### Ch 5

```Chapter 5
Gases and the Kinetic
Molecular Theory
5-1
Dr. Wolf’s CHM 101
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
5-2
Dr. Wolf’s CHM 101
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.
5-3
Dr. Wolf’s CHM 101
Figure 5.1
5-4
The three states of matter.
Dr. Wolf’s CHM 101
Measuring Gas Pressure
Barometer - A device to measure atmospheric
pressure. Pressure is defined as force divided by
area. The force is the force of gravity acting on
the air molecules.
Manometer - A device to measure gas
pressure in a closed container.
5-5
Dr. Wolf’s CHM 101
Figure 5.3
5-6
Dr. Wolf’s CHM 101
A mercury barometer.
Figure 5.4
5-7
Two types of manometer
Dr. Wolf’s CHM 101
Table 5.2 Common Units of Pressure
Unit
Scientific Field
pascal(Pa);
kilopascal(kPa)
1.01325x105Pa;
101.325 kPa
SI unit; physics, chemistry
atmosphere(atm)
1 atm
chemistry
millimeters of
mercury(Hg)
760 mmHg
chemistry, medicine, biology
torr
760 torr
chemistry
14.7lb/in2
engineering
1.01325 bar
meteorology, chemistry,
physics
pounds per square
inch (psi or lb/in2)
bar
5-8
Atmospheric Pressure
Dr. Wolf’s CHM 101
Sample Problem 5.1
PROBLEM:
Converting Units of Pressure
A geochemist heats a limestone (CaCO3) sample and collects
the CO2 released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature,
Dh = 291.4mmHg. Calculate the CO2 pressure in torrs,
atmospheres, and kilopascals.
PLAN: Construct conversion factors to find the other units of pressure.
SOLUTION:
291.4mmHg
1torr
= 291.4torr
1mmHg
291.4torr
1atm
= 0.3834atm
760torr
0.3834atm
101.325kPa
1atm
5-9
Dr. Wolf’s CHM 101
= 38.85kPa
Boyle’s Law - The relationship between volume and
the pressure of a gas.
(Temperature is kept constant.)
5-10
Dr. Wolf’s CHM 101
Boyle’s Law
V a
PV = constant
5-11
Dr. Wolf’s CHM 101
1
P
n and T are fixed
V = constant / P
Charles’s Law - The relationship between volume and the
temperature of a gas.
(Pressure is kept constant.)
5-12
Dr. Wolf’s CHM 101
V a
Boyle’s Law
= constant
T
P a T
Amonton’s Law
P
T
combined gas law
P
V a T
Charles’s Law
V
1
= constant
V a
n and T are fixed
P and n are fixed
V = constant x T
V and n are fixed
P = constant x T
T
P
V = constant x
T
PV
P
T
When the amount of gas, n, is constant
5-13
Dr. Wolf’s CHM 101
= constant
Avogadro’s Law - The volume of a gas is
directly proportionate to the amount of gas. (Pressure
and temperature kept constant.)
An experiment to study the relationship between the
volume and amount of a gas.
V a n
V
n
= constant
V = constant x n
Twice the amount
gives twice the volume
5-14
Dr. Wolf’s CHM 101
If the constant pressure is 1 atm and the constant temperature is 0o C,
1 mole of any gas has a volume of 22.4 L .
This is known as the Standard Molar Volume
5-15
Dr. Wolf’s CHM 101
IDEAL GAS LAW
fixed n and T
Boyle’s Law
V
= constant
fixed n and P
fixed P and T
Charles’s Law
V=
V = constant X T
constant X n
P
THE IDEAL GAS LAW
PV = nRT
R=
5-16
PV
nT
Dr. Wolf’s CHM 101
=
1atm x 22.414L
1mol x 273.15K
=
0.0821atm-L
Mol-K
Sample Problem 5.2
PROBLEM:
Applying the Volume-Pressure Relationship
Boyle’s apprentice finds that the air trapped in a J tube occupies
24.8cm3 at 1.12atm. By adding mercury to the tube, he increases
the pressure on the trapped air to 2.64atm. Assuming constant
temperature, what is the new volume of air (inL)?
