Chapter 10Chapter 10GasesGases

CHEMISTRY The Central Science

9th Edition

David P. White

• OCTOBER 5, 2010

• HOMEWORK PRESSURE CONVERSIONS

• P 433 Q 9, 17, 21

• READ SECTION 10.3 P 404

• Gases are highly compressible and occupy the full volume of their containers.

• When a gas is subjected to pressure, its volume decreases.

• Gases always form homogeneous mixtures with other gases.

• Gas molecules only occupy about 0.1 % of the volume of their containers.

• Have extremely low densities.

Characteristics of GasesCharacteristics of Gases

• Pressure is the force acting on an object per unit area:

• Gravity exerts a force on the earth’s atmosphere• A column of air 1 m2 in cross section exerts a force of 105

N.• The pressure of a 1 m2 column of air is 100 kPa.

PressurePressure

AF

P

Atmospheric pressure is the weight of air per unit of area.

F= m . aF= m . amass of air column = 10,000 kgmass of air column = 10,000 kg

acceleration due to gravity at the acceleration due to gravity at the surface of Earth = 9.8 m/s^2surface of Earth = 9.8 m/s^2

Atmosphere Pressure and the Barometer

Atmosphere Pressure and the Barometer• SI Units: 1 N = 1 kg.m/s2; 1 Pa = 1 N/m2.• Atmospheric pressure is measured with a barometer.• If a vacated tube is inserted into a container of mercury

open to the atmosphere, the mercury will rise 760 mm up the tube.

• Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.

Standard PressureStandard Pressure

• It is equal to– 1.00 atm

– 760 torr (760 mm Hg)– 101.325 kPa

•Normal atmospheric pressure at sea level.

Units of PressureUnits of Pressure

• 1 atm = 760 mmHg = 760 torr =

• 1.01325 10^5 Pa = 101.325 kPa.• Pascals

– 1 Pa = 1 N/m2

• Bar– 1 bar = 105 Pa = 100 kPa

Examples – Convert the following pressures:

1. 658.2 mm Hg to kPa

2. 1.85 atm to torr

3. 337.3 kPa to atm

Examples – Convert the following pressures:

1. 658.2 mm Hg to kPa 87.75 kPa

2. 1.85 atm to torr 1410 torr

3. 337.3 kPa to atm 3.329 atm

ManometerManometer

Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Pressure and the Manometer• The pressures of gases not open

to the atmosphere are measured in manometers.

• A manometer consists of a bulb of gas attached to a U-tube containing Hg:– If Pgas < Patm then Pgas = Patm.- Ph

– If Pgas > Patm then Pgas = Patm + Ph.

Example – The mercury in a manometer is 46 mm higher on the open end than on the gas bulb end. If atmospheric pressure is 102.2 kPa, what is the pressure of the gas in the bulb?

Example – The mercury in a manometer is 46 mm higher on the open end than on the gas bulb end. If atmospheric pressure is 102.2 kPa, what is the pressure of the gas in the bulb?813 mm Hg108 kPa

The Pressure-Volume Relationship: Boyle’s Law• Weather balloons are used as a practical consequence to

the relationship between pressure and volume of a gas.• As the weather balloon ascends, the volume increases.• As the weather balloon gets further from the earth’s

surface, the atmospheric pressure decreases.• Boyle’s Law: the volume of a fixed quantity of gas is

inversely proportional to its pressure (assuming all other variables are unchanged).

• Boyle used a manometer to carry out the experiment.

The Gas LawsThe Gas Laws

Boyle’s LawBoyle’s Law

The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

The Pressure-Volume Relationship: Boyle’s Law• Mathematically:

• A plot of V versus P is a hyperbola. T constant.• Similarly, a plot of V versus 1/P must be a straight line

passing through the origin.

PV

1constant constantPV

2211 VPVP

As As PP and and VV are areinversely proportionalinversely proportional

A plot of V versus P results in a curve.

Since

V = k (1/P)This means a plot of V versus 1/P will be a straight line.

PV = k

Examples

1. A gas that occupies 2.84 L has a pressure of 88.6 kPa. What would be the pressure of the gas sample if it only occupied 1.66 L (assuming the same temperature)?

2. A 10.0-L sample of argon gas has a pressure of 0.885 atm. At what volume would the sample have a pressure of 6.72 atm?

