Gases and the KineticMolecular Theory

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Chapter 5
Gases and the Kinetic-Molecular Theory
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Gases and the Kinetic Molecular Theory
5.1 An Overview of the Physical States of Matter
5.2 Gas Pressure and Its Measurement
5.3 The Gas Laws and Their Experimental
Foundations
5.4 Further Applications of the Ideal Gas Law
5.5 The Ideal Gas Law and Reaction Stoichiometry
5.6 The Kinetic-Molecular Theory A Model for
Gas Behavior
5.7 Real Gases Deviations from Ideal Behavior
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An Overview of the Physical States of Matter
The Distinction of Gases from Liquids and Solids
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with temperature.
3. Gas have relatively low viscosity.
4. Most gases have relatively low densites under
normal conditions.
5. Gases are miscible.
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Figure 5.1
The three states of matter.
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A mercury barometer.
Figure 5.3
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Two types of manometer
Figure 5.4
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Table 5.2 Common Units of Pressure
Atmospheric Pressure
Unit
Scientific Field
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Sample Problem 5.1
Converting Units of Pressure
SOLUTION
291.4mmHg
291.4torr
291.4torr
0.3834atm
0.3834atm
38.85kPa
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Figure 5.5
The relationship between volume and the pressure
of a gas.
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Boyles Law
V a
n and T are fixed
PV constant
V constant/P
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Figure 5.6
The relationship between volume and the
temperature of a gas.
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Boyles Law
n and T are fixed
Charless Law
V a T
P and n are fixed
V constant x T
Amontons Law
P a T
V and n are fixed
P constant x T
combined gas law
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An experiment to study the relationship between
the volume and amount of a gas.
Figure 5.7
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Standard Molar Volume
Figure 5.8
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THE IDEAL GAS LAW
PV nRT
R


fixed n and T
fixed n and P
fixed P and T
Boyles Law
Charless Law
Avogadros Law
V
constant X n
V
V
constant X T
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Sample Problem 5.2
Applying the Volume-Pressure Relationship
PLAN
SOLUTION
P and T are constant
V1 in cm3
P1 1.12atm
P2 2.64atm
1cm31mL
V1 24.8cm3
V2 unknown
V1 in mL
24.8cm3
0.0248L
103mL1L
V1 in L
xP1/P2
P1V1 P2V2

V2 in L
P1V1
P2
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Sample Problem 5.3
Applying the Temperature-Pressure Relationship
PROBLEM
A 1-L steel tank is fitted with a safety valve
that opens if the internal pressure exceeds
1.00x103 torr. It is filled with helium at 230C
and 0.991atm and placed in boiling water at
exactly 1000C. Will the safety valve open?
PLAN
SOLUTION
P1(atm)
T1 and T2(0C)
P1 0.991atm
P2 unknown
1atm760torr
K0C273.15
T1 230C
T2 100oC
P1(torr)
T1 and T2(K)
x T2/T1
P2(torr)
753torr
949torr
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Sample Problem 5.4
Applying the Volume-Amount Relationship
PROBLEM
A scale model of a blimp rises when it is filled
with helium to a volume of 55dm3. When 1.10mol
of He is added to the blimp, the volume is
26.2dm3. How many more grams of He must be added
to make it rise? Assume constant T and P.
PLAN
We are given initial n1 and V1 as well as the
final V2. We have to find n2 and convert it from
moles to grams.
SOLUTION
n1(mol) of He
P and T are constant
x V2/V1
n1 1.10mol
n2 unknown
n2(mol) of He
V1 26.2dm3
V2 55.0dm3
subtract n1
mol to be added
x M
g to be added
4.84g He
2.31mol
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Sample Problem 5.5
Solving for an Unknown Gas Variable at Fixed
Conditions
PROBLEM
A steel tank has a volume of 438L and is filled
with 0.885kg of O2. Calculate the pressure of O2
at 210C.
PLAN
V, T and mass, which can be converted to moles
(n), are given. We use the ideal gas law to find
P.
SOLUTION
V 438L
T 210C (convert to K)
n 0.885kg (convert to mol)
P unknown
210C 273.15 294K
27.7mol O2
1.53atm
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Sample Problem 5.6
Calculating Gas Density
PROBLEM
Calculate the density (in g/L) of carbon dioxide
and the number of molecules per liter (a) at STP
(00C and 1 atm) and (b) at ordinary room
conditions (20.0C and 1.00atm).
PLAN
Density is mass/unit volume substitute for
volume in the ideal gas equation. Since the
identity of the gas is known, we can find the
molar mass. Convert mass/L to molecules/L with
Avogardros number.
d mass/volume
PV nRT
V nRT/P
d

