GAS LAWSGAS LAWS

Kinetic Molecular TheoryKinetic Molecular Theory

• Particles in an ideal gas…– have no volume.– have elastic collisions. – are in constant, random, straight-line motion.– don’t attract or repel each other.– have an average KE directly related to Kelvin

temperature.

Real GasesReal Gases• Particles in a REAL gas…

– have their own volume– attract and repel each other

• Gas behavior is most ideal…– at low pressures– at high temperatures

***Most real gases act like ideal gases except under high pressure and low temperature.

Characteristics of GasesCharacteristics of Gases• Gases expand to fill any container.

– Take the shape and volume of their container.

• Gases are fluids (like liquids).– Little to no attraction between the particles

• Gases have very low densities.= lots of empty space between the particles

Characteristics of GasesCharacteristics of Gases• Gases can be compressed.

– lots of empty space between the particles– Indefinite density

• Gases undergo diffusion.– random motion – scatter in all directions

PressurePressure

area

forcepressure

Which shoes create the most pressure?

Pressure- how much a gas is pushing

on a container.

• Atmospheric pressure- atmospheric gases push on everything on Earth

• UNITS AT SEA LEVEL

1 atm =101.3 kPa (kilopascal)= 760 mmHg =760 torr

PressurePressure• Barometer

– measures atmospheric pressure

Mercury Barometer

Aneroid Barometer

PressurePressure• Manometer

– measures contained gas pressure

C. JohannessonU-tube Manometer Bourdon-tube gauge

C. Johannesson

Temperature= how fast the molecules are moving

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

K = ºC + 273

• Always use absolute temperature (Kelvin) when working with gases.

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.3 kPa

760 mm Hg

-OR-

STP

-OR-

Volume = how much space a gas occupies

Units– L, mL, cm3

• 1000 mL = 1 L • 1 mL = 1 cm3

BASIC GAS LAWS

Charles’ LawCharles’ Law

• T V (temperature is directly proportional to volume)

• T ↑ V↑ & T↓ V↓

• V1 = V2

T1 T2 T is always in K

– K = °C + 273

– P and n = constant

V

T

CharlesLaw.exe

Timberlake, Chemistry 7th Edition, page 259

(Pressure is held constant)

T1 T2

V1 V2=

Charles’ Law

Charles’ Law

Timberlake, Chemistry 7th Edition, page 254

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Charles’ LawCharles’ LawThe egg out of the bottle

• Mrs. Rodriguez inflates a balloon for a party. She is in an air-conditioned room at 27.0oC, and the balloon has a volume of 4.0 L. Because she is a curious and intrepid chemistry teacher, she heats the balloon to a temperature of 57.0oC. What is the new volume of the balloon if the pressure remains constant?

Charles’ Law ProblemCharles’ Law Problem

Given Unkown Equation

Substitute and Solve

T1 = 27.0oC +273= 300 K V1 = 4.0 L T2 = 57.0oC +273= 330 K

V2 = ? L P1V1 = P2V2

T1 V1T2

4.0 L = V2 =300 K 330K

4.4 L

• A 25 L balloon is released into the air on a warm afternoon (42º C). The next morning the balloon is recovered on the ground. It is a very cold morning and the balloon has shrunk to 22 L. What is the temperature in º C?

Charles’ Law Learning CheckCharles’ Law Learning Check

Given Unkown Equation

Substitute and Solve

V1 = 25 L T1 = 42 oC +273= 315 K V2 = 22 L

T2 = ? ºC P1V1 = P2V2

T1 V1T2

25 L = 22 L =315 K T2

277.2 K -273 = 4.2 ºC

Boyle’s LawBoyle’s Law

• P↓ V ↑ & P↑ V ↓• P 1/V (pressure is inversely proportional to volume)

• P1V1 = P2V2– T and n = constant

P

V

Boyle'sLaw.exe

Timberlake, Chemistry 7th Edition, page 253

P1V1 = P2V2

(Temperature is held constant)

Boyle’s LawBoyle’s Law

Timberlake, Chemistry 7th Edition, page 254

Marshmallows in a vacuum

Boyle’s LawBoyle’s Law

Mechanics of Breathing

Timberlake, Chemistry 7th Edition, page 254

Boyle’s LawBoyle’s Law

A balloon is filled with 30.L of helium gas at 1.00 atm. What is the volume when the balloon rises to an altitude where the pressure is only 0.25 atm?

