Properties of Gases...•We measure air pressure with a barometer •Column of mercury supported by air pressure •Force of the air on the surface of the mercury counter balances

Copyright © 2011 Pearson Education, Inc.

Properties of Gases

Mr. Matthew Totaro Legacy High School

AP Chemistry


The Structure of a Gas • Gases are composed of particles

that are flying around very fast in their container(s)

• The particles in straight lines until they encounter either the container wall or another particle, then they bounce off

• If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there

http://ch301.cm.utexas.edu/simulations/gas-laws/GasLawSimulator.swf


Gases Pushing • Gas molecules are constantly in

motion • As they move and strike a surface,

they push on that surface push = force

• If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting pressure = force per unit area


The Effect of Gas Pressure •The pressure exerted by a gas can cause

some amazing and startling effects •Whenever there is a pressure difference, a gas

will flow from an area of high pressure to an area of low pressure the bigger the difference in pressure, the stronger

the flow of the gas • If there is something in the gas’s path, the gas

will try to push it along as the gas flows


Air Pressure •The atmosphere exerts a

pressure on everything it contacts the atmosphere goes

up about 370 miles, but 80% is in the first 10 miles from the earth’s surface

•This is the same pressure that a column of water would exert if it were about 10.3 m high


Atmospheric Pressure Effects •Differences in air pressure

result in weather and wind patterns

•The higher in the atmosphere you climb, the lower the atmospheric pressure is around you at the surface the

atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi


Pressure Imbalance in the Ear

If there is a difference in pressure across

the eardrum membrane, the membrane will be

pushed out – what we commonly call a

“popped eardrum”


The Pressure of a Gas •Gas pressure is a result of the

constant movement of the gas molecules and their collisions with the surfaces around them

•The pressure of a gas depends on several factors number of gas particles in a given

volume volume of the container average speed of the gas

particles


Measuring Air Pressure

gravity

• We measure air pressure with a barometer

• Column of mercury supported by air pressure

• Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury


1. The height of the column increases because atmospheric pressure decreases with increasing altitude

2. The height of the column decreases because atmospheric pressure decreases with increasing altitude

3. The height of the column decreases because atmospheric pressure increases with increasing altitude

4. The height of the column increases because atmospheric pressure increases with increasing altitude

1. The height of the column increases because atmospheric pressure decreases with increasing altitude

2. The height of the column decreases because atmospheric pressure decreases with increasing altitude

3. The height of the column decreases because atmospheric pressure increases with increasing altitude

4. The height of the column increases because atmospheric pressure increases with increasing altitude

Practice – What happens to the height of the column of mercury in a mercury barometer as you climb to the top

of a mountain?

10


Common Units of Pressure


Example: A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?

because mmHg are smaller than psi, the answer makes sense

1 atm = 14.7 psi, 1 atm = 760 mmHg

132 psi mmHg

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

psi atm mmHg


Practice—Convert 45.5 psi into kPa

because kPa are smaller than psi, the answer makes sense

1 atm = 14.7 psi, 1 atm = 101.325 kPa

645.5 psi kPa

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

psi atm kPa


Manometers • The pressure of a gas trapped in a container can be

measured with an instrument called a manometer • Manometers are U-shaped tubes, partially filled with

a liquid, connected to the gas sample on one side and open to the air on the other

• A competition is established between the pressures of the atmosphere and the gas

• The difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere


Manometer

for this sample, the gas has a larger pressure than the atmosphere, so


Boyle’s Law Robert Boyle (1627–1691)

•Pressure of a gas is inversely proportional to its volume constant T and amount of gas graph P vs V is curve graph P vs 1/V is straight line

•As P increases, V decreases by the same factor

•P x V = constant •P1 x V1 = P2 x V2


Boyle’s Experiment • Added Hg to a J-tube

with air trapped inside • Used length of air column

as a measure of volume




Boyle’s Experiment, P x V


Boyle’s Law: A Molecular View • Pressure is caused by the molecules striking the sides

of the container • When you decrease the volume of the container with

the same number of molecules in the container, more molecules will hit the wall at the same instant

• This results in increasing the pressure


Boyle’s Law and Diving

Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you. This allows you to take in air even when the outside pressure is large.