Examples

1. A gas that occupies 2.84 L has a pressure of 88.6 kPa. What would be the pressure of the gas sample if it only occupied 1.66 L (assuming the same temperature)?

152 kPa

2. A 10.0-L sample of argon gas has a pressure of 0.885 atm. At what volume would the sample have a pressure of 6.72 atm?

1.32 L

OCTOBER 6OCTOBER 6GAS LAWS –Section 3-4GAS LAWS –Section 3-4

• Boyle’s Law• Charles’s Law• Avogadro’s Law • The ideal gas equation• HOMEWORK : PAGE 434 10.23 TO 10.41• 41 IS A CHALLENGE QUESTION

• ODD ONLY

The Temperature-Volume Relationship: Charles’s Law

• Charles’s Law: the volume of a fixed quantity of gas at constant pressure increases as the temperature increases (assuming pressure constant).

• If we express T in Kelvin degrees, P constant

TV constant constantTV

2

2

1

1

T

V

T

V

• A plot of V versus T is a straight line.• When T is measured in C, the intercept on the

temperature axis is -273.15C. • We define absolute zero, 0 K = -273.15C.

• Note the amount of gas and pressure remain constant.

Charles’s LawCharles’s Law

• The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

A plot of V versus T will be a straight line.

• i.e., VT

= k

Examples (remember to change T to K to use Charles’ Law!!!)

1. A gas sample occupies a volume of 1.89 L at 25°C. What would be the volume of the sample at 75°C?

2. A sample of carbon dioxide in a 5.00-L container has a temperature of 56.9°C. At what temperature will this sample of CO2 occupy a volume of 3.00 L?

Examples

1. A gas sample occupies a volume of 1.89 L at 25°C. What would be the volume of the sample at 75°C?

2.21 L

2. A sample of carbon dioxide in a 5.00-L container has a temperature of 56.9°C. At what temperature will this sample of CO2 occupy a volume of 3.00 L?

198 K or -75°C

The Quantity-Volume Relationship: Avogadro’s Law

• Gay-Lussac’s Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers.

• Avogadro’s Hypothesis: equal volumes of gas at the same temperature and pressure will contain the same number of molecules.

• Avogadro’s Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.

• Mathematically:

• We can show that 22.4 L of any gas at 0C contain 6.02 1023 gas molecules.

nV constant2

2

1

1

V

n

V

n

• Consider the three gas laws.

• We can combine these into a general gas law:

The Ideal Gas EquationThe Ideal Gas Equation

), (constant 1

TnP

V

), (constant PnTV ),(constant TPnV

• Boyle’s Law:

• Charles’s Law:

• Avogadro’s Law:

PnT

V

• If R is the constant of proportionality (called the gas constant), then

• The ideal gas equation is:

• R = 0.08206 L·atm/mol·K = 8.314 J/mol·K

The Ideal Gas EquationThe Ideal Gas Equation

PnT

RV

nRTPV

• We define STP (standard temperature and pressure) = 0C, 273.15 K, 1 atm.

• Volume of 1 mol of gas at STP is:

L 41.22

atm 000.1K 15.273KL·atm/mol· 0.08206mol 1

PnRT

V

nRTPV

Examples

1. What is the volume of a 15.0-g sample of neon gas at 130.0 kPa and 39°C?

14.8 L

2. At what temperature will 2.85 moles of hydrogen gas occupy a volume of 0.58 L at 1.00 atm?

2.5 K or -270.7°C

Relating the Ideal-Gas Equation and the Gas Laws• If PV = nRT and n and T are constant, then PV = constant

and we have Boyle’s law.• Other laws can be generated similarly.• In general, if we have a gas under two sets of conditions,

then

22

22

11

11TnVP

TnVP

Examples

1. A gas sample occupies a volume of 0.685 L at 38°C and 0.775 atm. What will be the temperature of the sample if it occupies 0.125 L at 1.25 atm?

91.6 K or -181.6°C

2. A sample of nitrogen gas in a 2.00 L container has a pressure of 1800.6 mmHg at 25°C. What would be the pressure on the sample in a 1.00 L container at 125°C?

