Transcript Document 7444094

Gases and the Kinetic-Molecular
Theory
Chapter 12
Common Properties of Gases
• Gases can be compressed into smaller volumes by
applying increased pressure
• Gases exert pressure on their surroundings
• Gases expand without limits
– Occupy the volume of any container
• Gases diffuse into one another
– Gases are miscible unless they react
• The amounts and properties of gases are described
in terms of temperature, pressure, the volume
occupied, and the number of molecules (moles)
present.
• Force per unit area
Pressure
– lb/in2 or psi
• Barometer – device for
measuring the pressure of the
atmosphere
– Measures in millimeters of Hg (mm
Hg)
– The Hg will rise until the pressure
exerted by the atmosphere equals
the pressure exerted by the column
of Hg
• 1 atmosphere = 760 Torr = 760
mm Hg (pressure at sea level)
– 1 Torr = 1 mm Hg
Note: Pressure in Rexburg is less. Why?
Pressure
• A manometer measures the pressure of a gas in a
sealed flask (page 437)
– In this case, one side of the tube is open to the
atmosphere
– If the gas pressure is greater than the atmospheric
pressure, then
Pgas(torr) = Patm(torr) + h torr
– If the gas pressure is lower than the the atmospheric
pressure, then
Pgas(torr)=Patm(torr) - h torr
Problems: On chalkboard
Illustration: Pressure apparatus
Boyle’s Law
• At constant temperature and mass, the volume of a
gas decreases when the pressure increases
– V1/P (Syringe demonstration)
• Boyle illustrated that the product of pressure and
volume were constant
– PV = k (temperature and mass are constant)
• If the amount of gas and temperature do not change,
two conditions of pressure and volume will be equal
– P1V1 = k and P2V2 = k, therefore, P1V1=P2V2
– Demo: syringe
Boyle’s Law
• Let’s do some problems with Boyle’s Law
– At 250C a sample of He has a volume of 400
mL under a pressure of 760 torr. What volume
would it occupy under a pressure of 2.00 atm at
the same T?
– A sample of oxygen occupies 15.8 L under a
pressure of 285 torr. At what pressure would it
occupy 27.9 L?
Charles’s Law
• When the pressure and the mass of a gas are
constant, the volume of a gas is proportional
to the temperature
– VT
– Charles and Gay-Lussac perfomed expansion
studies on different gases as a function of
temperature
• A quantitative relationship is not obvious on
the celsius scale. The Kelvin scale was
devised.
Charles’s Law
35
30
Volume (L) vs
Temperature (K)
25
20
15
Gases liquefy
before reaching 0K
10
5
0
0
50
100
150
200
250
300
350
400
Lord Kelvin noticed that the extension of the volume to 0,
produced a value of –273.15C. This was defined as absolute
zero on the Kelvin scale. Using this scale, a quantitative
relationship was established.
Charles’s Law
• At constant pressure and mass, the volume
occupied of a gas is directly proportional to its
absolute volume
– VT and V=kT (constant n and P)
• V/T = k
• Since V/T is constant, it can be expressed that
– V1/T1 = V2/T2 (V1/T1 = k, and V2/T2 = k)
– This is assuming that the pressure and amount of gas do
not change upon changing the conditions (volume or
temperature).
Demo: Balloon with liquid nitrogen
Charles’s Law
• Problems with Charles’s Law
– A sample of hydrogen, H2, occupies 100 mL at 250C
and 1.00 atm. What volume would it occupy at 500C
under the same pressure?
– At 112C, a sample of O2 occupies 154 mL. What
temperature would be required in C to increase the
volume to 215 mL?
• Remember to use the Kelvin scale with the gas
laws. You will get the wrong answer if you do
not.
Standard Temperature and
Pressure
• Standard temperature and pressure
– Temperature = 273.15 K (exactly 0C)
– Pressure = 1 atmosphere (760. Torr)
This serves as a reference point for discussing
gases.
Abbreviated STP
Combined Gas Law
• Boyle’s and Charles’ Laws combined into one
statement
Boyle’s
Charles’s
P1V1=P2V2
V1/T1=V2/T2
For a given sample of gas PV  k . From
T
P1 V1 P2 V2

this relationship,
. This is the combined
T1
T2
gas law.
Combined Gas Law
• Problems
– A sample of nitrogen gas, N2, occupies 750 mL
at 750C under a pressure of 810 torr. What
volume would it occupy at standard conditions?
– A sample of methane, CH4, occupies 260 mL at
32oC under a pressure of 0.500 atm. At what
temperature would it occupy 500 mL under a
pressure of 1200 torr?
Avogadro’s Law
• At the same temperature and pressure, equal
volumes of all gases contain the same
number of molecules. Therefore, at
constant temperature and pressure, the
volume of a gas is directly proportional to
the number of moles.
– Vn or V/n = k
Demo: Expanding a balloon (assume P and T
constant)
Avogadro’s Law
• Since V/n is always equal to the same
constant,
V1 V2