• Because water is more dense than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm at 20 m the total pressure

is 3 atm • If your tank contained air

at 1 atm of pressure, you would not be able to inhale it into your lungs you can only generate

enough force to overcome about 1.06 atm


Boyle’s Law and Diving • If a diver holds her breath and rises to the

surface quickly, the outside pressure drops to 1 atm

• According to Boyle’s law, what should happen to the volume of air in the lungs?

• Because the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs.

• Always Exhale When Rising!!


P1 ∙ V1 = P2 ∙ V2

Example: A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?

because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, P1, P2 V2


Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon

is now 2.78 x 103 mL, what was it originally?


P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)

A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78x

103 mL, what was it originally?

because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm

V1, mL

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V2, P1, P2 V1


Charles’s Law Jacques Charles (1746–1823)

•Volume is directly proportional to temperature constant P and amount of gas graph of V vs. T is straight line

•As T increases, V also increases •Kelvin T = Celsius T + 273 •V = constant x T if T measured in Kelvin


If the lines are extrapolated back to a volume of “0,” they all show the same temperature, −273.15 °C, called absolute zero

If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line


•The pressure of gas inside and outside the balloon are the same

•At high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger

•The pressure of gas inside and outside the balloon are the same

•At low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small

Charles’s Law – A Molecular View


Example: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature at 2.80 L?

because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C

t1, K and °C

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, T2 T1


Practice – The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was 10.0 L, what is the

volume of hot air?


The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was 10.0 L, what is the

volume of hot air?

when the temperature increases, the volume should increase, and it does

V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, T1, T2 V2

Tro: Chemistry: A Molecular Approach, 2/e


Avogadro’s Law Amedeo Avogadro (1776–1856)

•Volume directly proportional to the number of gas molecules V = constant x n constant P and T more gas molecules = larger

volume •Count number of gas

molecules by moles •Equal volumes of gases

contain equal numbers of molecules the gas doesn’t matter


Example:A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?

because n and V are directly proportional, when the volume increases, the moles should increase, and they do

V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol

n2, and added moles

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, n1 n2


Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy?


Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy?

because n and V are directly proportional, when the moles decrease, the volume should decrease, and it does

V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

V2

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, n1, n2 V2


• By combining the gas laws we can write a general equation • R is called the gas constant • The value of R depends on the units of P and V

we will use 0.08206 and convert P to atm and V to L

• The other gas laws are found in the ideal gas law if two variables are kept constant

• Allows us to find one of the variables if we know the other three

Ideal Gas Law


Example: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?

1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

V = 3.24 L, P = 24.3 psi, t = 25 °C n, mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given: Find:

P, V, T, R n



Standard Conditions •Because the volume of a gas varies with pressure

and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions STP

•Standard pressure = 1 atm •Standard temperature = 273 K 0 °C


Practice – A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?


A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?

1 mole at STP occupies 22.4 L, because there is more than 1 mole, we expect more than 22.4 L of gas

V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P1, V1, T1, R n P2, n, T2, R V2


Practice — Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C



Practice—Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C.

mSO2 = 637 g, P = 6.08 x 104 mmHg, t = −23 °C,

V, L

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, n, T, R V g n



Molar Volume •Solving the ideal gas equation for the volume of

1 mol of gas at STP gives 22.4 L 6.022 x 1023 molecules of gas notice: the gas is immaterial

•We call the volume of 1 mole of gas at STP the molar volume it is important to recognize that one mole measures of

different gases have different masses, even though they have the same volume


Molar Volume


Practice — How many liters of O2 @ STP can be made from the decomposition of 100.0 g of PbO2?

2 PbO2(s) → 2 PbO(s) + O2(g) (PbO2 = 239.2, O2 = 32.00)


Practice — How many liters of O2 @ STP can be made from the decomposition of 100.0 g of PbO2?

