1

GASESGASESChemistry I – Chapter 11Chemistry I – Chapter 11

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Importance of GasesImportance of Gases

• Airbags fill with NAirbags fill with N22 gas in an gas in an

accident. accident. • Gas is generated by the Gas is generated by the

decomposition of sodium decomposition of sodium azide, NaNazide, NaN33..

• 2 NaN2 NaN33 ---> 2 Na + 3 N ---> 2 Na + 3 N22

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THREE STATES OF THREE STATES OF MATTERMATTER

THREE STATES OF THREE STATES OF MATTERMATTER

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General Properties General Properties of Gasesof Gases

• There is a lot of “free” There is a lot of “free” space in a gas.space in a gas.

• Gases can be expanded Gases can be expanded infinitely. (they will fill infinitely. (they will fill whatever “container” they whatever “container” they are in.)are in.)

• Gases fill containers Gases fill containers uniformly and completely.uniformly and completely.

• Gases diffuse and mix Gases diffuse and mix rapidly.rapidly.

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PressuPressurerePressure is the Pressure is the

amount of force per amount of force per unit area.unit area.

1.1.Gas Pressure is the Gas Pressure is the pressure caused by pressure caused by particles of gas particles of gas striking an object.striking an object.

2.2.Air pressure is the Air pressure is the pressure of the pressure of the column of column of atmosphere above atmosphere above you ,pressing down you ,pressing down on you. on you.

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Atmospheric Pressure

• Measured with a Measured with a BAROMETERBAROMETER

(developed by (developed by Torricelli in 1643)Torricelli in 1643)

• Hg rises in tube Hg rises in tube until force of Hg until force of Hg (down) balances (down) balances

the force of the force of atmosphere atmosphere

(pushing up). (pushing up). (Just like a straw in a (Just like a straw in a

soft drink)soft drink)

• Why is Hg so Why is Hg so good for use in good for use in

barometer?barometer?

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Atm. Atm. PressurePressureAbout 34 feet of water = 1 atm About 34 feet of water = 1 atm pressure!pressure!

Column height measures Column height measures Pressure of atmospherePressure of atmosphere

• 1 standard atmosphere 1 standard atmosphere (atm) *(atm) *

= 760 mm Hg (or torr) *= 760 mm Hg (or torr) *

= 29.92 inches Hg *= 29.92 inches Hg *

= 14.7 pounds/in= 14.7 pounds/in2 2 (psi)(psi) **

= 101.3 kPa (SI unit is = 101.3 kPa (SI unit is PASCAL)PASCAL)

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Pressure Conversions

• Factor Label Method is how we will make conversions.Sample Problem, p 379 of textOne standard atmosphere (1 atm) is equivalent to 760 mm of Hg.

What is that height expressed in inches? (Hint: 1.00 in = 25.4 mm)Analysis:Identify1. Your “given”: _____2. Your unknown (what you’re solving for):_____3. the relationship (formula, etc) between the given & the unknown:

___Write an equation:1. Your given is on the left, including the units.2. The conversion factor is written next, but you must decide how to

write it: 3. Write an “=“4. Write a blank for your answer, and the units your answer will be

written in.

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Pressure Conversions

A. What is 475 mm Hg expressed in atm?

1 atm

760 mm Hg

B. The pressure of a tire is measured as 29.4 psi.

What is this pressure in mm Hg?

760 mm Hg

14.7 psi = 1.52 x 103 mm Hg

= 0.625 atm475 mm Hg x

29.4 psi x

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Pressure Conversions

A. What is 2 atm expressed in torr?

B. The pressure of a tire is measured as 32.0 psi.

What is this pressure in kPa?

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Properties of Properties of GasesGasesGas properties can be modeled Gas properties can be modeled

using math. Model depends onusing math. Model depends on——

• V = volume of the gas (L)V = volume of the gas (L)• T = temperature (K)T = temperature (K)

–ALL temperatures in the ALL temperatures in the entire chapter MUST be entire chapter MUST be in Kelvin!!! No Exceptions!in Kelvin!!! No Exceptions!

• n = amount (moles)n = amount (moles)• P = pressureP = pressure

(atmospheres) (atmospheres)

Chapter 14•The Behavior of Gases

• (new page of your permanent Chem

notebook)

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Properties of Gases

• Compressibility-a measure of how much the volume of matter decreases under pressure.

