GasesGases GasesGases

Characteristics of Gases

• Expand to fill and assume the shape of their container

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• Diffuse into one another and mix in all proportions.

• Particles move from an area of high concentration to an area of low concentration.

• Next

Invisible

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• Expand when heated

PropertiesPropertiesPropertiesProperties

that determine physical that determine physical behavior of a gasbehavior of a gas

Amount• Mass or moles

Volume l x w x h πr²h

Temperature

Pressure

The molecules in a gas are in constant motion. The purple balls represent gaseous atoms that collide with each other and the walls of the container. Gas molecules are in constant motion. "Pressure" is a measure of the collisions of the atoms with the container.

Pressure

• Force per unit area

• Equation: P = F/A

F = force A = area Next

Force• Pushing or pulling on something

Formulas For Surface Area

• Square: 6s²• rectangle: 2ab + 2bc + 2ac• Cylinder: 2πr²+ 2πrh• sphere: 4πr²

Units of Measurement

• N/m2 Newton per square meter• N/cm2 Newton per square centimeter• Pa Pascal• kPa kilopascal• Torr Torr• mmHg millimeters of mercury• lb/in2 pound per square inch

Calculating Pressure Using P = F/A

Suppose that a woman weighing 135 lb and wearing high-heeled shoes momentarily places all her weight on the heel of one foot. If the area of the heel is 0.50 in²., calculate the pressure exerted on the underlying surface.

P = F/A = 135 lb/0.50 in² P = 270 lb/in²

Liquid Pressure

The pressure depends height of liquid column and density of liquid.

Ρ = dgh = density x acceleration due to gravity x

height

Calculating Pressure Exerted by a Liquid

Calculate the height of a mercury column, in feet, that is equal to a column of water that is 110 ft. high.

Known valuesdwater = 1.0 g/cm3 dmercury = 13.6 g/cm3 g = 32 ft/s2

Ρ = dgh Pwater = Pmercury

dgh = dgh

(1.0 g/cm3)(32ft/s2)(110 ft) = (13.6 g/cm3)(32 ft/s2)(h) h = 8.1 ft. Hg

Barometer• Barometer is a device

used to measure the pressure exerted by the atmosphere.

• Height of mercury varies with atmospheric conditions and with altitude.

Aneroid barometer

Mercury Barometer

Measurement of Gas Pressure

Standard Atmosphere (atm)

• Pressure exerted by a mercury column of exactly 760 mm in height

• the density of Hg = 13.5951 g/cm3 (0oC)

• The acceleration due to gravity (g) is 9.80665 m/s2 exactly.

1 atm =

• 760 mmHg• 760 Torr• 14.696 lb/in2

• 101.325 kPa

Check textbook for more values

Converting Pressure to an Equivalent Pressure

A gas is at a pressure of 1.50 atm. Convert this pressure to

a. Kilopascals 1 atm = 101.325 kPa

1.50 atm x 101.325 kPa = 152 kPa 1 atmb. mmHg

1.50 atm x 760 mmHg = 1140 mmHg 1 atm

Manometers

• Used to compare the gas pressure with the barometric pressure.

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Types of Manometers

• Closed-end manometer The gas pressure is equal to the

difference in height (Dh) of the mercury column in the two arms of the manometer

Closed-end Manometer

Open-end Manometer

The difference in mercury levels (Dh) between the two arms of the manometer gives the difference between barometric pressure and the gas pressure

Three Possible Relationships

1. Heights of mercury in both columns are equal if gas pressure and atmospheric pressure are equal.

Pgas = Pbar

2. Gas pressure is greater than the barometric pressure.

∆P > 0

Pgas = Pbar + ∆P

3. Gas pressure is less than the barometric pressure.

∆P < 0

Pgas = Pbar + ∆P

Standard Temperature & Pressure (STP)

Temperature = 0ºC Pressure = 1 atm

The Simple Gas LawsThe Simple Gas LawsThe Simple Gas LawsThe Simple Gas LawsBoyle’s LawBoyle’s LawCharles’ LawCharles’ Law

Gay-Lussac’s LawGay-Lussac’s LawCombined Gas LawCombined Gas Law

Temperature & Gas LawsThree temperature scales Fahrenheit (ºF) Celsius (ºC) Kelvin K (no degree symbol)

Always use K when performing calculations with the gas law equations.

K = (ºC) + 273 Example: ºC = 20º K = 293 Next

Absolute Zero of Temperature

Temperature at which the volume of a hypothetical gas becomes zero

-273.14 ºC Next

Absolute or Kelvin Scale

Temperature scale that has -273.15 ºC as its zero.

