Transcript Document

Chapter 5: The Gaseous State
Bushra Javed
CHM 2045
1
Contents
1. Gas pressure & it’s units
2. Empirical Gas Laws
• Boyle’sLaw
• Charles’Law
• Combined gas Law
• Avogadro’s Law
• Dalton’s Law
• Ideal Gas Law
2
Contents
3. Kinetic-Molecular Theory
4. Molecular Speeds
5. Diffusion and Effusion
Graham’s Law
6. Real Gases
3
The Gaseous State
• Gases differ from liquids and solids:
• Have low densities, are compressible & can
expand
• Physical condition of any gas can be defined
• by four variables:
• Pressure, P
• Volume, V
• temperature, T
• amount or number of moles n.
4
Pressure , P of a Gas
Pressure is the force exerted per unit area.
It can be given by two equations:
F
P
A
kg  m
s 2  kg
m2
m  s2
 Pa (pascal)
The SI unit for pressure is the pascal, Pa.
5
Pressure , P of a Gas
Other Units
atmosphere, atm
mmHg
torr
bar
6
Atmospheric Pressure
• Atmospheric pressure results from the
exertion & pressure of air molecules in the
environment.
• A barometer is a device for measuring the
pressure of the atmosphere.
• A manometer is a device for measuring the
pressure of a gas or liquid in a vessel.
Atmospheric Pressure
5|8
Empirical Gas Laws
All gases behave quite simply with respect to
temperature, pressure, volume, and molar
amount.
By holding two of these physical properties
constant, it becomes possible to show a simple
relationship between the other two properties.
T
5|9
Boyle’s Law
Pressure –Volume Relationship
The volume of a sample of gas at constant
temperature varies inversely with the applied
pressure.
1
The mathematical relationship: V 
P
In equation form:
PV  constant
PiVi  PfVf
5 | 10
Boyle’s Law
Pressure –Volume Relationship
• When the volume decreases, the gas
molecules collide with the container more
often and the pressure increases.
• When the volume increases, the gas
molecules collide with the container less often
and the pressure decreases.
11
Boyle’s Law
plot of V versus P
for 1.000 g O2 at
0°C. This plot is
nonlinear.
12
Boyle’s Law
At one atmosphere
the volume of the
gas is 100 mL. When
pressure is doubled,
the volume is halved
to 50 mL. When
pressure is tripled,
the volume
decreases to onethird, 33 mL.
5 | 13
Boyle’s Law
Example 1
A volume of oxygen gas occupies 38.7 mL at 751 mmHg
and 21°C. What is the volume if the pressure changes
to 359 mmHg while the temperature remains constant?
Vi = 38.7 mL
Pi = 751 mmHg
Ti = 21°C
Vf = ?
Pf = 359 mmHg
Tf = 21°C
PiVi
Vf 
Pf
5 | 14
Boyle’s Law
Vf = ?
Pf = 359 mmHg
Tf = 21°C
Vi = 38.7 mL
Pi = 751 mmHg
Ti = 21°C
(38.7 mL)(751mmHg)
Vf 
(359 mmHg)
= 81.0 mL
5 | 15
Boyle’s Law
Example 2
A sample of methane, CH4, occupies a volume of
244.0 mL at 25°C and exerts a pressure of 1135.0
mmHg. If the volume of the gas is allowed to
expand to 720.0 mL at 298 K, what will be the
pressure of the gas? P2 = ?
a. 3350 mmHg
b. 385 mmHg
c. 4580 mmHg
d. 0.0149 mmHg
16
If you plot volume vs.
temperature for any gas at
constant pressure, the points
will all fall on a straight line
If the lines are
extrapolated back
to a volume of “0,”
they all show the
same temperature,
−273.15 °C, called
absolute zero
17
Absolute Zero
The temperature -273.15°C is called absolute
zero. It is the temperature at which the volume
of a gas is hypothetically zero.
This is the basis of the absolute temperature
scale, the Kelvin scale (K).
5 | 18
Charles’s Law
The volume of a sample of gas at constant
pressure is directly proportional to the absolute
temperature (K).
The mathematical relationship: V  T
In equation form:
V
 constant
T
Vi Vf