4810 mm Hg

OCTOBER 7OCTOBER 7Section 10-5Section 10-5

• Ideal Gas Equation Problems• Gas Densities and Molar Mass

Volumes of gases in chemical reactions

HW • 38 ideal gas equation

43-45-46 concepts

47-49-50 density and molar mass

52-54 to 56 Stoichiometry with gases

Gas Densities and Molar Mass• Density has units of mass over volume. • Rearranging the ideal-gas equation with M as molar mass

we get

Further Applications of the Further Applications of the Ideal-Gas EquationIdeal-Gas Equation

RTP

dV

nRTP

Vn

nRTPV

MM

Examples

1. What is the density of nitrogen gas at 35°C and 1.50 atm?

Examples

1. What is the density of nitrogen gas at 35°C and 1.50 atm?

1.67 g/L

• The molar mass of a gas can be determined as follows:

OR

The cat paws throw dirt over the P

Volumes of Gases in Chemical Reactions• The ideal-gas equation relates P, V, and T to number of

moles of gas.• The n can then be used in stoichiometric calculations.

PdRTM

PV

mRTM

Examples

1. 4.83 g of a gas occupy a 1.50 L flask at 25°C and 112.4 kPa. What is the molar mass of the gas?

71.0 g/mol

STOICHIOMETRY STOICHIOMETRY PROBLEMSPROBLEMS

• If an air bag has a volume of 36 L and is filled up with Nitrogen at a pressure of 1.15 atm at a temperature of 26C, how many grams of NaN3 must be decomposed

• 2 Na N3 (s) 2 Na (s) + 3 N2

(73.2 g)

• How many L of NH3 at 850 0C and 5 atm are required to react wih 32 g of O2?

• 4NH3+ 5 O2 ->4NO + 6 H2 O

• (14.7 L )

October 8October 8Section 10-6Section 10-6

• Gas Mixtures and Partial Pressures• Partial Pressures and Mole Fraction• Collecting Gases over Water

• HW page 436 Q 55 and 56 collecting gas over water• 10:57 to 67 odd numbers (red)

• Since gas molecules are so far apart, we can assume they behave independently.

• Dalton’s Law: in a gas mixture the total pressure is given by the sum of partial pressures of each component:

• Each gas obeys the ideal gas equation:

Gas Mixtures and Partial Gas Mixtures and Partial PressuresPressures

321total PPPP

VRT

nP ii

• Combining the equations

Partial Pressures and Mole Fractions

• Let ni be the number of moles of gas i exerting a partial pressure Pi, then

where i is the mole fraction (ni/nt).

VRT

nnnP 321total

totalPP ii

Examples:

1. Hydrogen gas is added to a 2.00-L flask at a pressure of 5.6 atm. Oxygen gas is added until the total pressure in the flask measures 8.4 atm. What is the mole fraction of hydrogen in the flask? What the pp of Oxygen?

2 .35 moles of argon and 2.75 moles of neon are placed in a 15.0-L tank at 35°C. What is the total pressure in the flask? What is the pressure exerted by neon?

Examples:

1. Hydrogen gas is added to a 2.00-L flask at a pressure of 5.6 atm. Oxygen gas is added until the total pressure in the flask measures 8.4 atm. What is the mole fraction of hydrogen in the flask?

0.67 2.8 atm

2. 1.35 moles of argon and 2.75 moles of neon are placed in a 15.0-L tank at 35°C. What is the total pressure in the flask? What is the pressure exerted by neon? Ptot = 6.91 atm

PNe = 4.63 atm

Collecting Gases over Water

• It is common to synthesize gases and collect them by displacing a volume of water.

• To calculate the amount of gas produced, we need to correct for the partial pressure of the water:

watergastotal PPP

Collecting Gases over Water

Example - 150.82 mL of an unknown gas is collected over water at 27°C and 1.032 atm. The mass of the gas is 0.1644 g. What is the molar mass of the gas? The vapor pressure of water at 27°C is 26.74 torr

OCTOBER 12OCTOBER 12Section 10-7 and 10-8Section 10-7 and 10-8

Kinetic-Molecular Theory

Application to the Gas Laws

Molecular Effusion – Graham’s Law

Diffusion and Mean Free Path

HW P 437 69 to 79 odd only and 80

LABORATORY BOOK – MUST GET BY WED

FOR THURSDAY MUST STUDY LAB # 8

DETERMINING THE MOLAR MASS OF A GAS

AND DO PRELAB QUESTIONS

PEARSON SITE – DO REGISTER

ONLY 7 PEOPLE REGISTERED

• Theory developed to explain gas behavior.• Theory of moving molecules.• Assumptions:

1.-Gases consist of a large number of molecules in continuous, random motion.