Temperature interval of one Kelvin equals one degree celsius.

1 K = 1 ºC

Boyle’s Law

For a fixed amount of gas at constant temperature, gas volume is inversely proportional to gas pressure.

P1V1 = P2V2

Charles’ Law

The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature.

V1 = V2

T1 T2

or V1T2 = V2T1

Example . Charles’ Law

A 4.50-L sample of gas is warmed at constant pressure from 300 K to

350 K. What will its final volume be?

Given: V1 = 4.50 L T1 = 300. K T2 = 350. K V2 = ?

Equation: V1 = V2

T1 T2

or V1T2 = V2T1

(4.50 L)(350. K) = V2 (300. K)

V2 = 5.25 L

Gay-Lussac’s Law

The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant.

P1 = P2

T1 T2

or P1T2 = P2T1

On the next slide (43)

The amount of gas and its volume are the same in either case, but if the gas in the ice bath (0 ºC) exerts a pressure of 1 atm, the gas in the boiling-water bath (100 ºC) exerts a pressure of 1.37 atm. The frequency and the force of the molecular collisions with the container walls are greater at the higher temperature.

Combined Gas Law

Pressure and volume are inversely proportional to each other and directly proportional to temperature.

P1V1 = P2V2

T1 T2

or P1V1T2 = P2V2T1

Example. Combined Gas Law

A sample of gas is pumped from a 12.0 L vessel at

27ºC and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is the final pressure?

Given:P1 = 760 Torr P2 = ?V1 = 12.0 L V2 = 3.5 LT1 = 300 K T2 = 325 K

Equation:

P1V1 = P2V2 T1 T2

or P1V1T2 = P2V2T1

(760 Torr)(12.0 L)(325 K) = ( P2)(3.5 L)(300 K)

P2 = 2.8 x 10³ Torr

Law of Combining Law of Combining VolumesVolumes

Law of Combining Law of Combining VolumesVolumes

Gay-Lussac’s LawGay-Lussac’s Law

Law of Combining Volumes

Gases react in volumes that are related as small whole numbers.

The small whole numbers are the stoichiometric coefficients.

Avogadro’s LawAvogadro’s LawAvogadro’s LawAvogadro’s Law

Volume & MolesVolume & Moles

Avogadro’s Explanation of Gay-Lussac’s Law

When the gases are measured at the same temperature and pressure, each of the identical flasks contains the same number of molecules. Notice how the combining ratio: 2 volumes H2 to 1 volume O2 to 2 volumes H2O leads to a result in which all the atoms present initially are accounted for in the product.

Avogadro’s Hypothesis

Different gases compared at same temperature and pressure

a. Equal - equal number of Volumes molecules

b. Equal number of – equal moleculess volumes

Avogadro’s LawAt a fixed temperature and pressure, the

volume of a gas is directly proportional to the amount of gas.

V = c · nV = volume c = constant n= # of moles

Doubling the number of moles will cause the volume to double if T and P are constant.

More STP Values For Gases

• 1 mol of a gas = 22.4 L

• Number of molecules contained in 22.4 L of a gas is 6.022 x 1023

A Molar Volume of Gas

Source: Photo by James Scherer. © Houghton Mifflin Company. All rights reserved.

Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law

Equation

Includes all four gas variables:• Volume• Pressure• Temperature• Amount of gas Next

PV = nRT• Gas that obeys this equation if said to be an ideal gas (or perfect gas).

• No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperatures.

• Ideal gas constant, R, is R = PV nT = 1 atm x 22.4 L 1 mol x 273.15 K

R = 0.082058 L·atm/mol· K

Applications of the Applications of the Ideal Gas LawIdeal Gas Law

Applications of the Applications of the Ideal Gas LawIdeal Gas Law

Molar MassMolar MassDensityDensity

Molar Mass Determination

• n = number of moles

• Moles = mass of sample m Molar mass M

Ideal Gas Equation: n = PV RT Substituting: m = PV M RT

M = mRT PV

Example: Molar MassThe density of carbon tetrachloride vapor at 714 torr and

125ºC is 4.43 g/L. What is its molar mass?