2 PbO2(s) → 2 PbO(s) + O2(g)

because less than 1 mole PbO2, and ½ moles of O2 as PbO2, the number makes sense

1 mol O2 = 22.4 L, 1 mol PbO2 = 239.2g, 1 mol O2 ≡ 2 mol PbO2

100.0 g PbO2, 2 PbO2 → 2 PbO + O2 L O2

Check:

Solution:

Conceptual Plan:

Relationships:

Given: Find:

L O2 mol PbO2 g PbO2 mol O2


Density at Standard Conditions

•Density is the ratio of mass to volume •Density of a gas is generally given in g/L •The mass of 1 mole = molar mass •The volume of 1 mole at STP = 22.4 L


Practice – Calculate the density of N2(g) at STP


Practice – Calculate the density of N2(g) at STP

the units and number are reasonable

N2, dN2, g/L

Check:

Solution:

Conceptual Plan:

Relationships:

Given: Find:

MM d



Gas Density

•molar mass = d x R x T P


Example: Calculate the density of N2 at 125°C and 755 mmHg

because the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is

P = 755 mmHg, t = 125 °C, dN2, g/L

Check:

Solution:

Conceptual Plan:

Relationships:

Given: Find:

P, MM, T, R d


Practice — Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g


m=9.988g, n=0.250 mol, P=775 mmHg, t=27 °C,

density, g/L

m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300.K

density, g/L

Solution:

Practice — Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g

the unit is correct and the value is reasonable Check:

Conceptual Plan:

Relationships:

Given:

Find:

V, m d P, n, T, R V


T(K) = 55°C + 273.15T = 328 K

m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,

molar mass, g/mol

m=0.311g, V=0.225 L, P=1.1658 atm, T=328 K,

molar mass, g/mol

Example: Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg

the unit is correct, the value is reasonable Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

n, m MM P, V, T, R n

55


Practice — What is the molar mass of a gas if 12.0 g occupies 197 L at 3.80 x 102 torr and 127 °C?



m=12.0 g, V= 197 L, P=380 torr, t=127°C,

molar mass, g/mol

m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,

molar mass, g/mol

Practice — What is the molar mass of a gas if 12.0 g occupies 197 L at 380 torr and 127 °C?

the unit is correct and the value is reasonable Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

n, m MM P, V, T, R n


Mixtures of Gases • When gases are mixed together, their molecules behave

independent of each other all the gases in the mixture have the same volume

all completely fill the container ∴ each gas’s volume = the volume of the container

all gases in the mixture are at the same temperature therefore they have the same average kinetic energy

• Therefore, in certain applications, the mixture can be thought of as one gas even though air is a mixture, we can measure the pressure, volume,

and temperature of air as if it were a pure substance we can calculate the total moles of molecules in an air sample,

knowing P, V, and T, even though they are different molecules


Composition of Dry Air


Partial Pressure • The pressure of a single gas in a mixture of gases is

called its partial pressure • We can calculate the partial pressure of a gas if we know what fraction of the mixture it composes and the

total pressure or, we know the number of moles of the gas in a container of

known volume and temperature • The sum of the partial pressures of all the gases in

the mixture equals the total pressure Dalton’s Law of Partial Pressures because the gases behave independently


The partial pressure of each gas in a mixture can be calculated using the ideal gas law


Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature

598 K, and 0.17 moles Xe.


Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe

the unit is correct, the value is reasonable

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol

PNe, atm

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

nXe, V, T, R PXe Ptot, PXe PNe


Mole Fraction The fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes The ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, Χ for gases, = volume % / 100% The partial pressure of a gas is equal to the mole fraction of that gas times the total pressure


Solution:

Ex: Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K

mHe = 24.2 g, mO2 = 4.32 g, V = 12.5 L, T = 298 K

χHe, χO2, PHe, atm, PO2, atm, Ptotal, atm

Conceptual Plan:

Relationships:

Given:

Find: ntot, V, T, R Ptot mgas ngas χgas

χgas, Ptotal Pgas

nHe = 6.05 mol, nO2 = 0.135 mol, V = 12.5 L, T = 298 K

χHe=0.97817, χO2=0.021827, PHe, atm, PO2, atm, Ptotal, atm


Practice – Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature

598 K, and 0.17 moles Xe



Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe

the unit is correct, the value is reasonable

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol PNe, atm

Check:

Solution:

Conceptual Plan:

Relationships:

Given: Find:

nXe, V, T, R PXe Ptot, PXe PNe Ptot, PNe χNe



Collecting Gases • Gases are often collected by having them displace

water from a container • The problem is that because water evaporates,

there is also water vapor in the collected gas • The partial pressure of the water vapor, called the

vapor pressure, depends only on the temperature so you can use a table to find out the partial pressure of

the water vapor in the gas you collect if you collect a gas sample with a total pressure of 758.2

mmHg* at 25 °C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg Table 5.4*


Collecting Gas by Water Displacement


Vapor Pressure of Water


Ex: 1.02 L of O2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find mass O2.