• Gases are compressible b/c of the space btwn particles in a gas

• At room temp, distance btwn particles in an enclosed gas ~ 10x the diameter of the particles

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Factors to Describe Gas Pressure

• Pressure in kPa

• Volume in L

• Temperature in K

• Number of moles, n

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Number of Moles

• Pressure is caused by the particles striking the walls of the container.

• If the gas is in a rigid container, the volume is constant

• If you triple the number of gas particles (n), you triple the amount of pressure. (see p 415 of text)

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Volume

• Reducing the volume in a container will raise the pressure (inverse relationship.)

• See p 416 If you press on a piston, no gas escapes, but the volume decreases.

• The same # of particles are striking a smaller area, so pressure increases.

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Temperature

• As a gas is heated, the temperature increases

• If temperature increases, particle speed increases

• IF the container is rigid, the pressure will increase.

• If the container is not rigid, the volume will increase & pressure will remain the same.

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Section 14.1 Assessmentp 417

1. Why is a gas easy to compress?

2. List 3 factors that can affect gas pressure.

3. Why does a collision with an air bag cause less damage than a collision with a steering wheel?

4. How does a decrease in temp affect the pressure of a contained gas?

5. If the temp is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to ¼ of its original value?

6. Assuming the gas in a container remains at a constant temp, how could you increase the gas pressure in a container a hundredfold?

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Properties of Gases, Properties of Gases, cont.cont.

• We can study the relationship between 2

variables if we keep the others the same.

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Boyle’s LawBoyle’s Law

• 2 variables we will study: P & V

• All other variables kept the same (T, n, etc.)

• Ex: a rising balloon

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Boyle’s LawBoyle’s LawP P αα 1/V 1/V

This means Pressure This means Pressure and Volume are and Volume are INVERSELY INVERSELY PROPORTIONALPROPORTIONAL

For example, P goes For example, P goes up as V goes down.up as V goes down.

PP11VV11 = P = P22 V V22

Robert Boyle Robert Boyle (1627-1691). (1627-1691). Son of Early of Son of Early of Cork, Ireland.Cork, Ireland.

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Boyle’s Law and Boyle’s Law and Kinetic Molecular Kinetic Molecular

TheoryTheory

Boyle’s Law and Boyle’s Law and Kinetic Molecular Kinetic Molecular

TheoryTheory

P proportional to 1/VP proportional to 1/V

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Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s LawA bicycle pump is a A bicycle pump is a

good example of good example of Boyle’s law. Boyle’s law.

As the volume of the As the volume of the air trapped in the air trapped in the pump is reduced, pump is reduced, its pressure goes its pressure goes up, and air is up, and air is forced into the tire.forced into the tire.

Sample Problem 14.1 (p 419)

• A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa?

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GivensP1= 103 kPa P2 = 25.0kPa

V1 = 30.0L

UnknownV2

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Charles’s LawCharles’s Law• 2 variables we will study: V & T

• All other variables kept the same (P, n, etc.)

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Charles’s Charles’s LawLaw

If n and P are If n and P are constant, constant, then V then V αα T T

V and T are directly V and T are directly proportional.proportional.

VV11 V V22

==

TT11 T T22

• If one temperature goes If one temperature goes

up, the volume goes up!up, the volume goes up!

Jacques Charles (1746-Jacques Charles (1746-1823). Isolated boron 1823). Isolated boron and studied gases. and studied gases. Balloonist.Balloonist.

Sample Problem 14.2 Using Charles Law (p 421)

• A balloon inflated in a room at 24˚C has a volume of 4.00L. The balloon is then heated to a temperature of 48˚C. What is the new volume if the pressure remains constant?

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Givens: T1 = 24˚C + 273 = 297K T2 = 48 + 273 =321K V1 = 4.00 L

Unknown = V2

4.00L = V2

297K 321K

4.00L*321K = V2

297K

CW/HWPractice Problems, pp 419-423

# 7-12

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Charles’s original balloonCharles’s original balloon

Modern long-distance balloonModern long-distance balloon

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Charles’s LawCharles’s Law

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Gay-Lussac’s LawGay-Lussac’s LawIf n and V are constant, If n and V are constant,

then P then P αα T TP and T are directly P and T are directly

proportional.proportional.

