25°C to 50°C?

32/78

V1 = 10.5 LT1 = 25°CV2 = ? LT2 = 50°C

V2 = 11.4 L

= 298 K

= 323 K

KV

KL

3232985.10 2=

22985.10323 V

KLK=

×



The Combined Gas LawThe combined gas law assumes that nothing is held constant.

P = PressureVPVP Each “pair” of unitsV = Volumen = # of molesT = Temperature22

22

11

11

TnVP

TnVP

=Each pair of units must match and temperature must be in Kelvin!

Example: What is the final volume if a 0.125 mole sample of gas at 1.7 atm, 1.5 L and 298 K is changed to STP and particles are added to 0.225 mole?

P1 = 1.7 atmV1 = 1.5 L

STP is standard temperature (273 K) and pressure (1 atm)

33/78

n1 = 0.125 moleT1 = 298 KP2 = 1.0 atmV2 = ? Ln2 = 0.225 moleT2 = 273 K V2 = 4.2 L

STP is standard temperature (273 K) and pressure (1 atm)

KmoleVatm

KmoleLatm

273225.00.1

298125.05.17.1 2

××

=××

2298125.00.15.17.1273225.0 V

KmoleatmLatmKmole=

×××××

Only Really Need One Law

2211 VPVP=

The combined gas law can be used for all “before” and “after” gas law problems!

2211 TnTn

22

12

11

11

TnVP

TnVP

=

For example, if volume is held constant, then

and the combined gas law becomes:

21 VV =

34/78

When two variables on opposites sides are the same, they cancel out and the rest of the equation can be used.

22

2

11

1

TnP

TnP

=



“Transforming” the Combined Gas Law

Watch as variables are held constant and the combined gas law “becomes” the other 3 laws.

VPVP

22

22

11

11

TnVP

TnVP

=Hold pressure and temperature constant Avogadro’s Law

22

22

11

11

TnVP

TnVP

=Hold moles and temperature constant Boyles’ Law

35/78

22

22

11

11

TnVP

TnVP

=Hold pressure and moles constant Charles’ Law

How to Memorize What’s Held Constant

How do you know what to hold constant for each law?

Hold Pressure and Temperature constantAvogadro’s Law

Hold moles and Temperature constantBoyles’ Law

Avogadro was a Professor at Turin University (Italy)

The last letter of his first name Robert is T

36/78

Hold Pressure and moles constantCharles’ Law

The last letter of his first name, Robert, is T

Charles was from Paris



Using only the Combined LawExample: What is the final volume if a 15.5 L sample of gas at 755

mm Hg and 298 K is changed to STP?

P 755 H “moles” is not mentioned in the problem therefore

STP is standard temperature (273 K) and pressure (1 atm).

P1 = 755 mm HgV1 = 15.5 LT1 = 298 KP2 = 1.0 atmV2 = ? LT2 = 273 K

“moles” is not mentioned in the problem—therefore it is being held constant.It is not needed in the combined law formula.

VHgmmLHgmm 7605.15755 2××Pressure units must match!

= 760 mm Hg

22

22

11

11

TnVP

TnVP

=

37/78V2 = 14.1 L

Kg

Kg

2732982=Pressure units must match!

1 atm = 760 mm Hg

22987605.15755273 V

KHgmmLHgmmK=

×××

Mixtures of Gases

38/78



Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure The sum of thePressure – The sum of the pressures of each type of gas equals the pressure of the total sample.

39/78

∑= gaseachofpartialtotal PP

Dalton’s Law in LabDalton’s Law of Partial Pressure is often used in labs where gases are collected.

Gases are often collected by bubbling through water.

And bubbles up to the top (less

dense)

40/78

This results in a mixture of gases - the one being collected and water vapor.

Reaction producing gas Gas travels

through tube

Through water



Hydrogen gas is collected by bubbling through water. If the total pressure of the gas is 0.970 atm, and the partial pressure of water at that temperature is 0.016 atm, find the pressure of the hydrogen gas.

Dalton’s Law in Lab - ExampleExample:

hydrogenwatertotal PPP +=

hydrogenPatmatm += 016.0970.0

Ptotal = 0.970 atmPwater = 0.016 atmPhydrogen = ?

hydrogenPatmatm =− 016.0970.0 Phydrogen = 0.954 atm

41/78

Definition: Mole Fraction

Mole Fraction (χ) –Ratio of moles (n) ofRatio of moles (n) of one type of gas to the total moles of gas.