Given Unkown Equation

• Substitute and Solve

Boyle’s Law ProblemBoyle’s Law Problem

V2 0.25 atm = 30 L x 1.0 atm = 120 L

V1 = 30 LP1 = 1 atmP2 = .25atm

V2 = ? L P1V1 = P2V2

T1 T2

A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.

Given Unkown Equation

•Substitute and Solve

Boyle’s Law Learning CheckBoyle’s Law Learning Check

V1 = 100. mL = 0.100 LP1 = 150. kPaP2 = 200. kPa

V2 = ? L P1V1 = P2V2

T1 T2

AVOGADRO’S LAWAVOGADRO’S LAW• Vn Vn • V n (direct)

• V1 = V2

n1 n2

– T & P Constant

Avogadro'sLaw.exe

 

 

n

V

A 3.0 liter sample of gas contains 7.0 moles. How much gas will there be, in order for the sample to be 2.3 liters? P & T do not change

Given Unkown Equation

• Substitute and Solve

Avogadro’s Law ProblemAvogadro’s Law Problem

V1 = 3.0 Ln1 = 7.0 molV2 = 2.3 L

n2 = ? mol P1V1 = P2V2

n1T1 n2T2

3.0 L = 2.3 L =7.0 mol n2 mol

5.4 mol

Gay-Lussac’s LawGay-Lussac’s Law

• P1 = P2

T1 T2

– V & n constant

• Direct relationship

• PT PT

P

T

Gay-Lussac'sLaw.exe

Gay-Lussac LawGay-Lussac Law• Collapsing Barrel

Gay-LussacGay-Lussac LawLaw• Tank car implosion

COMBINED IDEAL GAS LAWCOMBINED IDEAL GAS LAW

• P1V1 = P2V2

n1T1 n2T2

• If P, V, n, or T are constant then they cancel out of the equation.

• n usually constant (unless you add or remove gas), so

• P1V1 = P2V2

T1 T2

Combined Gas Law ProblemCombined Gas Law Problem• Ms. Evans travels to work in a hot air balloon from the Rocky

Mountains. At her launch site, the temperature is 5.00 °C, the atmospheric pressure is 0.801 atm, and the volume of the air in the balloon is 120.0 L. When she lands in Plano, the temperature is 28.0 °C and the atmospheric pressure is 101.3 kPa. What is the new volume of the air in the balloon?

Given Unkown Equation

Substitute and Solve V2 x 1 atm = 120.0 L x 0.801 atm = 104 L 301K 278 K

T1 = 5.0oC +273= 278 KP1 = 0.801 atmV1 = 120.0 LT2 = 28.0oC +273= 301 KP2 = 101.3 kPa = 1 atm

V2 = ? L V1 x P1 = V 2 x P2 T 1 T 2

Combined Gas Law Learning CheckCombined Gas Law Learning Check• Nitrogen gas is in a 7.51 L container at 5.C and 0.58 atm.

What is the new volume of the gas at STP?

Given Unkown Equation

Substitute and Solve

V2 x 1.0 atm = 7.51L x 0.58 atm = 4.3 L 273 K 278 K

T1 = 5.0oC +273= 278 KP1 = 0.58 atmV1 = 7.51 LT2 = 273 KP2 = 1 atm

V2 = ? L V1 x P1 = V 2 x P2 T 1 T 2

Ideal Gas Law (“Pivnert”)Ideal Gas Law (“Pivnert”)PV=nRT R = The Ideal Gas Constant R = 0.0821 (L*atm) R = 62.4 (L*mm Hg) (mol*K) (mol*K) R = 8.31 (L*kPa) (mol*K)•V has to be in Liters, n in Moles, T in Kelvin, •P can be in atm, kPa or mmHg•* Choose which R to used based on the units of your pressure.

P V = n R T (atm) (L) = (moles) (L*atm/mol*K) (K) (kPa) (L) = (moles) (L*kPa/mol*K) (K) mm Hg (L) = (moles) (L*mmHg/mol*K) (K)

Ideal Gas Law ProblemIdeal Gas Law Problem• A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen

gas to a final pressure of 200.0 atm at 27.0 oC. How many moles of gas does the cylinder hold?

Given Unkown Equation

Substitute and Solve

n 0821 atm L/K Mole x 300 K = 200.0 atm x 20.0L= 162 moles

V = 20.0 LP = 200.0 atmT =27.0oC +273= 300 K

moles of nitrogen?

