Gas lawschem

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The Gas Laws

HS Ch i t R id L i S i

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

HS Chemistry Rapid Learning Series

Wayne Huang, PhDKelly Deters, PhDRussell Dahl, PhD

Elizabeth James, PhD



Learning Objectives

How gases cause pressure.

By completing this tutorial you will learn…

The Kinetic Molecular Theory (KMT).

How properties of a gas are related.

How to use several gas laws.

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The difference between ideal and real gases.

Concept MapChemistry

Studies

Previous content

New content

Matter

Gas

One state is

Volume

Pressure

D it

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TemperatureTemperature

# of Moles Molar Mass

Density

Gas Laws

Have properties

Related to each other with



Kinetic Molecular Theory

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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

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p p gbehavior based upon the motion of molecules.



Assumptions of the KMTAll gases are made of particles.

Gas particles are in constant, rapid, random motion

1

2

6

random motion.

The temperature of a gas is proportional to the average kinetic energy of the particles (3RT/2).

Gas particles are not attracted nor repelled from one another.

3

4

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

Avg. KE = Average Kinetic Energy (in J)g g gy ( )R = Gas constant (8.31 J/K mol)T = Temperature (in Kelvin)

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

Avg. KE = (3/2) RT

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

Avg. KE = (3/2) x (8.31J/K mol) x (301K)



Gas Behavior

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KMT and Gas Behavior

The Kinetic MolecularThe Kinetic Molecular Theory and its assumptions can be used to explain gas behavior.

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Definition: Pressure

Pressure – Force of gas gparticles running into a surface.

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Pressure and Number of Molecules

A b f

If pressure is molecular collisions with the container…

As number of molecules

increases, there are more

molecules to collide with the

wall.

Collisions between

molecules and the wall increase.

Pressure increases.

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As # of molecules increases, pressure increases.

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

P ∝ n



Pressure and Volume

A l

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.

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As volume increases, pressure decreases.

Pressure (P) and volume (V) are inversely proportional.

P ∝ 1/v

Definition: Temperature

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

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

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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.

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As temperature increases, pressure increases.

Pressure (P) and temperature (T) are directly proportional (∝).

P ∝ T

Pressure Inside and Outside a Container

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Definition: Atmospheric Pressure

Atmospheric Pressure –Pressure due to the layers of

Lower atmospheric

Pressure due to the layers of air in the atmosphere.

Less layers of air

Climb in altitude

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pressure.air…altitude…

As altitude increases (less air), 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. Contraction will raise 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).

Hi h

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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



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

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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

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General Strategy for Gas Law Problems

Id tif titi b th i it1

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

Identify quantities by their units.

Make a list of known and unknown quantities in a 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

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Plug quantities in and solve.4

Pressure UnitsSeveral units are used when describing pressure.

Pressure Unit Symbol

atmospheres atm

Pascals, kiloPascals

millimeters of mercury

pounds per square inch

Pa, kPa

mm Hg

psi

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Pressure Unit Conversions:1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi



Definition: Kelvin Scale

Kelvin (K) – temperature scale with an b l tabsolute 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.

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g p

K = °C + 273

Definition: Standard Temp & Pressure

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

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

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this hidden information when making your list!



Gas Laws

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KMT and Gas Laws

The Gas Laws are the experimental b ti f th b h i th tobservations of the gas behavior that

the Kinetic Molecular Theory explains.

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“Before” and “After” in Gas Laws

This section has 4 gas laws which have “before” and “after” conditions.F lFor example:

2

2

1

1

nP

nP=

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

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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 moles (n) and Volume (V).

Where T and P are held constant..

V = Volumen = # of moles of gas2

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!

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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 2.50 L. What volume is it if the pressure is changed to 745 mm Hg?

P it d t t h t

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P1 = 1.05 atmV1 = 2.50 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…

VPVP 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?

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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 = Temperature2

2

1

1

TT=

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 !

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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 = Temperature2

2

1

1

TT=

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?

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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)

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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=

×××××

Why You Only Really Need 1 out of the 4 Laws!

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 =

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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 P and temperature T constant Avogadro’s Law

22

22

11

11

TnVP

TnVP

=Hold moles n and temperature T constant Boyles’ Law

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22

22

11

11

TnVP

TnVP

=Hold pressure P and moles n constant Charles’ Law

How to Memorize What’s Held Constant

How do you know what to hold constant for each law? Use the mnemonics.

Hold Pressure and Temperature constant.Avogadro’s Law

Hold moles and Temperature constant.Boyles’ Law

Avogadro was a Professor at Turin University (Italy).

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

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Hold Pressure and moles constant.Charles’ Law

Charles was from Paris.

Gas Laws Rhyme: Avogadro hold P, T; Bolyes hold T; Charles hold P = “ABC – PreTend To Pee!”



Example of Using only the Combined Law

Example: 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 thereforeSTP 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

=

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Kg

Kg

2732982=Pressure units must match!

1 atm = 760 mm Hg

22987605.15755273 V

KHgmmLHgmmK=

×××

Mixtures of Gases

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Definition: Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure – The sum of the pressures of each type of gas p yp gequals the pressure of the total sample.

Ptotal = ∑ Ppartial of each gas

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

G ft ll t d b b bbli th h tGases are often collected by bubbling through water.And bubbles up to the top (less

dense).

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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:

Ptotal = 0.970 atmPwater = 0.016 atmPhydrogen = ?

Phydrogen = 0.954 atm

Ptotal = Pwater + Phydrogen

0.970 atm = 0.016 atm + Phydrogen

0.970 atm - 0.016 atm = Phydrogen

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Definition: Mole Fraction

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

l fmoles of gas.

total

AA n

n=χ

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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

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sample

totaltotal

AA P

nnP ×=

Dalton’s Law - 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?

