Combined and ideal gas lawsCombined and ideal 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

T1V2 = T2V1T1V2 = T2V1

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

P1T2 = P2T1P1T2 = P2T1

A Little ReviewA Little Review

Combined gas lawCombined gas law

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 :

Amount is held constant Is used when you have a

change in volume, pressure, or temperature

Amount is held constant Is used when you have a

change in volume, pressure, or temperature

P1V1P1V1

T1T1

= K = = K = P2V2P2V2

T2T2

Combined gas lawCombined gas law

Combined gas lawCombined gas law Amount is held constant Is used when you have a

change in volume, pressure, or temperature

Amount is held constant Is used when you have a

change in volume, pressure, or temperature

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

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/n = kV/n = k

Avogadro’s LawAvogadro’s Law

Which leads us to the ideal gas law –

So far we have always held at least 1 of the variables constant.

We can set up a much more powerful eqn, which can be derived by combining the proportions expressed by the previous laws.

Which leads us to the ideal gas law –

So far we have always held at least 1 of the variables constant.

We can set up a much more powerful eqn, which can be derived by combining the proportions expressed by the previous laws.

Ideal Gas LawIdeal Gas Law

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 the 4 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 the 4 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=8.314R=8.314L•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)

= =

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?

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

water vapor at 1.00 atm pressure and 300°C?

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

Loose Ends of GasesLoose Ends of Gases

• There are a couple more laws that we need to address dealing with gases.– Dalton’s Law of Partial

Pressures– Graham’s Law of Diffusion and

Effusion.

• There are a couple more laws that we need to address dealing with gases.– Dalton’s Law of Partial

Pressures– Graham’s Law of Diffusion and

Effusion.

Dalton’s Law of Partial PressureDalton’s Law of Partial Pressure

• States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.

• States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.

PT=P1+P2+P3+…PT=P1+P2+P3+…• What that means is that each

gas involved in a mixture exerts an independent pressure on its containers walls

• What that means is that each gas involved in a mixture exerts an independent pressure on its containers walls

• Therefore, to find the pressure in the system you must have the total pressure of all of the gases involved.

• This becomes very important for people who work at high altitudes like mountain climbers and pilots.

• For example, at an altitude of about 10,000m air pressure is about 1/3 of an atmosphere.

• Therefore, to find the pressure in the system you must have the total pressure of all of the gases involved.

• This becomes very important for people who work at high altitudes like mountain climbers and pilots.

• For example, at an altitude of about 10,000m air pressure is about 1/3 of an atmosphere.

Dalton’s Law of Partial PressureDalton’s Law of Partial Pressure

• The partial pressure of oxygen at this altitude is less than 50 mmHg.

• By comparison, the partial pressure of oxygen in human alveolar blood needs to be about 100 mmHg.

• Thus, respiration cannot occur normally at this altitude, and an outside source of oxygen is needed in order to survive.

• The partial pressure of oxygen at this altitude is less than 50 mmHg.

• By comparison, the partial pressure of oxygen in human alveolar blood needs to be about 100 mmHg.

• Thus, respiration cannot occur normally at this altitude, and an outside source of oxygen is needed in order to survive.

Dalton’s Law of Partial PressureDalton’s Law of Partial Pressure

• Three of the primary components of air are CO2, N2, and O2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO2 and N2 are given as PCO2

= 0.285mmHg and PN2

= 593.525mmHg. What is the partial pressure of O2?

• Three of the primary components of air are CO2, N2, and O2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO2 and N2 are given as PCO2

= 0.285mmHg and PN2

= 593.525mmHg. What is the partial pressure of O2?

Simple Dalton’s Law Calculation

Simple Dalton’s Law Calculation

PT = PCO2 + PN2 + PO2PT = PCO2 + PN2 + PO2

Simple Dalton’s Law Calculation

Simple Dalton’s Law Calculation

760mmHg = .285mmHg +

593.525mmHg + PO2

760mmHg = .285mmHg +

593.525mmHg + PO2

PO2= 167mmHgPO2= 167mmHg

• Partial Pressures are also important when a gas is collected through water.—Any time a gas is collected

through water the gas is “contaminated” with water vapor.

—You can determine the pressure of the dry gas by subtracting out the water vapor

• Partial Pressures are also important when a gas is collected through water.—Any time a gas is collected

through water the gas is “contaminated” with water vapor.