2.-Volume of individual molecules negligible compared to volume of container.

3.-Intermolecular forces (forces between gas molecules) are negligible (no attraction or repulsion between molecules)

Kinetic Molecular TheoryKinetic Molecular Theory

4.- Energy can be transferred between molecules during collisions, but AVERAGE kinetic energy DOES NOT CHANGE at constant temperature (When molecules collide one speeds up the other slows down)

5.-Average kinetic energy of molecules is proportional to the ABSOLUTE temperature. At a given temperature molecules of all gases have same average KE

Kinetic molecular theory gives us an understanding of pressure and temperature on the molecular level.

Pressure of a gas results from the

number of collisions per unit time on the walls of container.

• Magnitude of pressure given by how often and how hard the molecules strike against the walls.

TemperatureTemperature

– MOLECULAR MOTION INCREASES WITH TEMPERATURE

– Molecules at 1 T at a given moment have a wide range of speed

– At higher temperature a larger fraction of molecules have a higher speed.

• Gas molecules have an average kinetic energy.• Each molecule has a different energy.

• If 2 different gases are at same T their molecules have the same Average Kinetic Energy.

• If T increases their motion increases too.• MOLECULAR MOTION INCREASES WITH T

• Individual molecules move at different speeds but the average kinetic energy is one at each temperature.

• When particles collide, the momentum is conserved. That means that one particle will move faster but the other will slow down to conserve the total amount of energy in the system.

• The diagram shows the distribution of molecular speeds for a sample of gas at different temperatures.

• Increasing T increases both the most probable speed ( curve maximum) and rms speed (u) –root mean square-

• At higher T, a larger fraction of molecules move a greater speeds.

• The peak shows the most probable speed ( speed of the largest number of molecules.

• 400m/s is approximately 1000miles/hour

• KE = ½ m x speed2

• Root mean square is an statistical expression that is close to the mean.

• Example , the average between 4 different speeds • V1,v2,v3, v4 vav = ¼ (v1+v2+v3+v4)• u =

• As kinetic energy increases, the velocity of the gas molecules increases.

• Root mean square speed, u, is the speed of a gas molecule having average kinetic energy. – For one molecule

• Average kinetic energy, , is related to root mean square speed:

221 mu

• m is the mass of the molecules• u is the root mean square speed

• T is proportional to the speeds of the particles.

Application to Gas Laws• As volume increases at constant temperature, the average

kinetic energy of the gas remains constant. Therefore, u is constant. However, volume increases so the gas molecules have to travel further to hit the walls of the container. Therefore, pressure decreases.

• If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases.

EffusionEffusion

The escape of gas molecules through a tiny hole.

DiffusionDiffusion

The spread of one substance throughout a space or throughout a second substance.

Spread of a gas.

Molecular Effusion and Diffusion• As kinetic energy increases,(T) the velocity of the gas

molecules increases.• Average kinetic energy of a gas is related to its mass:• (KE) • Consider two gases at the same temperature: both have

same KE therefore the lighter gas molecules have to have higher speeds (rms speed u) than the heavier gas.

• Mathematically: (M is molecular mass)

221 mu

MRT

u3

• The lower the molar mass, M, the higher the speed of the molecules (rms speed).

• Calculate the rms speed for O2 molecules at 25 0 C

Convert mass to kgunits of speed m/s

answer 482 m/s approximately 1100 miles /hour

Graham’s Law of Effusion• As kinetic energy increases,

the velocity of the gas molecules increases.

• Effusion is the escape of a gas through a tiny hole (a balloon will deflate over time due to effusion).

• The rate of effusion can be quantified.

• Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:

• Only those molecules that hit the small hole will escape through it.

• Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.

1

2

2

1MM

rr

• Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:

• Only those molecules that hit the small hole will escape through it.

• Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.

1

2

2

1

2

1

2

13

3

MM

M

M RT

RT

uu

rr

Examples – For each pair of gases, determine which will effuse faster, and by how much it will be faster.

1. CH4 and Xe

2. Cl2 and N2

3. F2 and He

Examples – For each pair of gases, determine which will effuse faster, and by how much it will be faster.