An=χ

42/78

totalA n=χ

Mole fraction has no units as it is “moles/moles”.



Dalton’s Law and Mole FractionsDalton’s Law of Partial Pressure calculations can be done with mole fractions.

P

totalAA PP ×= χPressure

of gas “A”

Mole fractionof gas “A”

Pressureof the whole

l

43/78

sample

totaltotal

AA P

nnP ×=

Example - 1Dalton’s Law of Partial Pressure calculations can be done with mole fractions.

Example: If the total pressure of the sample is 115.5 kPa, and thea p e If the total pressure of the sample is 115.5 kPa, and the pressure of hydrogen gas is 28.7 kPa, what is the mole fraction of hydrogen gas?

totalhydrogenhydrogen PP ×= χPtotal = 115.5 kPaPhydrogen = 28.7 kPaχhydrogen = ?

kPakPa hydrogen 5.1157.28 ×= χ

kPa7.28

44/78

χhydrogen = 0.248

hydrogenkPaχ=

5.115



Example - 2Another type of problem:

Example: How many moles of oxygen are present in a sample with a total of 0.556 moles, 1.23 atm and a partial pressure f f 0 87 t ?for oxygen of 0.87 atm?

totaloxygenoxygen PP ×= χPtotal = 1.23 atmPoxygen = 0.87 kPantotal = 0.556 molesnoxygen = ?

totaltotal

oxygenoxygen P

nn

P ×=

n

45/78noxygen = 0.39 moles

atmmoles

natm oxygen 23.1

556.087.0 ×=

oxygennatm

molesatm=

×23.1

556.087.0

Gas Stoichimetry:Gas Stoichimetry: Molar Volume of a Gas

46/78



Definition: Molar Volume of a Gas

Standard Temperature and Pressure (STP) – 1 atm essu e (S ) at(760 mm Hg) and 273 K (0°C)

Molar Volume of a Gas –at STP 1 mole of any gas

47/78

at STP, 1 mole of any gas = 22.4 liters

From balanced equation:

Example:

Mass-Volume Problems (Gases)If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP?2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

q1 mole Zn 1 mole H2

K

D

1.5 g Zn mole Zn1

Molar volume of a gas:1 mole H2 = 22.4 L

mole H21 L H222.4

Molar Mass of Zn:1 mole Zn = 65.39 g

48/78

0.51 is a reasonable answer for L (514 mL)“L H2” is the correct unit2 sf given 2 sf in answer

U

S O

g

g Zn

= ________ L H2

65.39

0.51

mole Zn

2

1 mole H2

L H2

1

22.4



Stoichiometry and the Gas LawsWhat if you want the volume of a gas not at STP?Example: If you need react 1.5 g of zinc completely, what volume of

gas will be produced at 2.5 atm and 273°C?2 HCl (aq) + Zn (s) ZnCl (aq) + H (g)

From balanced equation: 1 mole Zn 1 mole H2

2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

1.5 g Zn

g Zn

mole Zn

65.39

1


mole Zn

mole H2

1

1

mole H2

L H2

1

22.4


P1 = 1.0 atmV1 = 0.51 LP2 = 2.5 atmV2 = ? L

49/78

g Zn

= ________ L H2

65.39

0.51

mole Zn1 mole H21

This is volume at STP (1 atm & 273°)

V2 = 0.20 L25.251.00.1 V

atmLatm=

×25.251.00.1 VatmLatm ×=×

Another Example

From balanced equation:

Example: What volume of H2 gas is produced at 25° and 0.97 atm from reacting 5.5 g Zn?2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

P1 = 1.0 atm From balanced equation: 1 mole Zn 1 mole H2

5.5 g Zn mole Zn1


mole H21L H2

22.4


V1 = 1.88 LT1 = 273 KP2 = 0.97 atmV2 = ? LT2 = 25°C = 298 K

50/78

g Zn

= ________ L H2

65.39

1.88

mole Zn1 mole H21

This is volume at STP (1 atm & 273°)

V2 = 2.1 L227397.0

88.10.1298 VKatm

LatmK=

×××

KVatm

KLatm

29897.0

27388.10.1 2×

=×



Ideal Gas Law

51/78

Definition: Ideal Gas LawIdeal Gas – all of the assumptions of the Kinetic Molecular Theory (KMT) areMolecular Theory (KMT) are valid.Ideal Gas Law – Describes properties of a gas under a set of conditions.