—You can determine the pressure of the dry gas by subtracting out the water vapor

Dalton’s Law of Partial PressureDalton’s Law of Partial Pressure

Ptot = Patmospheric pressure = Pgas + PH2OPtot = Patmospheric pressure = Pgas + PH2O

AtmosphericPressure

AtmosphericPressure

—The water’s vapor pressure can be determined from a list and subtract-ed from the atmospheric pressure

—The water’s vapor pressure can be determined from a list and subtract-ed from the atmospheric pressure

WATER VAPOR PRESSURESWATER VAPOR PRESSURES

Temp (°C) Vapor pressure (kPa)

11 0.651760.6517655 .87260.872601010 1.22811.22811515 1.70561.70562020 2.33882.33882525 3.16913.16913030 4.24554.24553535 5.62675.62674040 7.38147.38144545 9.58989.58985050 12.34412.344

WATER VAPOR PRESSURESWATER VAPOR PRESSURES

Temp (°C) Vapor pressure (kPa)

5555 15.75215.752 6060 .19.932.19.9326565 25.02225.0227070 31.17631.1767575 38.56338.5638080 47.37347.3738585 57.81557.8159090 70.11770.1179595 84.52984.529

100100 101.32101.32105105 120.79120.79

• Determine the partial pressure of oxygen collected by water displace-ment if the water temperature is 20.0°C and the total pressure of the gases in the collection bottle is 730 mmHg.

• Determine the partial pressure of oxygen collected by water displace-ment if the water temperature is 20.0°C and the total pressure of the gases in the collection bottle is 730 mmHg.

Simple Dalton’s Law Calculation

Simple Dalton’s Law Calculation

PH2O at 20.0°C= 2.3388 kPaPH2O at 20.0°C= 2.3388 kPa

We need to convert to mmHg.We need to convert to mmHg.

PT = PH2O + PO2PT = PH2O + PO2

Simple Dalton’s Law Calculation

Simple Dalton’s Law Calculation

730mmHg = 17.5468 + PO2730mmHg = 17.5468 + PO2

PO2= 712.5 mmHgPO2= 712.5 mmHg

2.3388 kPa2.3388 kPa760 mmHg760 mmHg

101.3 kPa101.3 kPa

PH2O = 17.5468 mmHgPH2O = 17.5468 mmHg

Graham’s LawGraham’s Law

• Thomas Graham studied the effusion and diffusion of gases.– Diffusion is the mixing of gases

through each other.– Effusion is the process whereby

the molecules of a gas escape from its container through a tiny hole

• Thomas Graham studied the effusion and diffusion of gases.– Diffusion is the mixing of gases

through each other.– Effusion is the process whereby

the molecules of a gas escape from its container through a tiny hole

• Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule.– The bigger the molecule the

slower it moves the slower it mixes and escapes.

• Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule.– The bigger the molecule the

slower it moves the slower it mixes and escapes.

Graham’s LawGraham’s Law

Graham’s LawGraham’s Law

• Kinetic energy can be calculated with the equation ½ mv2 —m is the mass of the object —v is the velocity.

• If we work with two different at the same temperature their energies would be equal and the equation can be rewritten as:

• Kinetic energy can be calculated with the equation ½ mv2 —m is the mass of the object —v is the velocity.

• If we work with two different at the same temperature their energies would be equal and the equation can be rewritten as:

• “M” represents molar mass• “v” represents molecular

velocity• “A” is one gas• “B” is another gas

• “M” represents molar mass• “v” represents molecular

velocity• “A” is one gas• “B” is another gas

½ MAvA2 = ½ MBvB

2½ MAvA2 = ½ MBvB

2

• If we want to compare both gases velocities, to determine which gas moves faster, we could write a ratio of their velocities.—Rearranging things and taking

the square root would give the eqn:

• If we want to compare both gases velocities, to determine which gas moves faster, we could write a ratio of their velocities.—Rearranging things and taking

the square root would give the eqn:

• This shows that the velocities of two different gases are inversely propor-tional to the square roots of their molar masses.—This can be expanded to deal

with rates of diffusion or effusion

• This shows that the velocities of two different gases are inversely propor-tional to the square roots of their molar masses.—This can be expanded to deal

with rates of diffusion or effusion

vAvA

vBvB

==MBMB

MAMA

Rate of effusion of ARate of effusion of A==

Rate of effusion of BRate of effusion of B

MBMB

MAMA

• The way you can interpret the equation is that the number of times faster A moves than B, is the square root of the ratio of the molar mass of B divided by the Molar mass of A—So if A is half the size of B than

it effuses or diffuses 1.4 times faster.

• The way you can interpret the equation is that the number of times faster A moves than B, is the square root of the ratio of the molar mass of B divided by the Molar mass of A—So if A is half the size of B than

it effuses or diffuses 1.4 times faster.

Graham’s LawGraham’s Law

If equal amounts of helium and argon are placed in a porous

container and allowed to escape, which gas will escape faster and

how much faster?

If equal amounts of helium and argon are placed in a porous

container and allowed to escape, which gas will escape faster and

how much faster?

Graham’s Law Example Calc.Graham’s Law Example Calc.

Rate of effusion of ARate of effusion of A==

Rate of effusion of BRate of effusion of B

MBMB

MAMA

Graham’s Law Example Calc.Graham’s Law Example Calc.

Rate of effusion of HeRate of effusion of He==

Rate of effusion of ArRate of effusion of Ar

40 g40 g

4 g4 g

Helium is 3.16 times faster than Argon.Helium is 3.16 times faster than Argon.