PP11 P P22

==

TT11 T T22

• If one temperature If one temperature

goes up, the pressure goes up, the pressure goes up!goes up!

Joseph Louis Gay-Joseph Louis Gay-Lussac (1778-1850)Lussac (1778-1850)

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Gas Pressure, Gas Pressure, Temperature, and Kinetic Temperature, and Kinetic

Molecular TheoryMolecular Theory

Gas Pressure, Gas Pressure, Temperature, and Kinetic Temperature, and Kinetic

Molecular TheoryMolecular Theory

P proportional to TP proportional to T

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Combined Gas Law• Since all 3 gas laws are

related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!

P1 V1 = P2 V2

T1 T2

No, it’s not related to R2D2

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Combined Gas Law

If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!

= =

P1 V1

T1

P2 V2

T2

Boyle’s Law

Charles’ Law

Gay-Lussac’s Law

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Combined Gas Law Problem

A sample of helium gas has a volume of 180 mL, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Set up Data Table

P1 = 0.800 atm V1 = 180 mL T1 = 302 K

P2 = 3.20 atm V2= 90 mL T2 = ??

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CalculationP1 = 0.800 atm V1 = 180 mLT1 = 302KP2 = 3.20 atm V2= 90 mL T2 = ??P1 V1 P2 V2

= P1 V1 T2 = P2 V2 T1

T1 T2

T2 = P2 V2 T1

P1 V1

T2 = 3.20 atm x 90.0 mL x 302 K

0.800 atm x 180.0 mLT2 = 604 K - 273 = 331 °C

= 604 K

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Learning Check

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

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One More Practice Problem

A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

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And now, we pause for this commercial message from

STPOK, so it’s really not THIS kind

of STP…

STP in chemistry stands for Standard Temperature and

Pressure

Standard Pressure = 1 atm (or an equivalent)

Standard Temperature = 0 deg

C (273 K)

STP allows us to compare amounts of

gases between different pressures and temperatures

STP allows us to compare amounts of

gases between different pressures and temperatures

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Try This One

A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

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Avogadro’s Avogadro’s HypothesisHypothesis

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 and n are directly relatedV and n are directly related..

twice as many twice as many moleculesmolecules

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Avogadro’s Hypothesis Avogadro’s Hypothesis and Kinetic Molecular and Kinetic Molecular

TheoryTheory

Avogadro’s Hypothesis Avogadro’s Hypothesis and Kinetic Molecular and Kinetic Molecular

TheoryTheory

P proportional to nP proportional to n

The gases in this The gases in this experiment are all experiment are all measured at the measured at the same T and V.same T and V.

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IDEAL GAS LAWIDEAL GAS LAW

Brings together gas Brings together gas properties.properties.

BE SURE YOU KNOW BE SURE YOU KNOW THIS EQUATION!THIS EQUATION!

P V = n R TP V = n R T

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Using PV = nRTUsing PV = nRTP = PressureP = Pressure n = number of n = number of

molesmoles

V = VolumeV = Volume T = TemperatureT = Temperature

R is a constant, called the R is a constant, called the Ideal Gas Ideal Gas Constant= Constant= 8.31 L*kPa8.31 L*kPa

K*molK*mol

We must We must convert the units to match R.convert the units to match R.

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Using PV = nRTUsing PV = nRT• Ex: p 439, Q#55Ex: p 439, Q#55• 1.24 moles of gas at 35 C and 96.2 kPa 1.24 moles of gas at 35 C and 96.2 kPa

pressure. What is the volume the gas pressure. What is the volume the gas occupies?occupies?

• V=?V=?

• n= 1.24 moln= 1.24 mol

• T = 35 + 273 = 308KT = 35 + 273 = 308K

• P = 96.2kPaP = 96.2kPa

• R=8.31L*kPa/K*molR=8.31L*kPa/K*mol

• (96.2kPa)V = (1.24 mol ) (8.21L*kPa/K*mol) (308K)(96.2kPa)V = (1.24 mol ) (8.21L*kPa/K*mol) (308K)

• V=33LV=33L

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Learning Check

Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

47

Learning Check

A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

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Gases in the AirThe % of gases in air Partial pressure (STP)

78.08% N2 593.4 mm Hg

20.95% O2 159.2 mm Hg

0.94% Ar 7.1 mm Hg

0.03% CO2 0.2 mm Hg

PAIR = PN2 + PO2 + PAr + PCO2 = 760 mm Hg

(Total Pressure = 760mm Hg)

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Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

What is the total pressure in the flask?What is the total pressure in the flask?