52/78

nRTPV =

This law does not have “before” and “after”—there is no change in conditions taking place.



Definition: Gas Constant

nRTPV =

Gas Constant (R) – constant equal to the ratio of P×V to n×T for a gas.

Values for R

8 31kPaL×

Use this one when the P unit is “kPa”

53/78

8.31 Kmole×

0.0821 KmoleatmL×

×

Use this one when the P unit is “atm”

Use this one when the P unit is “mm Hg”

62.4 KmoleHgmmL

××

Memorizing the Ideal Gas Law

nRTPV =

Phony Vampires are not Real Things

54/78



Ideal Gas Law ExampleAn example of the Ideal Gas Law:

P = PressureV = Volume

Choose your “R” based upon your “P” units.nRTPV = n = # of moles

R = Gas constantT = Temperature

upon your P units.

T must be in Kelvin!

nRTPV =

What is the pressure (in atm) of a gas if it is 2.75 L, has 0.25 moles and is 325 K?

Example:

P = ?Choose the “0.0821” for “R” since the problem asks for “atm”.

55/78

( ) KKmoleatmLmolesLP 3250821.025.075.2 ××

××=×P ? V = 2.75 Ln = 0.25 molesT = 325 K

Phydrogen = 2.43 atmR = 0.0821 (L×atm) / (mol×K)

( )L

KKmoleatmLmoles

P75.2

3250821.025.0 ××××

=

Definition: Molar Mass

Molar mass (MM) – Mass (m) per moles (n) of a substancemoles (n) of a substance.

nmMM =

Th f m

56/78

Therefore:MMmn =



Ideal Gas Law and Molar MassThe Ideal Gas Law is often used to determine molar mass.

nRTPVmn =and RTmPVnRTPV = MM

n =and RTMM

PV =

A gas is collected. The mass is 2.889 g, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass.

Example:

P = 98.0 kPaChoose the “8.31” for “R” since the problem uses “kPa”.

( )kPLg8892

57/78

P 98.0 kPa V = 0.936 Lm = 2.889 gT = 304 KMM = ? g/mole

MM = 79.6 g/moleR = 8.31 (L×kPa) / (mol×K)

( ) KKmolekPaL

MMgLkPa 30431.8889.2936.00.98 ××

×=×

( ) KKmolekPaL

LkPagMM 30431.8936.00.98

889.2××

××

=

Definition: Density

D i R i fDensity – Ratio of mass to volume for a sample.

VmD =

58/78



Ideal Gas Law and DensityUsing the density equation with the Ideal Gas Law:

VmD =andRT

MMmPV =

MMRT

VmP =

MMRTDP =

A gas is collected. The density is 3.09 g/L, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass.

Example:

59/78

P = 98.0 kPa V = 0.936 LD = 3.09 g/LT = 304 KMM = ? g/mole

MM = 79.6 g/mole

Choose the “8.31” for “R” since the problem uses “kPa”.

R = 8.31 (L×kPa) / (mol×K)

( )8.31 30498.0 3.09

L kPa Kmole KgkPa L MM

× ××=

( )8.31 3043.09

98.0

L kPa Kmole KgMM L kPa

× ××=

Real Gases

60/78



Definition: Real Gas

Real Gas – 2 of the assumptions of the Kinetic Molecular Theory are not valid.Kinetic Molecular Theory are not valid.

Gas particles are not attracted nor repelled from one another.

Gas particles do have attractions and repulsions towards one another.

61/78

The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant.

Gas particles do take up space - thereby reducing the space available for other particles to be.

Real Gas LawThe Real Gas Law takes into account the deviations from the Kinetic Molecular Theory.

nRTPV =Ideal Gas Law nRTPV =

( ) nRTnbVV

anP =−⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

Ideal Gas Law

Real Gas LawAlso called “van der Waals equation”

Take into account the

62/78

Take into account the change in pressure due

to particle attractions and repulsions

Takes into account the space the particles take up

“a” and “b” are constants that you look up for each gas!