CH4 and Xe 2.8607

1. Cl2 and N2 1.59095

2. F2 and He 3.08027

Diffusion and Mean Free Path • Diffusion of a gas is the spread of the gas through space.• Diffusion is faster for light gas molecules.• Diffusion is significantly slower than rms speed (consider

someone opening a perfume bottle: it takes while to detect the odor but rms speed at 25C is about 1150 mi/hr).

• Diffusion is slowed by gas molecules colliding with each other.

• Average distance of a gas molecule between collisions is called mean free path.

• At sea level, mean free path is about 6 10-6 cm.

• Tetrafluoroethylene C2F4 effuses through a barrier at a rate of 4.6 x 10 -6 mol/h. An unknown gas, consisting only of boron and hydrogen, effuses at a rate

of 5.8 x 10 -6 mol/h under the same conditions. What is the molar mass of the unknown gas?

Use Graham’s Law

M= 63g/mol

OCTOBER 14OCTOBER 14SECTION 10-9SECTION 10-9

• REAL GASES – DEVIATIONS FROM IDEAL BEHAVIOR

• VAN DER WAALS EQUATION

• HW 10.8• 10.81,85,95,97

• From the ideal gas equation, we have

• For 1 mol of gas, PV/RT = 1 for all pressures.• BUT REAL GASES DO NOT BEHAVE IDEALLY

SPECIALLY AT HIGH PRESSURES!!!• In a real gas, PV/RT varies from 1 significantly.• The higher the pressure the more the deviation from ideal

behavior (For P<10 atm we could use ideal-gas equation)

Real Gases: Deviations Real Gases: Deviations from Ideal Behaviorfrom Ideal Behavior

nRTPV

• As the pressure on a gas increases, the molecules are forced closer together.

• As the molecules get closer together, the volume of the gas molecules begin to be significant, and the volume of the gases is greater that what is predicted by the ideal gas equation.

• Also at the higher the pressure, the attraction between gas particles begins to manifest and less particles hit the walls of the container, therefore there is a reduction in the Pressure due to the molecular attractions

• From the ideal gas equation, we have

• As temperature increases, the gases behave more ideally.• The assumptions in kinetic molecular theory show where

ideal gas behavior breaks down:

– the molecules of a gas have finite volume;

– molecules of a gas do attract each other.

nRTPV

• As the gas molecules get closer together, the smaller the intermolecular distance.

• The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.

• Therefore, the less the gas resembles and ideal gas.• As temperature increases, the gas molecules move faster

and further apart.• Also, higher temperatures mean more energy available to

break intermolecular forces.• Then an increase in T helps the gases behave more

ideally

• Therefore, the higher the temperature, the more ideal the gas.

• The T increase helps the molecules break the forces of attractions.

Real Gases: Deviations Real Gases: Deviations from Ideal Behaviorfrom Ideal Behavior

The van der Waals Equation• We add two terms to the ideal gas equation: one to

correct for volume of molecules and the other to correct for intermolecular attractions

• The correction terms generate the van der Waals equation:

where a and b are empirical constants.

2

2

V

annbV

nRTP

The van der Waals Equation

• General form of the van der Waals equation:

Real Gases: Deviations Real Gases: Deviations from Ideal Behaviorfrom Ideal Behavior

2

2

V

annbV

nRTP

nRTnbVV

anP

2

2

Corrects for molecular volume

Corrects for molecular attraction

• The constant b adjusts the volume, since gas particles do have volume the total volume that the particles have to move in is less than the whole container. That is why b is substracting V, and the units are L/mol

( V-n b)

The constant a accounts for the attraction between molecules which increases with the square of the number of molecules per unit of Volume.

Units for a L2 atm/mol2

It indicates HOW STRONGLY THE MOLECULES ATTRACT EACH OTHER!!!

• If 1.00 mol of an ideal gas is confined to a 2.4 L at 0.0 C, it would exert a pressure of 1.00 atm. Use the Van Der Waals equation and the contants for Cl2 to estimate the pressure exerted by 1 mol of Cl2 in 22.4 L at 0.0 C

• Monday October 18 Test on unit 10. One period – 2 problems and multiple choice.

• Friday October 15. Experiment # 8• Determining the Molar Volume of a Gas.• MUST BRING LAB NOTEBOOK (a marble book)

AND LAB BOOK• KNOW THE PROCEDURE!!! You have to know what

are you going to do BEFORE walking into the lab room.