n1 n 2
at constant T and P
• A balloon with a volume of 2.35 L is filled with
1.82 moles of He gas. How many more moles of
He gas would have to be added to the balloon to
bring the volume to 5.49 L? Assume the T and P
are constant.
Standard Molar Volume
•
•
One mole of any gas has the same V at
STP.
This would be the standard molar volume
–
Standard molar volume = 22.4 L (this is one
mole)
What would be the volume of 0.25 moles of O2
gas be at STP?
– Deviations from the standard molar volume
(Table 12-3) indicate that the gases do not
behave ideally.
Standard Molar Volume
•
•
Using the standard molar volume densities
at various temperatures can be converted
to densities at STP
One mole of a gas occupies 36.5 L and its
density is 1.36 g/L at a given T & P.
– What is the molecular weight (g/mol)?
– What is the density at STP?
Another problem?
Deriving the Ideal Gas Law
•
•
•
•
•
•
•
Boyle’s Law - V 1/P
Charles’ Law - V T
Avogadro’s Law - V n
combination of these three - V nT/P
Ideal Gas Law V = nRT/P or PV = nRT
R = proportionality constant
R called the universal gas constant
Deriving R in the Ideal Gas Law
• One mole of an ideal gas occupies 22.414 liters at
STP (1.000 atm and 273.15 K).
1 . 0000 atm  22 . 414 L
PV
atm  L
R 