Real Gas Law Example

Example: At what temperature would a 0.75 mole sample of CO2be 2.75 L at 3.45 atm?

( ) nRTnbVV

anP =−⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

be 2.75 L at 3.45 atm? (van der Waals constants for CO2: a = 3.59 L2atm/mol2

b = 0.0427 L/molP = 3.45 atmV = 2.75 Ln = 0.75 moleT = ? Ka = 3.59 L2atm/mol2

b = 0.0427 L/mol

Choose the “0.0821” for “R” since the problem uses “atm”.

63/78T = 164 K

( ) ( ) ( ) TKmolatmLmolmol

LmolLL

molatmLmol

atm ××××=×−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ×+ 0821.075.00427.075.075.2

)75.2(

59.3)75.0(45.3 2

222

R = 0.0821 (L×atm) / (mol×K)

Diffusion and Effusion

64/78



Definition: Diffusion & Effusion

Diffusion – A gas spreads throughout a space.

Effusion A gas escapes through a

Perfume is sprayed in one corner of the room and a person on the other side smells it after a moment.

65/78

Effusion – A gas escapes through a tiny hole.Air leaks out of a balloon overnight and is flat the next day.

Diffusion, Effusion and Mass

Temperature is A heavier object Heavy molecules

The mass of a particle affects the rate of diffusion and effusion.

Temperature is proportional to average kinetic energy

A heavier object with the same kinetic energy as a lighter object moves slower than the lighter object

Heavy molecules move slower than smaller molecules

66/78

Diffusion: If molecules move slower, it will take them longer to reach the other side of the room.

Effusion: If molecules move slower, it will take them longer to find the hole to escape through.

Both rates of diffusion and effusion are inversely proportional to molecular mass.



Effusion and Graham’s Law - 1Effusion rates are related by Graham’s Law.

r1 = Rate of Effusion for molecule 1r2 = Rate of Effusion for molecule 2MM M l l f l l 1

21 MMr= MM1 = Molecular mass for molecule 1

MM2 = Molecular mass for molecule 212 MMr

Example: A gas molecule effuses 0.355 times as fast as O2. What is the molecular mass of the molecule?

Molecule 1 = O2

Molecule 2 = unknown moleculeIf the unknown molecule is 0.355 times as fast as O2,then make the rate of O2 = 1 and

67/78

r1 = 1r2 = 0.355MM1 = 32.00 g/moleMM2 = ? g/mole

then make the rate of O2 1 and the rate of unknown = 0.355

Molecular mass of O2:O 2 × 16.00 = 32.00 g/mole

Effusion and Graham’s Law - 2Effusion rates are related by Graham’s Law.

r1 = Rate of Effusion for molecule 1r2 = Rate of Effusion for molecule 2MM M l l f l l 1

21 MMr= MM1 = Molecular mass for molecule 1

MM2 = Molecular mass for molecule 212 MMr

Example: A gas molecule effuses 0.355 times as fast as O2. What is the molecular mass of the molecule?

Molecule 1 = O2

Molecule 2 = unknown molecule molegMM

/00.32355.01 2=

68/78

r1 = 1r2 = 0.355MM1 = 32.00 g/moleMM2 = ? g/mole

molegMM

/00.32355.01 2

2

=⎟⎠⎞

⎜⎝⎛

( ) 2

2

355.01/00.32 MMmoleg =⎟

⎠⎞

⎜⎝⎛×

MM = 254 g/mole



Diffusion and Graham’s LawDistances traveled during diffusion:

d1 = Distance traveled for molecule 1d2 = Distance traveled for molecule 2MM M l l f l l 1

21 MMdd

= MM1 = Molecular mass for molecule 1MM2 = Molecular mass for molecule 212 MMd

Example: A gas molecule is 4 times as heavy as O2. How far does it travel in the time that oxygen travels 0.25 m?

