The GAS LAWS Gases have mass Gases diffuse Gases expand to fill containers Gases exert pressure...

The GAS LAWS The GAS LAWS

Gases have mass Gases diffuse Gases expand to fill

containers Gases exert pressure Gases are compressible Pressure & temperature are

dependent

Gases have mass Gases diffuse Gases expand to fill

containers Gases exert pressure Gases are compressible Pressure & temperature are

dependent

Gas PropertiesGas Properties

Volume (V)• Units of volume (L)

Amount (n)• Units of amount (moles)

Temperature (T)• Units of temperature (K)

Pressure (P)•Units of pressure (mmHg)•Units of pressure (KPa)•Units of pressure (atm)

Volume (V)• Units of volume (L)

Amount (n)• Units of amount (moles)

Temperature (T)• Units of temperature (K)

Pressure (P)•Units of pressure (mmHg)•Units of pressure (KPa)•Units of pressure (atm)

Gas VariablesGas Variables

P1V1 = P2V2P1V1 = P2V2

Boyle’s law•pressure & volume•as P then V•at constant T, n

Boyle’s law•pressure & volume•as P then V•at constant T, n

Charles’ law: •Temperature & volume•As T then V•At constant P, n

Charles’ law: •Temperature & volume•As T then V•At constant P, n

V1 = V2

T1 T2

V1 = V2

T1 T2

A Little ReviewA Little Review

Gay-Lussac’s law: •Temperature & pressure•As P then T•At constant V, n

Gay-Lussac’s law: •Temperature & pressure•As P then T•At constant V, n

P1 = P2

T1 T2

P1 = P2

T1 T2

A Little ReviewA Little Review

A Problem

Blow up a balloon, measure its volume then take it to the top of Mt. Davis.

Explain what would happen to its volume. What variables would you need to use to

measure the change? How do those variables change as you go

up to the top of Mt. Davis?

Blow up a balloon, measure its volume then take it to the top of Mt. Davis.

Explain what would happen to its volume. What variables would you need to use to

measure the change? How do those variables change as you go

up to the top of Mt. Davis?

Now What??Now What??

PV=k1PV=k1 V/T=k2V/T=k2 P/T=k3P/T=k3

If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change :

If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change :

Combined gas lawCombined gas law The law expresses the

relationship between changes in volume, pressure, &

temperature

# of moles is held constant

The law expresses the relationship between changes in

volume, pressure, & temperature

# of moles is held constant

P1V1T2 = P2V2T1P1V1T2 = P2V2T1

P1V1P1V1

T1T1

= = P2V2P2V2

T2T2

A gas with a volume of 4.0L at STP. What is its volume at 2.0atm and at 30°C?

A gas with a volume of 4.0L at STP. What is its volume at 2.0atm and at 30°C?

Example problemExample problem

- P1 - P1

- V1 - V1

- T1 - T1

- P2 - P2 - V2 - V2

- T2 - T2

1atm1atm

4.0 L4.0 L

273K273K

2.0 atm2.0 atm??

30°C + 27330°C + 273= 303K= 303K

2.22L = V22.22L = V2

P1V1P1V1

T1T1

= = P2V2P2V2

T2T2

Example problemExample problem

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

Gay-Lussac’s Law of combining volumes of

gases

The law states thatThe law states that at constant at constant temperature and pressure, the temperature and pressure, the volumesvolumes of gaseous reactants of gaseous reactants and products can be and products can be expressed as ratios ofexpressed as ratios of small small whole numberswhole numbers..

Gay-Lussac’s Law of combining volumes of

gases

The law states thatThe law states that at constant at constant temperature and pressure, the temperature and pressure, the volumesvolumes of gaseous reactants of gaseous reactants and products can be and products can be expressed as ratios ofexpressed as ratios of small small whole numberswhole numbers..

Gay-Lussac’s other LawGay-Lussac’s other LawIn the early 1800s, French chemist Joseph Gay-Lussac studied

gas volume relationships involving a chemical reaction between hydrogen and oxygen. He observed that 2 L of hydrogen can react with 1 L of oxygen to form 2 L of water vapor at constant temperature and pressure.

Hydrogen gas + Oxygen gas Water vapor

2 L (2 volumes) 1 L (1 volume) 2 L (2 volumes)

This reaction showed a simple and definite 2:1:2 relationship between the volumes of the reactants and the product.