PLAN:
V1 in cm3
1cm3=1mL
V1 in mL
unit
conversion
103mL=1L
V1 in L
xP1/P2
P1 = 1.12atm
P2 = 2.64atm
V1 = 24.8cm3
V2 = unknown
24.8cm3
gas law
calculation
P1V1
V2 in L
P1V1
P2
Dr. Wolf’s CHM 101
1mL
L
1cm3
103mL
=R=
n1T1
V2 =
5-17
n and T are constant
SOLUTION:
P2V2
P1V1 = P2V2
n2T2
1.12atm
= 0.0248L
= 0.0248L = V1
2.46atm
= 0.0105L
Sample Problem 5.3
PROBLEM:
Applying the Temperature-Pressure Relationship
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:
T1 and T2(0C)
P1(atm)
1atm=760torr
P1(torr)
K=0C+273.15
T1 and T2(K)
x T2/T1
0.991atm
5-18
P2 = unknown
T1 = 230C
T2 = 100oC
P1V1
=
n1T1
P2(torr)
P2 = P1
P1 = 0.991atm
Dr. Wolf’s CHM 101
n2T2
760 torr = 753 torr
1atm
T2
T1
P2V2
= 753 torr
373K
296K
= 949 torr
P1
T1
=
P2
T2
Sample Problem 5.4
PROBLEM:
Applying the Volume-Amount Relationship
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.
n1(mol) of He
P and T are constant
SOLUTION:
x V2/V1
n1 = 1.10mol
n2 = unknown
n2(mol) of He
V1 = 26.2dm3
V2 = 55.0dm3
subtract n1
V1
n1
=
V2
n2
xM
n2 = 1.10mol
n2 = n1
55.0dm3
5-19
n1T1
=
P2V2
n2T2
V2
V1
= 2.31mol
26.2dm3
n2 - n1 = 2.31 -1.10 = 1.21 mol He
Dr. Wolf’s CHM 101
P1V1
4.003g He
mol He
= 4.84g He
Sample Problem 5.5
Conditions
A steel tank has a volume of 438L and is filled with 0.885kg of
O2. Calculate the pressure of O2 at 210C.
PROBLEM:
PLAN:
Solving for an Unknown Gas Variable at Fixed
V, T and mass, which can be converted to moles (n), are given. We
use the ideal gas law to find P.
SOLUTION:
0.885kg
V = 438L
T = 210C (convert to K)
n = 0.885kg (convert to mol)
P = unknown
103g
mol O2
kg
32.00g O2
= 27.7mol O2
27.7mol x 0.0821
P=
5-20
nRT
V
atm*L
mol*K
=
Dr. Wolf’s CHM 101
210C + 273.15 = 294K
x 294K
= 1.53atm
438L
Sample Problem 5.6
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).
PROBLEM:
PLAN:
Calculating Gas Density
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 Avogardro’s number.
d = mass/ V
PV = nRT
mult. both sides by M
n / V = P / RT
Mass = n M
44.01g/mol
SOLUTION:
(a)
x 1atm
atm*L
MxP
RT
= 1.96g/L
d=
0.0821
d =
x 273K
mol*K
5-21
1.96g
mol CO2
L
44.01g CO2
Dr. Wolf’s CHM 101
6.022x1023molecules
mol
= 2.68x1022molecules CO2/L
Sample Problem 5.6
Calculating Gas Density
continued
(b)
44.01g/mol
d=
0.0821
5-22
x 1atm
1.83g
mol CO2
L
44.01g CO2
Dr. Wolf’s CHM 101
= 1.83g/L
atm*L x 293K
mol*K
6.022x1023molecules
mol
= 2.50x1022molecules CO2/L
Calculating the Molar Mass, M,
of a Gas
Since PV = nRT
Then n = PV / RT
And n = mass / M
So m / M = PV / RT
And M = mRT / PV
Or M = d RT / P
5-23
Dr. Wolf’s CHM 101
Sample Problem 5.7
PROBLEM:
Finding the Molar Mass of a Volatile Liquid
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:
T = 100.00C
Mass of flask + gas = 78.416g
P = 754 torr
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
0.582g x 0.0821
M=
m RT
=
PV
atm*L
mol*K
x 373K
= 84.4g/mol
0.213L x 0.992atm
M of C6H12 is 84.16g/mol and the calculated value is within experimental error.