V=1.02 L, P=755.2 mmHg, T=293 K

mass O2, g

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Ptot, PH2O PO2

V=1.02 L, PO2=0.97059 atm, T=293 K

mass O2, g PO2,V,T nO2 gO2


Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.


V=10.0 L, nH2 = 0.12 mol, T = 323 K

Ptotal, mmHg

0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

PH2, PH2O Ptotal nH2,V,T PH2


Reactions Involving Gases

P, V, T of Gas A mole A mole B P, V, T of Gas B

• The principles of reaction stoichiometry from Chapter 4 can be combined with the gas laws for reactions involving gases

• In reactions of gases, the amount of a gas is often given as a volume instead of moles as we’ve seen, you must state pressure and temperature

• The ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio

• When gases are at STP, use 1 mol = 22.4 L


Ex: What volume of H2 is needed to make 35.7 g of CH3OH at 738 mmHg and 355 K?

CO(g) + 2 H2(g) → CH3OH(g) mCH3OH = 37.5g, P=738 mmHg, T=355 K

VH2, L

Solution:

Conceptual Plan:

Relationships:

Given:

Find: P, n, T, R V g CH3OH mol CH3OH mol H2

nH2 = 2.2284 mol, P=0.97105 atm, T=355 K

VH2, L



Practice – What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?

2 HgO(s) → 2 Hg(l) + O2(g) (MMHgO = 216.59 g/mol)



What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s) → 2 Hg(l) + O2(g)

mHgO = 10.0g, P=0.750 atm, T=313 K

VO2, L

Solution:

Conceptual Plan:

Relationships:

Given:

Find: P, n, T, R V g HgO mol HgO mol O2

nO2 = 0.023085 mol, P=0.750 atm, T=313 K

VO2, L


Properties of Gases

78

•Expand to completely fill their container •Take the shape of their container •Low density much less than solid or liquid state

•Compressible •Mixtures of gases are always homogeneous •Fluid


Kinetic Molecular Theory

79

•The particles of the gas (either atoms or molecules) are constantly moving

•The attraction between particles is negligible

•When the moving gas particles hit another gas particle or the container, they do not stick; but they bounce off and continue moving in another direction like billiard balls


Kinetic Molecular Theory

80

•There is a lot of empty space between the gas particles compared to the size of the particles

•The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature as you raise the temperature

of the gas, the average speed of the particles increases but don’t be fooled

into thinking all the gas particles are moving at the same speed!!


Gas Properties Explained – Pressure

•Because the gas particles are constantly moving, they strike the sides of the container with a force

•The result of many particles in a gas sample exerting forces on the surfaces around them is a constant pressure

81


Gas Properties Explained – Indefinite Shape and Indefinite Volume

Because the gas molecules have enough kinetic energy to overcome attractions, they keep moving around and spreading out until they fill the container.

As a result, gases take the shape and the volume of the container they are in.

82


Gas Properties Explained - Compressibility

Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can be squeezed closer together

83


Gas Properties Explained – Low Density

Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume; the result is they have low density

84


Density & Pressure

85

• Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them

•When more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure also higher density


Gas Laws Explained – Boyle’s Law

• Boyle’s Law says that the volume of a gas is inversely proportional to the pressure

• Decreasing the volume forces the molecules into a smaller space

• More molecules will collide with the container at any one instant, increasing the pressure

86


Gas Laws Explained – Charles’s Law

• Charles’s Law says that the volume of a gas is directly proportional to the absolute temperature

• Increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently on average

• To keep the pressure constant, the volume must then increase

87


Gas Laws Explained – Avogadro’s Law

•Avogadro’s Law says that the volume of a gas is directly proportional to the number of gas molecules

• Increasing the number of gas molecules causes more of them to hit the wall at the same time

•To keep the pressure constant, the volume must then increase

88


Gas Laws Explained – Dalton’s Law of Partial Pressures

• Dalton’s Law says that the total pressure of a mixture of gases is the sum of the partial pressures

• Kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact

• Therefore the molecules behave independently of each other, each gas contributing its own collisions to the container with the same average kinetic energy

• Because the average kinetic energy is the same, the total pressure of the collisions is the same