PPtotaltotal in gas mixture = P in gas mixture = PAA + P + PBB + ... + ...

Therefore, Therefore,

PPtotaltotal = P = PHH22OO + P + POO22 = 0.48 atm = 0.48 atm

Dalton’s Law: total P is sum of PARTIAL Dalton’s Law: total P is sum of PARTIAL pressures. pressures.

2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)

0.32 atm 0.32 atm 0.16 0.16 atmatm

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Dalton’s Dalton’s LawLaw

John DaltonJohn Dalton1766-18441766-1844

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Health NoteWhen a scuba diver is several hundred feet under water, the high pressures cause N2 from the

tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in

the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in

scuba tanks used for deep descents.

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GAS DENSITYGAS DENSITYGAS DENSITYGAS DENSITY

HighHigh densitydensity

Low Low densitydensity

22.4 L of ANY gas AT STP = 1 mole

53

Gases and Gases and StoichiometryStoichiometry

2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)

Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a in a flask with a

volume of 2.50 L. What is the volume of Ovolume of 2.50 L. What is the volume of O22 at at

STP?STP?

Bombardier beetle Bombardier beetle uses decomposition uses decomposition of hydrogen peroxide of hydrogen peroxide to defend itself.to defend itself.

54

Gases and Gases and StoichiometryStoichiometry

2 H2 H22OO2 2 (l) ---> 2 H(l) ---> 2 H22O (g) + OO (g) + O2 2 (g)(g)

Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume in a flask with a volume

of 2.50 L. What is the volume of Oof 2.50 L. What is the volume of O22 at STP? at STP?

SolutionSolution1.1 g1.1 g HH22OO22 1 mol H 1 mol H22OO22 1 mol O 1 mol O22 22.4 L 22.4 L

OO22

34 g H34 g H22OO22 2 mol H 2 mol H22OO22 1 mol O 1 mol O22 = 0.36 L O2 at STP

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Gas Stoichiometry: Practice!

A. What is the volume at STP of 4.00 g of CH4?

B. How many grams of He are present in 8.0 L

of gas at STP?

56

What if it’s NOT at STP?

• 1. Do the problem like it was at STP. (V1)

• 2. Convert from STP (V1, P1, T1) to the stated conditions (P2, T2)

57

Try this one!

How many L of O2 are needed to react 28.0 g NH3 at 24°C and 0.950 atm?

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

58

GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION

• diffusiondiffusion is the is the gradual mixing of gradual mixing of molecules of different molecules of different gases.gases.

• effusioneffusion is the is the movement of movement of molecules through a molecules through a small hole into an small hole into an empty container.empty container.

HONORS HONORS onlyonly

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GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION

Graham’s law governs Graham’s law governs effusion and diffusion effusion and diffusion of gas molecules.of gas molecules.

Thomas Graham, 1805-1869. Thomas Graham, 1805-1869. Professor in Glasgow and London.Professor in Glasgow and London.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

M of AM of B

Rate for B

Rate for A

HONORS HONORS onlyonly

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GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION

Molecules effuse thru holes in a Molecules effuse thru holes in a rubber balloon, for example, at a rubber balloon, for example, at a rate (= moles/time) that israte (= moles/time) that is

• proportional to Tproportional to T• inversely proportional to M.inversely proportional to M.

Therefore, He effuses more rapidly Therefore, He effuses more rapidly than Othan O22 at same T. at same T.

HeHe

HONORS HONORS onlyonly

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Gas DiffusionGas Diffusionrelation of mass to rate of relation of mass to rate of

diffusiondiffusion

Gas DiffusionGas Diffusionrelation of mass to rate of relation of mass to rate of

diffusiondiffusion• HCl and NH3 diffuse

from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.

• HCl and NH3 diffuse from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.

HONORS HONORS onlyonly