5-24
Dr. Wolf’s CHM 101
Partial Pressure of a Gas in a Mixture of Gases
Gases mix homogeneously.
Each gas in a mixture behaves as if it is the only gas present,
e.g. its pressure is calculated from PV = nRT with n equal to the
number of moles of that particular gas......called the Partial Pressure.
The gas pressure in a container is the sum of the partial pressures of
all of the gases present.
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + ...
P1= c1 x Ptotal
where c1 is the mole fraction
c1 =
5-25
Dr. Wolf’s CHM 101
n1
n1 + n2 + n3 +...
=
n1
ntotal
Collecting Gas over Water
Often in gas experiments the gas is collected “over water.” So
the gases in the container includes water vapor as a gas. The
water vapor’s partial pressure contributes to the total pressure
in the container.
5-26
Dr. Wolf’s CHM 101
Sample Problem 5.8
PROBLEM:
Applying Dalton’s Law of Partial Pressures
In a study of O2 uptake by muscle at high altitude, a physiologist
prepares an atmosphere consisting of 79 mol% N2, 17 mol%
16O and 4.0 mol% 18O . (The isotope 18O will be measured to
2,
2
determine the O2 uptake.) The pressure of the mixture is
0.75atm to simulate high altitude. Calculate the mole fraction
and partial pressure of 18O2 in the mixture.
18
PLAN: Find the c 18Oand P18O from Ptotal and mol% O2.
2
mol%
18O
2
SOLUTION:
2
divide by 100
c
18O
P18
2
multiply by Ptotal
partial pressure P
18O
5-27
2
Dr. Wolf’s CHM 101
O2
c
18O
=
2
4.0mol% 18O2
= 0.040
100
= c18 x Ptotal = 0.040 x 0.75atm = 0.030atm
O2
Sample Problem 5.9
PROBLEM:
Calculating the Amount of Gas Collected Over Water
Acetylene (C2H2), an important fuel in welding, is produced in
the laboratory when calcium carbide (CaC2) reaction with water:
CaC2(s) + 2H2O(l)
C2H2(g) + Ca(OH)2(aq)
For a sample of acetylene that is collected over water, the total
gas pressure (adjusted to barometric pressure) is 738torr and
the volume is 523mL. At the temperature of the gas (230C), the
vapor pressure of water is 21torr. How many grams of
acetylene are collected?
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.
P
SOLUTION:
C2H2 = (738-21)torr = 717torr
Ptotal
P
C2H2
atm
P
= 0.943atm
PV
717torr
H2O
n=
760torr
RT
n
g
C2H2
C2H2
5-28
Dr. Wolf’s CHM 101
xM
Sample Problem 5.9
Calculating the Amount of Gas Collected Over Water
continued
n
C2H2
0.943atm x
=
0.0821
atm*L
0.523L
= 0.203mol
x 296K
mol*K
0.203mol
26.04g C2H2
mol C2H2
5-29
Dr. Wolf’s CHM 101
= 0.529 g C2H2
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).
P,V,T
of gas A
ideal
gas
law
5-30
Dr. Wolf’s CHM 101
amount
(mol)
amount
(mol)
P,V,T
of gas A
of gas B
of gas B
molar ratio from
balanced equation
ideal
gas
law
Sample Problem 5.10
PROBLEM:
Using Gas Variables to Find Amount of
Reactants and Products
A laboratory-scale method for reducing a metal oxide is to heat it
with H2. The pure metal and H2O are products. What volume of
H2 at 765torr and 2250C is needed to form 35.5g of Cu from
copper (II) oxide?
PLAN: Since this problem requires stoichiometry and the gas laws, we have
to write a balanced equation, use the moles of Cu to calculate mols
and then volume of H2 gas.
CuO(s) + H2(g)
mass (g) of Cu
divide by M
mol of Cu
SOLUTION:
mol Cu
1mol H2
35.5g Cu
= 0.559mol H2
63.55g Cu 1 mol Cu
molar ratio
mol of H2
0.559mol H2 x 0.0821
use known P and T to find V
L of H2
5-31
Dr. Wolf’s CHM 101
Cu(s) + H2O(g)
atm*L
mol*K
1.01atm
x
498K
= 22.6L
Sample Problem 5.11
Using the Ideal Gas Law in a Limiting-Reactant
Problem
PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)]
to form ionic metal halides. What mass of potassium chloride forms
when 5.25L of chlorine gas at 0.950atm and 293K reacts with 17.0g of
potassium?