Ptotal = 115.5 kPaPhydrogen = 28.7 kPaχhydrogen = ?

Phydrogen = χhydrogen x Ptotal

28.7 kPa = χhydrogen x 115.5 kPa

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χhydrogen = 0.248

(28.7 kPa)/(115.5 kPa) = χhydrogen



Dalton’s Law - 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?

Ptotal = 1.23 atmPoxygen = 0.87 kPantotal = 0.556 molesnoxygen = ?

Poxygen = χoxygen x Ptotal

Poxygen = (noxygen/ntotal) x Ptotal

Poxygen = (noxygen/0.556mol) x Ptotal

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noxygen = 0.39 moles

(0.87atm x 0.556mol)/1.23atm = noxygen

Molar Volume of a Gas

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Definition: Molar Volume of a Gas

Standard Temperature and Pressure (STP)– 1 atm (760 mm Hg) and 273 K (0°C).

Molar Volume of a Gas – at STP, 1 mole of any gas = 22.4 liters.

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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

The KUDOS Method of Gas Stoichiometry – the calculation of the amounts of reactants and products in gaseous reaction.

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0.51 is a reasonable answer for L (514 mL).“L H2” is the correct unit.2 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



Gas Stoichiometry – Example #1What 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

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g Zn

= ________ L H2

65.39

0.51

mole Zn1 mole H21

This is volume at STP (1 atm & 273°)

V2 = 0.20 L

1.0atm x 0.51L = 2.5atm x V2

1.0atm x 0.51L)/(2.5atm) = V2

Gas Stoichiometry – Example #2

From balanced equation:

Example: What volume of H2 gas is produced at 25°C 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 H21 L H222.4


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

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g Zn

= ________ L H2

65.39

1.88

mole Zn1 mole H21

This is volume at STP (1 atm & 273°)

V2 = 2.1 L(1.0atm x 1.88L)/273K = (0.97atm x V2)/298K

(298K x 1.0atm x 1.88L)/(0.97atm x 273K) = V2



Ideal Gas Law

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Definition: Ideal Gas Law

Ideal Gas – All of the assumptions of the Kinetic Molecular Theory (KMT) lid(KMT) are valid.

Ideal Gas Law – Describes properties of a gas under a set of conditions.

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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 31 L•kPa•mol-1K-1

Use this one when the P unit is “kPa”.

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8.31 L•kPa•mol-1K-1

0.0821 L•atm•mol-1K-1

Note: J = L•kPa then R = 8.31 J/mole•K

Use this one when the P unit is “atm”.

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

62.4 L•mmHg•mol-1K-1

Memorizing the Ideal Gas Law

nRTPV =

Phony Vampires Are(=) not Real Things

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Ideal Gas Law - ExampleAn example of the Ideal Gas Law:

P = PressureV = Volume

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

R = Gas constantT = Temperature

upon your P units.

T must be in Kelvin!

What is the pressure (in atm) of a gas if it is 2.75 L, has 0.250 moles and is 325 K?

Example:

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

PV = nRT

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P ? V = 2.75 Ln = 0.250 molesT = 325 K Phydrogen = 2.43 atmR = 0.0821 (L×atm) / (mol×K)

P x 2.75L = 0.250mol x (0.0821 L•atm•mol-1•K-1) x 325KP = 0.250mol x (0.0821 L•atm•mol-1•K-1) x 325K/2.75L

Definition: Molar Mass

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

nmMM =

Th f m

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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”

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P 98.0 kPa V = 0.936 Lm = 2.889 gT = 304 KMM = ? g/mol MM = 79.6 g/molR = 8.31 (L×kPa) / (mol×K)

98.0kPa x 0.936L = (2.889g/MM) x (8.31 L•kPa•mol-1•K-1) x 304K

MM = 2.889g x (8.31 L•kPa•mol-1•K-1) x 304K/(98.0kPa x 0.936L )

Definition: Density

Density – Ratio of mass to volume for a sample.

VmD =

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

VmD =andRT

MMmPV =

MMRT

VmP =

RTP DMM

=

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:

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P = 98.0 kPa V = 0.936 LD = 3.09 g/LT = 304 KMM = ? g/mol MM = 79.6 g/mol

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

R = 8.31 (L×kPa) / (mol×K)

98.0kPa = (3.09g/L) x (8.31 L•kPa•mol-1•K-1) x 304K/MM

MM = (3.09g/L) x (8.31 L•kPa•mol-1•K-1) x 304K/(98.0kPa)

Real Gases

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Definition: Real Gas

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

#1 - Gas particles are not attracted nor repelled from one another.

Gas particles do have attractions and repulsions towards one another.

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#2 - 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 Law - EquationThe 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

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

b = 0.0427 L/mol)P = 3.45 atmV = 2.75 Ln = 0.75 molT = ? Ka = 3.59 L2atm/mol2

b = 0.0427 L/mol( ) nRTnbV

VanP =−⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

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

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T = 164 K

R = 0.0821 (L×atm) / (mol×K)

[3.45atm + (0.75mol)2 x (3.59 L2•atm•mol-2)/(2.75L)2] x [2.75L – (0.75mol x 0.0427L•mol)]= 0.75mol x (0.0821L•atm•mol-1•K-1) x T

Real gases do not use 2 Real gases do not use 2 Gas particles causeGas particles cause

Learning Summary

Ideal gases follow theIdeal gases follow the

gof the assumptions of

the KMT.

gof the assumptions of

the KMT.

Gas particles cause pressure.

Gas particles cause pressure.

Several Gas Laws areSeveral Gas Laws are

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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.



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The Gas Laws

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