Gay-Lussac also noticed simple and definite proportions by volume in other reactions of gases, such as in the reaction between hydrogen gas and nitrogen gas.

Hydrogen gas + Nitrogen gas Ammonia gas (NH3) (3 volumes) + (1 volume) = (2 volumes)

In the early 1800s, French chemist Joseph Gay-Lussac studied gas volume relationships involving a chemical reaction between hydrogen and oxygen. He observed that 2 L of hydrogen can react with 1 L of oxygen to form 2 L of water vapor at constant temperature and pressure.

Hydrogen gas + Oxygen gas Water vapor

2 L (2 volumes) 1 L (1 volume) 2 L (2 volumes)

This reaction showed a simple and definite 2:1:2 relationship between the volumes of the reactants and the product.

Gay-Lussac also noticed simple and definite proportions by volume in other reactions of gases, such as in the reaction between hydrogen gas and nitrogen gas.

Hydrogen gas + Nitrogen gas Ammonia gas (NH3) (3 volumes) + (1 volume) = (2 volumes)

Gay-Lussac ContributionsGay-Lussac Contributions

This simple observation, combined with the insight of Avogadro, provided more understanding of how gases react and combine with each other. In fact it lead to the discovery of three vital chemistry concepts.

1) Diatomic Molecules 2) Gas Density and molecular mass

comparison and determination(Led to determination of Atomic mass)

3) Avogadro's Principle The Mole

This simple observation, combined with the insight of Avogadro, provided more understanding of how gases react and combine with each other. In fact it lead to the discovery of three vital chemistry concepts.

1) Diatomic Molecules 2) Gas Density and molecular mass

comparison and determination(Led to determination of Atomic mass)

3) Avogadro's Principle The Mole

Avogadro’s Contribution

If the amount of gas in a container is increased, the volume increases.

If the amount of gas in a container is decreased, the volume decreases.

This can be proven simple by blowing up a balloon.

If the amount of gas in a container is increased, the volume increases.

If the amount of gas in a container is decreased, the volume decreases.

This can be proven simple by blowing up a balloon.

Amedeo Avogadro

Amedeo Avogadro lived in Italy,

And his work means an awful lot to chemistry.

Amedeo Avogadro may be dead – but,

We all should remember what he said.

Amedeo Avogadro lived in Italy,

And his work means an awful lot to chemistry.

Amedeo Avogadro may be dead – but,

We all should remember what he said.

6.022 x1023 The Mole

ChorusChorusEqual Volumes of Gases, at the same temp and pressure,

Have the same number of molecules.

Amedeo…..Avogadro, that’s his hypothesis.

Equal Volumes of Gases, at the same temp and pressure,



Amedeo Avogadro

You got two balloons filled with different gases,

Equal volumes but they both have different masses,

If conditions are the same everywhere,

Tell me how the number of molecules compare.

You got two balloons filled with different gases,

Equal volumes but they both have different masses,

If conditions are the same everywhere,

Tell me how the number of molecules compare.

Amedeo Avogadro

Why is his hypothesis so great?

You can use it to find atomic weights.

Its based on the ratio of the masses,

Of equal volumes of those gases.

Why is his hypothesis so great?

You can use it to find atomic weights.

Its based on the ratio of the masses,

Of equal volumes of those gases.

Mendeleev used it

Atomic Mass

The GAS LAWS AT WORKThe GAS LAWS AT WORK

Amedeo Avogadro

Avogadro helped shaped atomic theory

But when I think about it I get sorta teary.

Cause no one believed in what he said,

Till four years after he was dead.

Avogadro helped shaped atomic theory

But when I think about it I get sorta teary.

Cause no one believed in what he said,

Till four years after he was dead.

Amedeo Avogadro

When you speak your mind and everyone ignores you,

No one’s standing up and rooting for you,

So what if they snicker and they laugh,

Someday they will want your autograph.

When you speak your mind and everyone ignores you,

No one’s standing up and rooting for you,

So what if they snicker and they laugh,

Someday they will want your autograph.

So far we’ve compared all the variables except the amount of a gas (n). There is a lesser known law called

Avogadro’s Law which relates V & n.

It turns out that they are directly related to each other.

As # of moles increases then V increases.

So far we’ve compared all the variables except the amount of a gas (n). There is a lesser known law called

Avogadro’s Law which relates V & n.

It turns out that they are directly related to each other.

As # of moles increases then V increases.