Ti Tf
19
A balloon was immersed in
liquid nitrogen (black
container) and is shown
immediately after being
removed. It shrank
because air inside
contracts in volume.
As the air inside warms,
the balloon expands to its
orginial size.
5 | 20
Charles’s Law
Example 3
You prepared carbon dioxide by adding HCl(aq) to
marble chips, CaCO3. According to your calculations,
you should obtain 79.4 mL of CO2 at 0°C and 760
mmHg. How many milliliters of gas would you obtain at
27°C?
Vi = 79.4 mL
Pi = 760 mmHg
Ti = 0°C = 273 K
Vf = ?
Pf = 760 mmHg
Tf = 27°C = 300. K
TfVi
Vf 
Ti
Charles’s Law
Vi = 79.4 mL
Pi = 760 mmHg
Ti = 0°C = 273 K
TfVi
Vf 
Ti
Vf = ?
Pf = 760 mmHg
Tf = 27°C = 300. K
(300. K)(79.4 mL)
Vf 
(273 K)
= 87.3 mL
Charles’s Law
Example 4
The volume of a sample of gas measured at 10.0°C
and 1.00 atm pressure is 6.00 L. What must the final
temperature be in order for the gas to have a final
volume of 7.00 L at 1.00 atm pressure?
T2 = ?
Convert °C into Kelvin then back into °C.
a. –30.4°C
b. 8.6°C
c. 11.7°C
d. 57.2°C
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Combined Gas Law
The volume of a sample of gas at constant
pressure is inversely proportional to the
pressure and directly proportional to the
absolute temperature.
T
The mathematical relationship: V 
P
PV
In equation form:
 constant
T
PiVi PfVf

Ti
Tf
Combined Gas Law
Example 5
Divers working from a North Sea drilling
platform experience pressure of 5.0 × 101 atm at
a depth of 5.0 × 102 m. If a balloon is inflated to
a volume of 5.0 L (the volume of the lung) at
that depth at a water temperature of 4°C, what
would the volume of the balloon be on the
surface (1.0 atm pressure) at a temperature of
11°C? Vf or V2 = ?
5 | 25
Vi = 5.0 L
Pi = 5.0 × 101 atm
Ti = 4°C = 277 K
Vf = ?
Pf = 1.0 atm
Tf = 11°C = 284 K
Tf PiVi
Vf 
Ti Pf
(284 K)(5.0  10 atm)(5.0 L)
Vf 
(277 K)(1.0 atm)
1
= 2.6  102 L
Combined Gas Law
Example 6
If 7.75 L of radon gas is at 1.55 atm and –19 °C,
what is the volume at STP?
Standard Temperature and Pressure (STP)
The reference condition for gases, chosen by
convention to be exactly 0°C and 1 atm pressure
(a)4.65 L
(b)5.37 L
(c)8.33 L
(d)12.9 L
27
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
28
Avogadro’s Law
Example 6
If 1.00 mole of a gas occupies 22.4 L at STP, what
volume would 0.750 moles occupy?
Ans: 16.8 L
29
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
30
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
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Tro: Chemistry: A Molecular Approach, 2/e
Molar Volume
32
Tro: Chemistry: A Molecular Approach, 2/e
Ideal Gas Law/Ideal gas equation
Combining the three Laws
V α 1/ P
(Boyle’s Law)
V α T
(Charles’ Law)
V αn
(Avogadro’s Law)
V α T.n/P
or V = R . T . n / P
Where R = molar gas constant
33
Ideal Gas Equation
• Rearranging the above equation gives:
• PV = nRT
R = molar gas constant = 0.0821L .atm/ (mol.K)
R = 0.0821 atm.L/(K.mol)
R = 8.3145 J/(K.mol)
R = 8.3145kg. m2 /(.K.mol)
34
Ideal Gas Law Equation
Example 7
A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1
atm at 23°C. What is the mass of nitrogen in the
cylinder? Hint: Find mol first.
V = 50.0 L
PV
n
RT
(17.1atm )(50.0L)
n
L  atm 