M = dRT P = (4.43 g/L)(0.0821 L-atm/mol-k)(398 K) (714 torr x 1 atm/760 torr)

M = 154 g/mol

Gas Density (d)

• d = m/V so V = m/d

• Ideal Gas Equation: PV = nRT Substituting P m = mRT d M

“m” cancels out d = PM RT

Finding the Vapor Density of a Substance

Example. Vapor Density

The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressue is 1.6 Earth atm. Assuming ideal behavior, calcuate the density of Titan’s atmosphere.

d = PM RT

= (1.6)(1 atm)(28.6 g/mol) (0.0821 L-atm/mol-K)(95 K)

d = 5.9 g/L

Gases in Chemical Gases in Chemical ReactionsReactions

Gases in Chemical Gases in Chemical ReactionsReactions

Ideal Gas Law Ideal Gas Law & &

Balanced Chemical EquationBalanced Chemical Equation

Example. Reaction Stoichiometry

How many moles of nitric acid can be prepared using 450 L of NO2 at a pressure of 5.00 atm and a temperature of 295 K?

Step 1: Use Ideal Gas law equation n = PV (5.00 atm)(450 L) RT (0.0821 L-atm/mol-K)(295 K)

n = 92.9 mol Next ---- >

Step 2 3NO2 (g) + H2O --- > 2HNO3(aq) + NO(g)

92.9 mol NO2 x 2 mol HNO3

3 mol NO2

= 61.9 mol HNO3

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

Mixture of GasesMixture of Gases

Total Pressure

The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture.

Ptotal = PA + PB + ……

Total Pressure: Mixture of Gases

Total Volume

Vtotal = VA + VB + …….

The expression for percent by volume

(VA / Vtotal) x 100 %

Total Volume: Mixture of Gases

Mole Fraction of Compound in Mixture

• Fraction of all molecules in the mixture contributed by that component.

• Sum of all the mole fractions in a mixture is 1.

• Expression for mole fraction of a substance in terms of P and V

na = Pa = Va

ntot Ptot Vtot

Total Volume: Mixture of Gases

Example: Gas Mixtures & Partial Pressure

A gaseous mixture made from 6.00 g O2 and 9.00 g CH4 is placed in a 15.0 L vessel at 0ºC. What is the partial pressure of each gas, and what is the total pressure in the vessel?

Step 1: nO2 = 6.00 g O2 x 1 mol O2

32 g O2

= 0.188 mol O2

Next ---- >

nCH4 = 9.00 g CH4 x 1 mol CH4

16.0 g CH4

= 0.563 mol CH4

Step 2: Calculate pressure exerted by each

PO2 = nRT V = (0.188 mol O2)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.281 atm

PCH4 = (0.563 mol)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.841 atm

Step 3: Add pressures

Ptotal = PO2 + PCH4

= 0.281 atm + 0.841 atm

Ptotal = 1.122 atm

Mole fraction: na/ntotal

nO2 = 0.188 mol (0.188 + 0.563) mol

nO2 = 0.250

nCH4 = 0.563 mol (0.188 + 0.563) mol

nCH4 = 0.750

0.250 + 0.750 = 1.000

Volume of each Gas

PO2 = VO2

Ptot Vtot

0.281 atm = VO2

(0.281 + 0.841) atm = 15 L

VO2 = 3.8 L

VCH4 = 15 L – 3.8 L = 11 L

Gas Collected Over Water Wet Gas: mixture of the desired gas and

water vapor

Pbar = Pgas + Pwater

Pgas = Pbar – Pwater

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Pbar = Pgas + Pwater

Pgas = Pbar - Pwater

Gas Collected Over WaterA sample of KClO3 is partially

decomposed, producing O2 gas that is collected over water. The volume of gas collected is 0.250 L at 26ºC and 765 torr total pressure.

2KClO3(s) ----- > 2 KCl(s) + 3O2(g)

Next ---- >

a. How many moles of O2 are collected?

Step 1: Calculate pressure of O2 in mixture

PO2 = Ptot – Pwater vapor

= 765 torr – 25 torr

PO2 = 740 torr

Next ---- >

nO2 = PV

RT

= (740 torr)(1atm/760 torr)(0.250 L) (0.0821 L-atm/mol-K)(299 K)

nO2 = 9.92 x 10-3 mol

Next ----- >

b. How many grams of KClO3 were decomposed?

(9.92 x 10-3 mol O2) x 2 mol KClO3 x 123 g KCLO3

3 mol O2 1 mol KClO3

= 0.811 g KClO3

c. When dry, what volume would the collected O2 gas occupy at the same temperature and pressure?

Remove water vapor. P2 = 765 torr Temperature is the same.

P1V1 = P2V2

V2 = (740 torr)(0.250 L) (765 torr)

V2 = 0.242 L

Kinetic-Molecular Kinetic-Molecular Theory of GasesTheory of Gases

Kinetic-Molecular Kinetic-Molecular Theory of GasesTheory of Gases

Assumptions

• The molecules of gases are in rapid random motion.