V = knV = kn

Avogadro’s LawAvogadro’s Law

Avogadro’s LawAvogadro’s LawEqual volumes of gases at the Equal volumes of gases at the

same T and P have the same same T and P have the same number of molecules.number of molecules.

V = n (T/P) = knV = n (T/P) = knV and n are directly related.V and n are directly related.

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

V = n (T/P) = knV = n (T/P) = knV and n are directly related.V and n are directly related.

twice as many twice as many moleculesmolecules

Avogadro’s LawAvogadro’s Law

Example : 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? 5.00 L = V2

.0.965 mol 1.80 mol

V2 = 9.33 L

V1V1

n1n1

= = V2V2

n2n2

IF the MOLE is BACK… IF the MOLE is BACK… STOICHIOMETRY is BACK !!STOICHIOMETRY is BACK !!

Ammonia burns in oxygen to form nitric oxide Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of H(NO) and water vapor. How many liters of H22O O

are obtained from 3.5 liter of ammonia at STP? are obtained from 3.5 liter of ammonia at STP?

4NH3 + 5O2 4NO + 6H2O

3.5 L NH3 Y mol NH3 6 mol H2O X L H2O= 5.25 L H2O

1 X L NH3 4 mol NH3 Y mol H2O

If conditions are the same the volumes are the same!

Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L.

What is Standard molar Volume?

How can we use 22.4L?

Example #1 : What is the volume of 10 moles of CO2 gas at STP?

Ex. #2 What is the density of H2O vapor at STP?

D = mass/volume Molar mass H2O = 18.0 g/mol

Molar volume of H2O = 22.4 L/mol

D = 18.0g/mol 22.4 L/mol

Example #1 : What is the volume of 10 moles of CO2 gas at STP?

Ex. #2 What is the density of H2O vapor at STP?

D = mass/volume Molar mass H2O = 18.0 g/mol

Molar volume of H2O = 22.4 L/mol

D = 18.0g/mol 22.4 L/mol

10 mol CO2 22.4 L of CO2 = 224 L CO21 1 mole CO2

= 0.8036 g/L

Complete worksheet

problems 1- 6

Would it not be Grand if we could

have only one PERFECT law

Would it not be Grand if we could

have only one PERFECT law

So far we have held at least one So far we have held at least one variable constant for every law we variable constant for every law we

have usedhave used.

What if we combined all the variables and created a constant to hold them

together.

What would that formula look like?

How would it be arranged?

How about another song?

PV = nRT

THE IDEAL GAS THE IDEAL GAS LAWLAW

Pressure is P

Volume is V

n is number of mole

Temperature is T

P & VTemp is Temp is

measured measured on the on the Kelvin Kelvin Scale!Scale!

THE IDEAL GAS LAWTHE IDEAL GAS LAW

BOYLES + CHARLES + Gay-LUSSAC

+ AVOGADRO LAWS ------------------------------

-COMBINED TOGETHER

PV = PV = nRTnRT

If we combine all of the laws together including Avogadro’s Law mentioned earlier we get:

If we combine all of the laws together including Avogadro’s Law mentioned earlier we get:

Where R is the universal gas constant

Where R is the universal gas constant

Normallywritten asNormallywritten as

Ideal Gas LawIdeal Gas Law

PVPV

nTnT= R= R

PV = nRTPV = nRT

R is a constant that connects all

of the four gas variables

R is dependent on the units of the variables for P, V, & T• Temp is always in Kelvin• Volume is in liters• Pressure is in either atm or

mmHg or kPa

R is a constant that connects all

of the four gas variables

R is dependent on the units of the variables for P, V, & T• Temp is always in Kelvin• Volume is in liters• Pressure is in either atm or

mmHg or kPa

Ideal Gas Constant (R)Ideal Gas Constant (R)

Because of the different pressure units there are 3 possibilities for our ideal gas constant

Because of the different pressure units there are 3 possibilities for our ideal gas constant

R=.0821R=.0821L•atmL•atmmol•Kmol•K

•If pressure is given in mmHg

•If pressure is given in mmHgR=62.4R=62.4L•mmHgL•mmHg

mol•Kmol•K• If pressure

is given in kPa

• If pressure is given in kPa

R=8.31R=8.31 L•kPaL•kPamol•Kmol•K

•If pressure is given in atm

•If pressure is given in atm

Ideal Gas ConstantIdeal Gas Constant

Using the Ideal Gas LawUsing the Ideal Gas Law

What volume does 9.45g of C2H2 occupy at STP?