PLAN: After writing the balanced equation, we use the ideal gas law to find the
number of moles of reactants, the limiting reactant and moles of product.
2K(s) + Cl2(g)
2KCl(s)
P = 0.950atm
V = 5.25L
SOLUTION:
T = 293K
n = unknown
PV
0.950atm x 5.25L
n =
=
= 0.207mol
Cl2
RT
atm*L
0.0821
x 293K
2mol KCl
mol*K
0.207mol Cl2
= 0.414mol
1mol Cl2
KCl formed
17.0g mol K = 0.435mol K
39.10g K
2mol KCl = 0.435mol
Cl2 is the limiting reactant.
0.435mol K
KCl formed
2mol K
74.55g KCl
0.414mol KCl
= 30.9 g KCl
mol KCl
5-32
Dr. Wolf’s CHM 101
Postulates of the Kinetic-Molecular Theory
Postulate 1: Particle Volume
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.
Postulate 2: Particle Motion
Gas particles are in constant, random, straight-line
motion except when they collide with each other or with
the container walls.
Postulate 3: Particle Collisions
Collisions are elastic therefore the total kinetic
energy(Kk) of the particles is constant.
5-33
Dr. Wolf’s CHM 101
A molecular description of Boyle’s Law
Boyle’s Law
5-34
Dr. Wolf’s CHM 101
V a
1
P
n and T are fixed
A molecular description of Charles’s Law
Charles’s Law
5-35
Dr. Wolf’s CHM 101
V a T
n and P are fixed
A molecular description of Dalton’s law of partial pressures.
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + ...
5-36
Dr. Wolf’s CHM 101
A molecular description of Avogadro’s Law
5-37
Dr. Wolf’s CHM 101
V a
n
P and T are fixed
Kinetic-Molecular Theory
Gas particles are in motion and have a molecular speed, .
But they are moving at various speeds, some very slow, some very fast,
but most near the average speed of all of the particles,  (avg) .
Since kinetic energy is defined as
½ mass x (speed)2, we can define
the average kinetic energy,
Ek(avg) = ½m 2 (avg)
Since energy is a function of
temperature, the average energy
and, hence, average molecular
speed will increase with
temperature.
5-38
Dr. Wolf’s CHM 101
An increase in temperature results in an
increase in average molecular kinetic energy.
The relationship between average kinetic energy and temperature
is given as, Ek(avg) = 3/2 (R/NA) x T
(where R is the gas constant in energy units, 8.314 J/mol-K
NA is Avogadro’s number, and T temperature in K.)
To have the same average kinetic energy, heavier atoms must have
smaller speeds.
The root-mean-square speed,  (rms) , is the speed where a molecule has
the average kinetic energy. The relationship between  (rms) and molar
mass is:
 (rms) = (3RT/M ) ½
So the speed (or rate of movement) is: rate a 1 / (M ) ½
5-39
Dr. Wolf’s CHM 101
Relationship between molar mass and
molecular speed.
5-40
Dr. Wolf’s CHM 101
Graham’s Law of Effusion
Effusion is the process by which a gas in a closed container moves
through a pin-hole into an evacuated space. This rate is proportional to
the speed of a molecule so.....
Graham’s Law of Effusion
The rate of effusion of a gas is inversely related to the
square root of its molar mass.
Rate of effusion a 1 / (M ) ½
So doing two identical effusion experiments measuring the rates of two
gases, one known, one unknown, allows the molecular mass of the
unknown to be determined.
rate A
rate B
5-41
Dr. Wolf’s CHM 101
=
( MB / MA ) ½
Sample Problem 5.12
Applying Graham’s Law of Effusion
PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4).
PLAN: The effusion rate is inversely proportional to the square root of the
molar mass for each gas. Find the molar mass of both gases and find
the inverse square root of their masses.
SOLUTION:
M of CH4 = 16.04g/mol
M of He = 4.003g/mol
rate
He
rate
5-42
CH4
Dr. Wolf’s CHM 101
= ( 16.04/ 4.003 ) ½
= 2.002
End of Chapter 5
5-43
Dr. Wolf’s CHM 101
```