Molecule 1 = unknown moleculeMolecule 2 = O2 moleg

molegm

d/00.128/00.32

25.01 =

69/78

d1 = ?d2 = 0.25 mMM1 = 4×32.00 g/mole = 128 g/moleMM2 = 32.00 g/mole Molecular mass of O2:

O 2 × 16.00 = 32.00 g/mole

41

25.01 =m

d md 25.021

1 ×=

d1 = 0.125 m

Gases & The AP Exam

70/78



Gases in the Exam

Gas stoichiometry at STP (1 mole = 22.4 L)

Common Gases problems:

Using the ideal gas law

Using the combined gas law

Dalton’s law of partial pressure (and using with mole fractions).

71/78

Multiple Choice QuestionsDalton’s Law of Partial Pressure is a common question topic.

Example: A flask contains 2.5 moles gas particles and has a total pressure of 1.5 atm. How many moles of O2 are in the flask if the partial pressure of O2 is 0.3 atm?

A. 0.20 moles O2B. 0.50 moles O2C. 2.5 moles O2D. 12.5 moles O2E 0 l O

72/78

E. 0 mole O2

Partial pressure = mole fraction × total pressure0.30 atm = mole fraction × 1.5 atmMole fraction O2 = 1/5Total moles = 2.5 moles1/5 of 2.5 = 0.5 moles O2

Answer: B



Free Response QuestionsGases are often used in conjunction with other topics in Free Response questions.Given the hydrocarbon butane (C4H10), answer the following:

A. Write the balanced equation for the combustion of butane to produce carbon dioxide and water.

B. What volume of dry carbon dioxide, at 30°C and 0.97 atm will result from the complete combustion of 3.75 g butane?

C. Under identical conditions, a sample of an unknown gas effuses at three times the rate of butane. What is the molecular mass of the unknown gas?

73/78

gD. Draw 2 structural isomers of butane.

These are the gases sub-questions in this problem.

Answering Free Response QuestionsAnswer sub-question B:

It’s stoichiometry, which requires a balanced equation, which you would have written in sub-question A:you ou d a e tte sub quest o2 C4H10 + 13 O2 8 CO2 + 10 H2O

B. What volume of dry carbon dioxide, at 30°C and 0.97 atm will result from the complete combustion of 3.75 g butane?

3.75 g C4H10 1 mole C4H10 8 mole CO2 22.4 L CO2 = _____ L CO2 at STP58 14 g C H 2 mole C H 1 mole CO

74/78

STP58.14 g C6H10 2 mole C4H10 1 mole CO2

5.78 L CO2 at STP

2

22

1

11

TVP

TVP

=K

VatmK

Latm303

)97.0(273

)78.5)(1( 2= V = 6.61 L CO2



Answering Free Response QuestionsAnswer sub-question C:

C. Under identical conditions, a sample of an unknown gas effuses at three times the rate of butane. What is the molecular mass ofat three times the rate of butane. What is the molecular mass of the unknown gas?

1

2

2

1

MMMM

rr=

MM C4H10 = 58.14 g/moleRate C4H10 = 1Rate ? = 3MM ? = ? g/mole

1 2MM=

75/78

14.583=

14.5831 2

2 MM=⎟

⎠⎞

⎜⎝⎛ MM2 = 6.49 g/mole

Real gases do not use 2 of the

Real gases do not use 2 of the Gas particles

ca se press reGas particles

ca se press re

Rates of Effusionand Diffusion are

inversely

Rates of Effusionand Diffusion are

inversely

Learning Summary

Ideal gases follow theIdeal gases follow the

assumptions of the KMT

assumptions of the KMT

cause pressurecause pressure yproportional to molecular mass

yproportional to molecular mass

Several Gas Laws areSeveral Gas Laws are

76/78

Ideal gases follow the assumption of the Kinetic Molecular

Theory (KMT)

Ideal gases follow the assumption of the Kinetic Molecular

Theory (KMT)

Several Gas Laws are used to determine

properties under a set of conditions

Several Gas Laws are used to determine

properties under a set of conditions



Congratulations

You have successfully completed the core tutorial

The Gas Laws

77/78

Rapid Learning Center

Rapid Learning Center

Wh t’ N t

Chemistry :: Biology :: Physics :: Math

What’s Next …

Step 1: Concepts – Core Tutorial (Just Completed)

Step 2: Practice – Interactive Problem Drill

Step 3: Recap – Super Review Cheat Sheet

78/78

Go for it!

http://www.RapidLearningCenter.com

Technology

Gases