What volume does 9.45g of C2H2 occupy at STP?

P P

V V T T

1atm1atm

?? 273K273K

R R

n n = .3635 mol

= .3635 mol

.0821 .0821L•atmL•atm

mol•Kmol•K

9.45g9.45g

26g26g

PV = nRTPV = nRT(1.0atm)(1.0atm)(V)(V)

(.3635mol

)(.3635mol

)(273K)(273K)

V = 8.15LV = 8.15L

==

(.0821 )(.0821 )L•atmmol•KL•atmmol•K

(1.0atm)(1.0atm)(V)(V) (8.147L•atm)(8.147L•atm)==

A camping stove propane tank holds 3000g of C3H8. How large a container would be needed to

hold the same amount of propane as a gas at 25°C and a

pressure of 303 kpa?

A camping stove propane tank holds 3000g of C3H8. How large a container would be needed to

hold the same amount of propane as a gas at 25°C and a

pressure of 303 kpa? P P

V V T T

303kPa303kPa

?? 298K298K

R R

n n = 68.2 mol= 68.2 mol

8.31 8.31 L•kPaL•kPa

mol•Kmol•K

3000g3000g

44g44g

PV = nRTPV = nRT(303kPa)(303kPa) (V) (V)

(68.2 mol)(68.2 mol) (298K)(298K)

V = 557.7LV = 557.7L

==

(8.31 )(8.31 )L•kPamol•KL•kPamol•K

(303kPa) (303kPa) (V) (V) (168,970.4 L•kPa)

(168,970.4 L•kPa)

= =

The Ideal Gas Equations

RTM

mPV )2

nRTPV )1

RT

PMD )3

Try These 5 problems

44.0 g/mol= 1.96 g/L

22.4 L/mol

D = m/V Molar Mass = 44.0 g/mol :: Molar volume = 22.4

L/mol D =

D = PM/RT = (1.02 atm)(92g/mol)(0.0821L*atm/mol*K)(298K)

= 3.83 g/L

1) Calculate the density of one mole of CO2 gas at STP. {D = m/V}

2) What is the density of laughing gas(N2O4) released at 25°C and 1.02 atm?

Try These 5 problems

3) A 0.519 g gas has a volume of 200 ml at STP. Is this gas propane (C3H8), butane (C4H10) or something else?

Using the Ideal Gas Law PV = mRT/M M = mRT/PV = (0.519 g)(0.0821 L atm/mol K)(273K)(1 Atm) (.200 L)

= 58.1 g/mol the molar mass of butane

4) A 1.25 g sample of a gaseous product of a chemical reaction was found to have a volume of 350 ml at 20.0°C and 750 mm Hg. What is the molar mass of this gas?

5) What is the volume of a gas whose molar mass is 90.2

g/mol. 1.39 g of the gas is contained at 755 mm Hg and 22 °C

Try These 5 problems4) Using the Ideal Gas Law PV = mRT/M M = mRT/PV

= (1.25 g)(62.4 L mm Hg/mol K)(293K)(750 mm Hg) (.350 L) = 87.0

g/mol

5) Using the Ideal Gas Law PV = mRT/M V = mRT/PM

= (1.39 g)(62.4 L mm Hg/mol K)(295K)(755 mm Hg) (90.2 g/mol) =.376 L

Ideal Gas Law & StoichiometryIdeal Gas Law & Stoichiometry

What volume of hydrogen gas must be burned to form 1.00 L of

water vapor at 1.00 atm pressure and 300°C?

(hint: Solve for mole of H2O then convert)

What volume of hydrogen gas must be burned to form 1.00 L of

water vapor at 1.00 atm pressure and 300°C?

(hint: Solve for mole of H2O then convert) PV = nRTPV = nRT(1.00 atm)(1.00 atm)

(1.00 L)(1.00 L)nH2O=nH2O=

(.0821L atm/mol K)

(.0821L atm/mol K)

(573K)(573K)

nH2O= .021257 mols

nH2O= .021257 mols

Ideal Gas Law & StoichiometryIdeal Gas Law & Stoichiometry

2 mol H2O 2 mol H2O

2 mol H22 mol H2

==1mol H21mol H2

22.4 L H222.4 L H2

2H2 + O2 2H2O2H2 + O2 2H2O

.021257 mol.021257 mol

.476 L H2.476 L H2

The GAS LAWS Gases have mass Gases diffuse Gases expand to fill containers Gases exert pressure...

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