89


Dalton’s Law & Pressure

•Because the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side

90


Kinetic Energy and Molecular Velocities

• Average kinetic energy of the gas molecules depends on the average mass and velocity KE = ½mv2

• Gases in the same container have the same temperature, therefore they have the same average kinetic energy

• If they have different masses, the only way for them to have the same kinetic energy is to have different average velocities lighter particles will have a faster average velocity than

more massive particles 91


Molecular Speed vs. Molar Mass •To have the same average kinetic energy, heavier

molecules must have a slower average speed

92


Temperature and Molecular Velocities _ • KEavg = ½NAmu2

NA is Avogadro’s number • KEavg = 1.5RT R is the gas constant in energy units, 8.314 J/mol∙K 1 J = 1 kg∙m2/s2

• Equating and solving we get NA∙mass = molar mass in kg/mol

• As temperature increases, the average velocity increases

93


Molecular Velocities • All the gas molecules in a sample can travel at

different speeds • However, the distribution of speeds follows a

statistical pattern called a Boltzman distribution • We talk about the “average velocity” of the

molecules, but there are different ways to take this kind of average

• The method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities

94


Boltzman Distribution Distribution Function

Molecular Speed

Frac

tion

of M

olec

ules

O2 @ 300 K

95


Temperature vs. Molecular Speed

•As the absolute temperature increases, the average velocity increases the distribution function

“spreads out,” resulting in more molecules with faster speeds

96


Mean Free Path • Molecules in a gas travel in

straight lines until they collide with another molecule or the container

• The average distance a molecule travels between collisions is called the mean free path

• Mean free path decreases as the pressure increases

97


Diffusion and Effusion • The process of a collection of molecules spreading

out from high concentration to low concentration is called diffusion

• The process by which a collection of molecules escapes through a small hole into a vacuum is called effusion

• The rates of diffusion and effusion of a gas are both related to its rms average velocity

• For gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of its molar mass

98


Effusion

99


Graham’s Law of Effusion

•For two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation:

Thomas Graham (1805–1869)

100


Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen

101 Tro: Chemistry: A Molecular Approach, 2/e

Copyright © 2011 Pearson Education, Inc. 102

Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen

O2, 32.00 g/mol; H2 2.016 g/mol

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

MMO2, MMH2 rateA/rateB

This means that, on average, the O2 molecules are traveling at ¼ the speed of H2 molecules

Copyright © 2011 Pearson Education, Inc. 103

Ex: Calculate the molar mass of a gas that effuses at a rate 0.462 times N2

MM, g/mol

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

rateA/rateB, MMN2 MMunknown


Ideal vs. Real Gases

104

• Real gases often do not behave like ideal gases at high pressure or low temperature

• Ideal gas laws assume 1. no attractions between gas molecules 2. gas molecules do not take up space

based on the kinetic-molecular theory • At low temperatures and high pressures these

assumptions are not valid


Real Gas Behavior •Because real

molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures

105


The Effect of Molecular Volume

•At high pressure, the amount of space occupied by the molecules is a significant amount of the total volume

•The molecular volume makes the real volume larger than the ideal gas law would predict

• van der Waals modified the ideal gas equation to account for the molecular volume b is called a van der Waals constant and is different

for every gas because their molecules are different sizes

Johannes van der Waals (1837–1923)

106


Real Gas Behavior •Because real molecules attract each other, the

molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures

107


The Effect of Intermolecular Attractions •At low temperature, the attractions between the

molecules is significant •The intermolecular attractions makes the real

pressure less than the ideal gas law would predict • van der Waals modified the ideal gas equation to

account for the intermolecular attractions a is another van der Waals constant and is different

for every gas because their molecules have different strengths of attraction

108


van der Waals’ Equation

•Combining the equations to account for molecular volume and intermolecular attractions we get the following equation used for real gases

109


Real Gases •A plot of PV/RT vs. P for 1 mole of a gas

shows the difference between real and ideal gases

• It reveals a curve that shows the PV/RT ratio for a real gas is generally lower than ideal for “low” pressures – meaning the most important factor is the intermolecular attractions

• It reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideal for “high” pressures – meaning the most important factor is the molecular volume

110


PV/RT Plots

111

Documents

Properties of Gases...•We measure air pressure with a barometer •Column of mercury supported by air pressure •Force of the air on the surface of the mercury counter balances