 0.08206
(296 K)
m ol  K 

28.02 g
mass 35.20 mol
mol
P = 17.1
atm
T = 23 C =
296 K
Ans
mass = 986 g
5 | 35
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
36
Tro: Chemistry: A Molecular Approach, 2/e
Density at Standard Conditions
Example 8:
Calculate the density of N2(g) at STP
37
Gas Density and Molar Mass
Gas Density
Using the ideal gas law, it is possible to calculate
the moles in 1 L at a given temperature and
pressure.
The number of moles can then be converted to
grams (per liter).
Molar Mass
To find molar mass, find the moles of gas, and
then find the ratio of mass to moles.
38
Gas Density and Molar Mass
Example 8
What is the density of methane gas (natural gas),
CH4, at 125°C and 3.50 atm?
Note: Since density of gases is measured in g/L, you
can assume the volume is 1L.
Hint: Find mols first, then convert to grams ;Mm =
16.04 g/mol
a)
b)
c)
d)
1.71g/L
1.72g/L
4.5g/L
0.04g/L
39
Molar Mass of a Gas
• One of the methods chemists use to determine the
molar mass of an unknown substance is to heat a
weighed sample until it becomes a gas, measure
the temperature, pressure, and volume, and use
the ideal gas law
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Tro: Chemistry: A Molecular Approach, 2/e
Gas Density and Molar Mass
Example 9
An unknown gas occupies a volume of 4.75 L at
1227 °C and 5.00atm. If the mass is 5.45 g, what
is the molar mass of gas?
(R = 0.0821 atm•L/mol•K)
Hint: Find mols first n = 0.1928
(a)21.5 g/mol
(b) 23.8 g/mol
(c )28.3 g/mol
(d)141 g/mol
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Gas Mixtures
Dalton found that in a mixture of unreactive
gases, each gas acts as if it were the only gas in
the mixture as far as pressure is concerned.
5 | 42
Originally (left), flask A contains He at 152 mmHg and flask B contains
O2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask
B (right). The new pressure in flask B is 760 mmHg.
5 | 43
Dalton’s Law of Partial Pressures
Partial Pressure
The pressure exerted by a particular gas in a
mixture.
Dalton’s Law of Partial Pressures
The sum of the partial pressures of all the different
gases in a mixture is equal to the total pressure of
the mixture:
P = PA + PB + PC + . . .
5 | 44
Dalton’s Law of Partial Pressures
Example 10
A 100.0-mL sample of air exhaled from the lungs
is analyzed and found to contain 0.0830 g N2,
0.0194 g O2, 0.00640 g CO2, and 0.00441 g water
vapor at 35°C. What is the partial pressure of
each component and the total pressure of the
sample?
Hint: Convert grams to mols
Then use ideal gas equation, PV= nRT to find
partial pressure of each gas.
5 | 45