PV=nRT R= .0821 atm L/K Mole

Ideal Gas Law Learning CheckIdeal Gas Law Learning Check• A balloon contains 2.00 mol of nitrogen at a pressure of 0.980

atm and a temperature of 37C. What is the volume of the balloon?

Given Unkown Equation

Substitute and Solve

0.980 atm x V= 2.00 mol x .0821 atm L/K Mole x 310 K = 51.9 L

n = 2.00 molP = 0.980 atmT =37.0oC +273= 310 K

V in L? PV=nRT R= .0821 atm L/K Mole

Dalton’s Law of Partial PressureDalton’s Law of Partial Pressure• The total pressure of a mixture of gases is

equal to the sum of the partial pressures of the component gases.

• Ptotal = Pgas 1 + Pgas 2 + P gas 3 + …

A metal container holds a mixture of 2.00 atm of nitrogen, 1.50 atm of oxygen and 3.00 atm of helium. What is the total pressure in the canister?

6.5 atm

Welcome to Mole Welcome to Mole IslandIsland

Welcome to Mole Welcome to Mole IslandIsland

Welcome to Mole Welcome to Mole IslandIsland

Gas StoichiometryGas StoichiometryMoles Moles Liters of a Gas: Liters of a Gas:

– 22C4H10 (g) + 13O2(g) O8CO2(g) + 10H2O(g)

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2 mol + 13 mol ? 8 mol + 10 mol

Avogadro’s principle states that one mole of any gas occupies 22.4 L at STP.

Thus when gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts but also relatives volumes

2 L + 13 L ? 8 L + 10 LRecall:The coefficients in a chemical reaction represent molar amounts of substances taking part in the reaction.

Gas Stoichiometry ProblemGas Stoichiometry ProblemIn the following combustion reaction, what volume of methane (CH4) is needed to produce 26 L of

water vapor?

– CH4 (g) + 2O2(g) O CO2(g) + 2H2O(g)

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x L ? 26 L

1 L ? 2 L

x L = 26 L 1L 2L

x= 13 L

1 mol ? 2 mol

Gas StoichiometryGas Stoichiometry use ideal gas lawuse ideal gas law

– Looking for grams or moles of gas? • Step 1: start with ideal gas law to find moles of gas• Step 2: 1change to grams of gas

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PV=nRT

Grams/mol? 1) Use Ideal Gas Law 2) Do stoichiometry calculations

Example 1Example 1How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? PV=nRT

4 Al(s) + 3 O2(g) 2 Al2O3(s)

Given liters: Start with Ideal Gas Law and calculate moles of O2.

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Given Unkown V O2 = 15.0 L O2 grams of Al2O3? R= .0821 atm L/K MoleP O2 = 97.3 kPa= 0.9605 atm T O2 =21oC +273= 294 K

Step 1: Calculate moles of O2 n = PV = 0.9605 atm x 15.0 L = 0.5969 mol O2

RT 0.0821 atm L/K Mole 294 K

Step 2: Calculate mass of Al2O3

0.5969 mol O2 = X mol Al2O3 = 0.3979 mol Al2O3 3moleO2 2 mole Al2O3

Use stoich to convert moles of O2 to grams Al2O3.

41 g Al2O3 0.3979 mol Al2O3 x 101.96 g Al2O3=

1 mol Al2O3

Gas StoichiometryGas Stoichiometry use ideal gas lawuse ideal gas law

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

PV=nRT

• Looking for volume of gas?•Step 1: start with stoichiometry conversion to find moles of gas•Step 2: use ideal gas law to find the volume

Liters ? 1) Do stoichiometry calculations 2) Use Ideal Gas Law

What volume of CO2 forms from 5.25 g of CaCO3 at 101.3 kPa & 25ºC?

CaCO3 CaO + CO2

Looking for liters: Start with stoich and calculate moles of CO2.

Plug this into the Ideal Gas Law to find volume.

Given Unkown PV=nRTm = 5.25 g CaCO3 volume of CO2? R= .0821 atm L/K MoleP = 101.3 kPa = 1 atm T =25.0oC +273= 298 K

Step 1: Calculate moles of CO2 5.25 g CaCO3 x 1 mole CaCO3 = 0.0525 mol CaCO3

100 g CaCO3

1 mole CO2 = 1mole CaCO3 ; 0.0525 mol CO2

Step 2: Calculate volume of CO2

V = nRT = 0.0525 mol CO2 x .0821 atm L/K Mole x 298 K = 1.28 L P 1 atm

Example 2Example 2