• Their average velocities (speeds) are proportional to the absolute (Kelvin) temperature.

• At the same temperature, the average kinetic energies of the molecules of different gases are equal.

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Relationship Between Temperature and Average

Kinetic Energy

Higher temperature means greater motion

(KE)av = 3 RT 2

Obtained from PV = RT = 2 (KE)av

n 3

PV = RT = 2 (KE)av = ½ mv² n 3

Apply to all gases whether alone or mixed with other gases

1. Constant V a. T changes – KE changes b. T increases – pressure increases c. KE changes – speed of molecules changes

2. Constant temperature a. V increases, P decreases b. KE doesn’t change – T didn’t change c. Speed of molecules doesn’t change-KE didn’t change

Root-Mean Square Speed (u 2)

• u 2 represents the average of the squares of the particles velocities.• The square root of u 2 is called the root mean square velocity. • The unit of measurement is m/s.

• u 2 = 3RT Units: R = 8.3145 J K-1mol-1

M or 8.3145 kgm2/s2 K-1mol-1

Joule = kg·m2/s2

Joule is a unit of measurement for energy.

Example

How is the root-mean square speed (rms) of F2 molecules in a gas sample changed by

a) An increase in temperature.b) An increase in volumec) Mixing with a sample of Ar at the same

temperature

Answers

a. Increasesb. No effectc. No effect

Example. Root-Mean Square Speed

Calculate the root mean square speed for the atoms in a sample of helium gas at 25ºC.

Given:T = 25ºC = 298 K

Known:R = 8.3145 J/K·mol or 8.3145 kgm2/s2 / K·mol

MHe : Change g to kg 4.00 g/mol = 4.00 x 10-3 kg/mol

Equation: u 2 = 3RT M

u 2= 3 (8.3145 kgm2/s2 /K·mol)(298K)

4.00 x 10-3 kg/mol

u = 1.36 x 103 m/s

Effusion

The escape of gas molecules from their container through a tiny orifice or pin hole.

Model of Gaseous Effusion

Diffusion

Diffusion is the moving or mixing of molecules of different substances as a result of random molecular motion.

Factors That Influence Effusion and Diffusion

• The greater the temperature the faster the ga will move or vise versa/

• If the mass increases, the kinetic energy will also increase if the speed remains constant.

• If the mass increases, the speed decreases if the kinetic energy remains constant.

Example. Rate of Effusion

Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process fuel for nuclear reactors.

Known:Molar Masses H2 = 2.016 g/mol UF6 = 352.02 g/mol

(Rate of effusion)² = MU compound

MH gas

= 352.02 2.016

Rate of effusion = 13.21

Graham’s Law

• The rate of effusion (or diffusion) of two different gases are inversely proportional to the square roots of their molar masses.

• (rate of effusion of A)2 = MB

• (rate of effusion of B)2 MA

Ratios of Rates the square root of two molar

masses is also equal to the ratio

• Molecular speeds• Effusion times• Distance traveled by molecules• Amount of gas effused

Based on Assumptions of Kinetic Molecular Theory

• Rates of diffusion of different gases are inversely proportional to the square roots of their densities.

• Rates of diffusion are inversely proportional to the square roots of their molar masses.

Nonideal (Real) GasesNonideal (Real) GasesNonideal (Real) GasesNonideal (Real) Gases

Ideal vs. Nonideal (Real) Gases

Ideal

• High pressure- compressible, volume approaches zero

• Force of collisions with container wall is great

Nonideal

• High pressure – molecules are practically incompressible

• Force of collision with container wall is less due to attractive force among the molecules.

Behavior of Gases

Behave ideally at

• High temperatures

• Low pressures

Behave nonideally at

• Low temperatures• High pressures

van der Waals van der Waals EquationEquation

van der Waals van der Waals EquationEquation

van der Waals EquationEquation corrects for volume and intermolecular forces

(P + n²a/V²)(V-nb) = nRT

• n²a/V² = related to intermolecular forces of attraction

• n²a/V² is added to P = measured pressure is lower than expected

• a & b have specific values for particular gases

• V - nb = free volume within the gas

Intermolecular Force of Attraction

Attractive forces of the orange molecules for the purple molecule cause the purple molecule to exert less force when it collides with the wall than it would if these attractive forces did not exist.

Source of Sketches/problems

• http://cwx.prenhall.com/bookbind/pubbooks/hillchem3/medialib/media_portfolio/04.html

• Chemistry, 7th ed. Brown, LeMay & Bursten, Prentice Hall