1 mol N2
 0.0830g N2
28.01g N2


L  atm 
 0.08206
308 K 
mol  K 

 0.749 atm
PN2 


100.0mL 13L 
 10 mL 

1 mol O 2 
L  atm 
 0.0194g O 2
 0.08206
308 K 
32.00 g O 2 
mol  K 

 0.153 atm
PO2 


100.0mL 13L 
 10 mL 

1 mol CO 2 
L  atm 
 0.00640g CO 2
 0.08206
308 K 
44.01g CO 2 
mol  K 

 0.0368atm
PCO 2 


100.0mL 13L 
 10 mL 

1 mol H2 O 
L  atm 
 0.00441g H2 O
 0.08206
308 K 
18.01
g
H
O
mol

K

2


 0.0619atm
PH2O 
 1L 
100.0mL 3 
 10 mL 
5 | 46
Dalton’s Law of Partial Pressures
PN2  0.749atm
PO2  0.153atm
PCO 2  0.0368atm
PH2O  0.0619atm
P  PN2  PO2  PCO2  PH2O
P = 1.00 atm
5 | 47
Dalton’s Law of Partial Pressures
Collecting Gas Over Water
Gases are often collected over water. The result
is a mixture of the gas and water vapor.
The total pressure is equal to the sum of the gas
pressure and the vapor pressure of water.
48
Dalton’s Law of Partial Pressures
The partial pressure of water depends only on
temperature .(See Table 5.6).
The pressure of the gas can then be found using
Dalton’s law of partial pressures.
49
Collecting Gas by Water Displacement
50
Tro: Chemistry: A Molecular Approach, 2/e
Dalton’s Law of Partial Pressures
Example 10
If carbon dioxide gas is collected over water at
25 °C and775 torr, what is the partial pressure of
the CO2?
The vapor pressure of water at 25 °C is 23.8 torr.
a) 23.8 torr
b) 750 torr
C) 751 torr
d) 775 torr
e) 799 torr
51
mole fraction(X )
The concentration of any individual gas in a gas
mixture can be expressed as a mole fraction(X )
Mole fraction(X) = Moles of component
Total Moles in the mixture
Mole fraction of component 1,for example ,is
X = n1 / n1 + n2 + n3 + ... = n1 / n total
52
mole fraction(X )
Example 11
The partial pressure of air in the alveoli (the air sacs in the
lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0
mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0
mmHg. What is the mole fraction of each component of
the alveolar air?
PN2  570.0mmHg
PO2  103.0 mmHg
PCO 2  40.0 mmHg
PH2O  47.0 mmHg
5 | 53
mole fraction(X )
P  PN2  PO2  PCO 2  PH2O
570.0 mmHg
103.0 mmHg
40.0 mmHg
47.0 mmHg
P = 760.0 mmHg
5 | 54
Mole fraction of N2
Mole fraction of O2
570.0 mmHg

760.0 mmHg
Mole fraction ofCO2
103.0 mmHg

760.0 mmHg
Mole fraction of H2O
47.0 mmHg

760.0 mmHg
40.0 mmHg

760.0 mmHg
5 | 55
mole fraction(X )
X N2 = 0.7500
X O2 = 0.1355
X CO2 = 0.0526
X H2O = 0.0618
5 | 56
Kinetic-Molecular Theory
A theory, developed by physicists, that is based on
the assumption that a gas consists of molecules in
constant random motion.
Kinetic energy is related to the mass and velocity:
EK
1

mv
2
2
m = mass
v = velocity
5 | 57
Kinetic-Molecular Theory
58
Kinetic-Molecular Theory
Postulates of the Kinetic Theory
1. Gases are composed of molecules whose sizes
are negligible.
2. Molecules move randomly in straight lines in all
directions and at various speeds.
3. The forces of attraction or repulsion between
two molecules (intermolecular forces) in a gas
are very weak or negligible, except when the
molecules collide.
.
5 | 59
Kinetic-Molecular Theory
4. When molecules collide with each other, the
collisions are elastic.
5. The average kinetic energy of a molecule is
proportional to the absolute temperature
60
Kinetic-Molecular Theory
• An elastic collision occurs when the two
objects "bounce" apart when they collide.
• In an elastic collision, both momentum and
kinetic energy are conserved. Almost no
energy is lost to sound, heat, or deformation.
61
Kinetic-Molecular Theory
The Kinetic-Molecular Theory explains Boyle’s
Law
• Compressing a gas makes the V smaller but
does not alter the KE avg of the molecules
since T is constant.
• Though the speed of the particles remains
constant, the frequency of collisions increases
because the container is smaller.
• Therefore, P increases as V decreases.
62
Molecular Speeds
According to kinetic theory, molecular speeds
vary over a wide range of values. The
distribution depends on temperature, so it
increases as the temperature increases.
Root-Mean Square (rms) Molecular Speed, u
A type of average molecular speed, equal to the
speed of a molecule that has the average
molecular kinetic energy
3RT
u
Mm
5 | 63
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
64
Molecular Speed vs. Molar Mass
• To have the same average kinetic energy,
heavier molecules must have a slower average
speed
65
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
66
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
• Ee 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
67
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
68
Molecular Speeds
When using the equation
u
3RT
Mm
R = 8.3145 J/(mol · K).
T must be in Kelvins
Mm must be in kg/mol
69
Molecular Speeds
Example 12
What is the rms speed of carbon dioxide
molecules in a container at 23°C?
T = 23°C = 296 K
CO2 molar mass =
0.04401 kg/mol
u rms 
3RT
Mm
5 | 70
Recall
kg  m 2
J
s2
u rms 