SOLUTION
1.96g/L
d
(a)
2.68x1022molecules CO2/L
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Sample Problem 5.6
Calculating Gas Density
continued
(b)
1.83g/L
2.50x1022molecules CO2/L
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Figure 5.11
Determining the molar mass of an unknown volatile
liquid
based on the method of J.B.A. Dumas (1800-1884)
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Sample Problem 5.7
Finding the Molar Mass of a Volatile Liquid
PROBLEM
An organic chemist isolates from a petroleum
sample a colorless liquid with the properties of
cyclohexane (C6H12). She uses the Dumas method
and obtains the following data to determine its
molar mass
Is the calculated molar mass consistent with the
liquid being cyclohexane?
PLAN
Use unit conversions, mass of gas and density-M
relationship.
SOLUTION
m (78.416 - 77.834)g
0.582g
x
M

M of C6H12 is 84.16g/mol and the calculated value
is within experimental error.
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Daltons Law of Partial Pressures
Ptotal P1 P2 P3 ...
P1 c1 x Ptotal
where c1 is the mole fraction
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Figure 5.12
Collecting a water-insoluble gaseous reaction
product and determining its pressure.
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Sample Problem 5.8
Applying Daltons Law of Partial Pressures
PLAN
SOLUTION
mol 18O2
0.040
divide by 100
c 18O2
0.030atm
multiply by Ptotal
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The Molar Mass of a Gas
n
M
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Sample Problem 5.9
Calculating the Amount of Gas Collected Over Water
PLAN
The difference in pressures will give us the P
for the C2H2. The ideal gas law will allow us to
find n. Converting n to grams requires the molar
mass, M.
SOLUTION
(738-21)torr 717torr
Ptotal
0.943atm
717torr
H2O
x M
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Sample Problem 5.9
Calculating the Amount of Gas Collected Over Water
continued
0.943atm
0.523L
x
0.203mol
x
296K
0.529 g C2H2
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Figure 15.13
Summary of the stoichiometric relationships among
the amount (mol,n) of gaseous reactant or product
and the gas variables pressure (P), volume (V),
and temperature (T).
amount (mol) of gas B
amount (mol) of gas A
P,V,T of gas A
P,V,T of gas B
ideal gas law
ideal gas law
molar ratio from balanced equation
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Sample Problem 5.10
Using Gas Variables to Find Amount of Reactants
and Products
SOLUTION
mass (g) of Cu
divide by M
35.5g Cu
0.559mol H2
mol of Cu
molar ratio
0.559mol H2
22.6L
mol of H2
use known P and T to find V
L of H2
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Sample Problem 5.11
Using the Ideal Gas Law in a Limiting-Reactant
Problem
SOLUTION
x
5.25L
n
Cl2
0.414mol KCl formed
0.435mol K
0.435mol KCl formed
Cl2 is the limiting reactant.
0.414mol KCl
30.9 g KCl
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Postulates of the Kinetic-Molecular Theory
Because the volume of an individual gas particle
is so small compared to the volume of its
container, the gas particles are considered to
have mass, but no volume.
Gas particles are in constant, random,
straight-line motion except when they collide
with each other or with the container walls.
Collisions are elastic therefore the total
kinetic energy(Kk) of the particles is constant.
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Distribution of molecular speeds at three
temperatures.
Figure 5.14
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A molecular description of Boyles Law
Figure 5.15
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Figure 5.16
A molecular description of Daltons law of
partial pressures.
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Figure 5.17
A molecular description of Charless Law
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Avogadros Law
V a n
Ek 1/2 mass x speed2
R 8.314Joule/molK
Grahams Law of Effusion
The rate of effusion of a gas is inversely
related to the square root of its molar mass.
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Figure 5.18
A molecular description of Avogadros Law
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Figure 5.19
Relationship between molar mass and molecular
speed.
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Sample Problem 5.12
Applying Grahams Law of Effusion
SOLUTION
M of CH4 16.04g/mol
M of He 4.003g/mol
2.002
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