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
Distinction Between Gases and Liquids/Solids
(condensed phases)
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 densities
under normal conditions.
5. Gases are miscible.
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States of Matter
Figure 5.1
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A Mercury Barometer
Pressure force/area
A device used to measure atmospheric pressure
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
engineering
14.7 lb/in2
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Sample Problem 5.1
Converting Units of Pressure
SOLUTION
291.4 mm Hg x
291.4 torr
291.4 torr x
0.3834 atm
0.3834 atm x
38.85 kPa
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Three laws (Boyles, Charless and Avogadros)
are combined to describe a universal relationship
among the key gas variables (volume, pressure,
temperature, amount). This universal
relationship is known as the Ideal Gas Law.
Lets examine the three individual laws first,
and then see how they are combined to generate
the Ideal Gas Law.
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Relationship between volume and pressure of a gas
Boyles Law
Figure 5.5
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Boyles Law
V a
n and T are fixed
(volume is inversely proportional to pressure)
PV constant
V constant / P
or
P pressure V volume n number of moles of
gas T temperature
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Relationship between volume and temperature of a
gas
Charless Law
Figure 5.6
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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
(Boyles Charless)
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An experiment to study the relationship between
volume and amount of a gas
Avogadros Law
Figure 5.7
V a n (P and T fixed)
At fixed T and P, equal volumes of any ideal gas
contain equal numbers of particles (or moles).
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Standard Molar Volume
STP 0 oC (273.15 K) 1 atm (760 torr)
Standard Molar Volume 22.4141 L or 22.4 L
Figure 5.8
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THE IDEAL GAS LAW
PV nRT
R


At fixed n and T
At fixed n and P
At 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
n and T are constant
V1 in cm3
P1 1.12 atm
P2 2.64 atm
1 cm3 1 mL
V1 24.8 cm3
V2 unknown
V1 in mL
24.8 cm3
0.0248 L
103 mL 1 L
x
x
V1 in L

P1V1 P2V2
or
x P1/P2
V2 in L
P1V1
x
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 1.00
x 103 torr. It is filled with helium at 23 oC
and 0.991 atm and placed in boiling water at
exactly 100 oC. Will the safety valve open?
PLAN
SOLUTION
n and V are constant
P1 (atm)
T1 and T2 (oC)
P1 0.991 atm
P2 unknown
1 atm 760 torr
K oC 273.15
T1 23 oC
T2 100 oC
P1 (torr)
T1 and T2 (K)
or
x T2/T1
P2 (torr)
753 torr
949 torr
(valve will not open)
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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 55 dm3 (V2). When
1.10 mol of He (n1) are added to the blimp, the
volume is 26.2 dm3 (V1). How many more grams of
He must be added to make it rise? Assume
constant T and P.
PLAN
We are given the initial n1 and V1 and the final
V2. We need to find n2 and convert it from moles
to grams.
n1(mol) of He
SOLUTION
P and T are constant
x V2 / V1
n1 1.10 mol
n2 unknown
n2(mol) of He
V1 26.2 dm3
V2 55.0 dm3
subtract n1
or
mol to be added
x M
2.31 mol
x
g He to add
4.84 g He
2.31 mol - 1.10 mol 1.21 mol x
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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 438 L and is filled
with 0.885 kg of O2. Calculate the pressure of
O2 at 21 oC.
PLAN
V, T and mass, which can be converted to moles
(n), are given. Use the ideal gas law to find P.
SOLUTION
V 438 L
T 21 oC (convert to K)
n 0.885 kg (convert to mol)
P unknown
21 oC 273.15 294 K
27.7 mol O2
x
1.53 atm
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The Density of a Gas
PV nRT or PV m/M x RT where m mass and
M molar mass
m/V d (M x P)/RT where d density
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Sample Problem 5.6
Calculating the Density of a Gas
PROBLEM
Calculate the density (in g/L) of carbon dioxide
and the number of molecules per liter (a) at STP
(0 oC and 1 atm) and (b) at ordinary room
conditions (20. oC and 1.00 atm).
PLAN
Density is mass/unit volume substitute for
volume in the ideal gas equation. Since the
identity of the gas is known, the molar mass can
be determined. Convert mass/L to molecules/L
using Avogadros number.
d mass/volume
PV nRT
V nRT/P
d