kg  m 2 


2
296 K 
3 8.3145 s
mol  K 




kg 

 0.04401

mol

urms
2
m
 1.68  105 2
s
urms
m
 4.10  10
s
2
5 | 71
root-mean-square (rms) Speed
Example 13
Calculate the root-mean-square velocity for the
O2 molecules in a sample of O2 gas at 22.5°C.
(R = 8.3145 J/Kmol)
a) 132.4 m/s
b) 15.18 m/s
c) 479.9 m/s
d) 277.1 m/s
72
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
73
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Effusion
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Tro: Chemistry: A Molecular Approach, 2/e
Graham’s Law of Effusion
Thomas Graham (1805–1869)
• For two different gases at the same
temperature, the ratio of their rates of effusion
is given by the following equation:
75
Tro: Chemistry: A Molecular Approach, 2/e
Graham’s Law of Effusion
Example 14
Both hydrogen and helium have been used as
the buoyant gas in blimps. If a small leak were to
occur, which gas would effuse more rapidly and
by what factor?
1
Rate H2

Rate He
2.016

1
4.002
2.016
4.002
Hydrogen will diffuse more quickly by a factor of 1.4.
5 | 76
Graham’s Law of Effusion
Example 15
Which of the following gases will have the
slowest rate of effusion at constant
temperature?
a)CF4
b)F2
c)H2
d)Ne
77
Ideal vs. Real Gases
• 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
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Tro: Chemistry: A Molecular Approach, 2/e
Real Gases
At high pressure, some of the assumptions of
the kinetic theory no longer hold true:
1. At high pressure, the volume of the gas
molecule (Postulate 1) is not negligible.
2. At high pressure, the intermolecular forces
(Postulate 3) are not negligible.
5 | 79
The Effect of Molecular Volume
Johannes van der Waals (1837–1923)
• 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
80
Tro: Chemistry: A Molecular Approach, 2/e
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
81
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
82
Van der Waals Equation
Example 16
a)Use the van der Waals equation to calculate
the pressure exerted by 2.00 mol CO2 that has a
volume of 10.0 L at 25°C.
b)Compare this with value with the pressure
obtained from the ideal gas law.
n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
For CO2:
a = 3.658 L2 atm/mol2
b = 0.04286 L/mol
5 | 83
Van der Waals Equation
n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
Ideal gas law:
nRT
P
V
2.00 mol 0.08206 L  atm (298 K)
P

mol  K 
10.0 L
= 4.89 atm
5 | 84
n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
For CO2:
a = 3.658 L2 atm/mol2
b = 0.04286 L/mol
nRT
n2a
P
 2
V  nb V
L2  atm 
2
L  atm 


2.00 mol 0.08206
298 K  2.00 mol  3.658
2

mol
mol

K




P

L 


10.0 L 2
10.0 L  2.00 mol 0.04286

mol

P  4.933 atm  0.146 atm
Pactual = 4.79 atm
5 | 85