SOLUTION
d
1.96 g/L
(a)
2.68 x 1022 molecules CO2/L
x
x
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Sample Problem 5.6
(continued)
(b)
1.83 g/L
2.50 x 1022 molecules CO2/L
x
x
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The Molar Mass of a Gas
n
M
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Determining the molar mass of an unknown volatile
liquid
based on the method of J.B.A. Dumas (1800 -1884)
M mRT/PV or M dRT/P
Figure 5.11
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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.
d
SOLUTION
m (78.416 - 77.834)g
0.582 g of gas
M
x
M

M of C6H12 is 84.16 g/mol - the calculated
value is within experimental error
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Daltons Law of Partial Pressures
In a mixture of unreacting gases, the total
pressure is equal to the sum of the partial
pressures of the individual gases.
Ptotal P1 P2 P3 ...
where
P1 c1 x Ptotal
and c1 is the mole fraction
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Sample Problem 5.8
Applying Daltons Law of Partial Pressures
PLAN
mol 18O2
0.040
SOLUTION
divide by 100
c 18O2
0.030 atm
18O2
multiply by Ptotal
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Collecting a water-insoluble gaseous reaction
product and determining its pressure
Figure 5.12
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Sample Problem 5.9
Calculating the Amount of Gas Collected Over Water
PLAN
The difference in pressures will give P for C2H2.
The ideal gas law allows a determination of n.
Converting n to grams requires the molar mass, M.
SOLUTION
(738 - 21) torr 717 torr
Ptotal
717 torr x
H2O
0.943 atm
x M
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Sample Problem 5.9
(continued)
0.943 atm
0.523 L
x
0.0203 mol C2H2
x
296 K
0.529 g C2H2
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Summary of the stoichiometric relationships
between 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
Figure 15.13
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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
35.5 g Cu x
0.559 mol H2
x
mol of Cu
molar ratio
0.559 mol H2
22.6 L
mol of H2
use known P and T to find V
L of H2
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Using the Ideal Gas Law in a Limiting Reactant
Problem
Sample Problem 5.11
SOLUTION
x
5.25 L
n

Cl2
0.414 mol KCl formed
0.435 mol K x
0.435 mol KCl formed
0.414 mol KCl x
30.9 g KCl
Cl2 is the limiting reactant
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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(Ek) of the particles is constant.
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Distribution of molecular speeds at three
temperatures
The most probable speed increases as the
temperature increases
For N2 gas
The average kinetic energy, Ek, is proportional
to the absolute temperature
Figure 5.14
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A molecular description of Boyles Law
Figure 5.15
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A molecular description of Daltons law of
partial pressures
Figure 5.16
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A molecular description of Charless Law
Figure 5.17
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A molecular description of Avogadros Law
Figure 5.18
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Why do equal numbers of molecules of two
different gases, such as O2 and H2, occupy the
same volume (c.f. standard molar volume)?
At constant T, two gases possess the same kinetic
energy thus, the heavier gas must be moving more
slowly.
Ek 1/2 mass x speed2
root-mean-square speed a molecule moving at this
speed has the average kinetic energy

R 8.314 joule/mol K
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Relationship between molar mass and molecular
speed
Figure 5.19
At a given temperature, gases with lower molar
masses have higher most probable speeds
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EFFUSION the process by which a gas escapes
from its container through a tiny hole into an
evacuated space
Grahams Law of Effusion
The rate of effusion of a gas is inversely
related to the square root of its molar mass.
(related to the rms speed)
rateA/rateB MB1/2/MA1/2
The same relationships pertain to gaseous
diffusion rates!
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Sample Problem 5.12
Applying Grahams Law of Effusion
SOLUTION
M of CH4 16.04 g/mol
M of He 4.003 g/mol
2.002
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Diffusion of a gas particle through a space
filled with other particles
Distribution of Molecular Speeds Mean Free Path
the average distance a molecule travels between
collisions at a given T and P Collision
Frequency the average number of collisions per
second (has implications for chemical reaction
rates)
Figure 5.20
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Real Gases
Molecules are not points of mass. There are
attractive and repulsive forces between molecules.
Real gases approach ideal behavior at high T and
low P.
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Figure 5.21
The behavior of several real gases with
increasing external pressure
At moderately high P intermolecular attractions
dominate At very high P molecular volume
effects dominate
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The effect of intermolecular attractions on
measured gas pressure
Figure 5.22
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The effect of molecular volume on measured gas
volume
Figure 5.23
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The van der Waals equation for n moles of a real
gas
(adjusts V down)
(P n2a/V2) (V - nb) nRT
(adjusts P up)
a and b are the van der Waals constants