 0 . 082057
nT 1 . 0000 mole  273 .15 K
mole  K
• R can has different values depending on the units.
Make sure you are using the units that match the
gas constant.
Remember, the ideal gas law is used to describe one
set of conditions.
The Ideal Gas Law
• Problems with the ideal gas law
– What volume would 50.0 g of ethane, C2H6,
occupy at 140oC under a pressure of 1820 torr?
– Calculate the number of moles in, and the mass
of, an 8.96 L sample of methane, CH4,
measured at standard conditions.
– Calculate the pressure exerted by 50.0 g of
ethane, C2H6, in a 25.0 L container at 25oC.
Determining Molecular Weights
and Formulas
• We observed in chapter 2 that the molecular
weight must be known in order to determine
molecular formulas.
• For gases, molecular weights and formulas can
commonly be determined with the help of the
ideal gas law.
A 1.502-gram sample of a pure gaseous compound
occupies 852 mL at 34.2C and 845 Torr. What is
the molecular weight?
Determining Molecular Weights
and Formulas
• A 250.0-mL flask contains 0.585 grams of a
gaseous compound at 115C and 0.950 atm
pressure. What is the molecular weight of the
compound?
• A compound that contains only carbon &
hydrogen is 80.0% C and 20.0% H by mass. At
STP 546 mL of the gas has a mass of 0.732 g .
What is the molecular (true) formula for the
compound?
Determining Molecular Weights
and Formulas
• A 1.74 g sample of a compound that
contains only carbon & hydrogen contains
1.44 g of C and 0.300 g of H. At STP 101
mL of the gas has a mass of 0.262 gram.
What is its molecular (true) formula?
Dalton’s Law of Partial Pressure
• It can be shown utilizing the ideal gas law
that pressure exerted by a mixture of gases
is the sum of the partial pressures of the
individual gases.
– Ptotal = PA + PB + PC +…. (constant T and V)
• A, B, and C are the individual gases that are present
Look at page 456 for the derivation of Dalton’s Law
Dalton’s Law of Partial Pressures
• A 25.0-liter flask contains 0.450 moles of
ethane, 0.350 moles of oxygen, and 0.200
moles of nitrogen at 28C. What is the
pressure, in atmospheres, inside the flask?
Mole Fractions and Partial
Pressures
• The mole fraction of a component in a
mixture can be expressed as
no . mol A
XA 
total no . mol of all components
• Where XA indicates the mole fraction of A
XA 
no . mol A
no . mol A  no . mol B  .....
• Using the ideal gas law (page 458), the gas
mole fraction can be expressed in terms of P
XA 
PA
P
and X B  B
Ptotal
Ptotal
PA  X A  Ptotal and PB  X B  Ptotal
Mole Fractions and Partial
Pressures
• A sample of air was analyzed. The total
number of moles in the sample was 0.582.
The number of moles of O2 in the sample
was 0.185. If the total pressure is 692 torr.
What is the partial pressure of oxygen?
• Example 12-18 in books
Vapor Pressure
The total pressure in a sealed container containing water is
affected by the pressence of water. Gaseous molecules
leaving the liquid phase produces a contribution to the total
pressure. This contribution is called the vapor pressure of
water. The vapor pressure of water increases with
increasing temperature (Table 12-4). Actually, every liquid
has a unique vapor pressure.
Any gas in contact with water soon becomes saturated with
water vapor.
Producing Gases over Water (by
displacement)
• The pressure inside a sealed container is the
sum of the partial pressure of H2O(l) (or
vapor pressure) and the partial pressure of
the other gas(es) that is present.
Patm  Pgas  PH 2 O or Pgas  Patm  PH 2 O
– A sample of hydrogen was collected over water
at 250C. The volume in the collection flask is
250.0 mL and the pressure is 748 torr. What is
the partial pressure of the hydrogen gas in the
collection flask? How many moles of H2 and
H2O(g) are in the container?
Producing Gases over Water
• Demonstration: 0.292 grams of magnesium is
reacted with 14.0 mL of 1.00 M HCl. Determine
the limiting reactant. The pressure inside the flask
(Patm) is 642 torr. Determine the total volume
occupied.
• A sample of oxygen was collected by
displacement of water. The volume in the
collection flask was 742 mL at 27oC. The
barometric pressure was 753 torr. What volume
would the dry oxygen occupy at STP?
Mass-Volume Relationships in
Reactions Involving Gases
• Many chemical reactions produce gases.
– NaHCO3(s) + CH3COOH(aq) 
NaCH3COO(aq) + CO2(g) + H2O(l)
• 0.790 grams of NaHCO3 is reacted with
3.00 mL of 6.00 M CH3COOH. Determine
the limiting reactant. How much CO2(g) is
produced? Is this a displacement reaction?
Mass-Volume Relationships in
Reactions Involving Gases
• What volume of CO2(g) is produced when 1.519
grams of C4H8 is reacted with excess O2 (this is a
combustion reaction)? The volume of gas is
measured at STP.
• How many L of O2(g) is produced when 252
moles of KClO3 is decomposed according to the
reaction below? The amount of O2(g) was
measured at 353 K and 692 Torr.
– KClO3(s)  KCl(s) + O2(g)
The Kinetic-Molecular Theory
Assumptions
• Gases consists of small discrete molecules that are far
apart
– Molecule size is very small compared to the total volume
• The gas molecules are in constant straight-line motion
with varying velocities
• The collisions experienced by the gas molecules are
elastic
– No energy loss
• There are no attractive or repulsive forces acting
between the gas molecules
The Kinetic-Molecular Theory
• The average kinetic energy of molecules is
directly proportional to the temperature.
1
– Can be expressed as mv 2
2
• The average KE of molecules of different
gases are equal at any given temperature
– This means that smaller molecules will have
greater velocities at a given temperature
Average molecular KE  T and Average molecular velocity 
T
molecular weight
The Kinetic-Molecular Theory
• Only the average temperature of the gas
molecules has been addressed. There will
actually be a distribution of molecule
speeds that depends on both the temperature
and the mass of the gaseous molecules
(refer to page 465)
– An excellent discussion on the kineticmolecular theory is presented in the enrichment
section and 12.11 on the CD.
The Kinetic-Molecular Theory
and Boyle’s Law
• The pressure is a result of gas molecules colliding
with the walls of the container
– Rate of molecules colliding with the walls
– Velocity of molecules colliding with the wall
• Decreasing the volume increases the pressure
since there will be a higher rate of collisions in the
reduced volume.
– The average velocity of molecules remains the same.
The Kinetic-Molecular Theory
and Dalton’s Law
• Molecules are far apart and do not interact
with each other significantly.
– Each gas acts independently of the other
molecules when colliding with the container
• Frequency and velocity of collisions
Therefore, each gas exerts a partial
pressure, and the total pressure is a sum of
all the molecule-wall collisions.
The Kinetic-Molecular Theory
and Charles’s Law
• Average KE is proportional to the
temperature
– Increasing the temperature causes the speed of
the collisions to increase. Normally, this would
increase the pressure. Volume has to increase
to keep the pressure constant.
– The reverse occurs when the temperature
decreases.
Diffusion and Effusion of Gases
• Diffusion – intermingling or mixing of
gases
• Effusion – gases passing through porous
(possessing small holes) containers
• Rate of effusion or diffusion is inversely
proportional to the square roots of the
molecular weights
rate of diffusion of A

rate of diffusion of B
MW of B
MW of A
Diffusion and Effusion of Gases
• Demo: U-tube and balloons (N2 and He)
• Calculate the ratio of the rate of effusion of
He to that of sulfur dioxide, SO2, at the
same T & P.
• The diffusion rate of CH4 is 4.5 m/s in air.
What is the diffusion rate of C3H8 in air?
• Another problem on CD.
Diffusion and Effusion of Gases
•
The kinetic-molecular theory can be used
to calculate an average speed of gaseous
molecules that behave ideally.
u rms
–
3 RT

M
urms is root-mean square speed of the gaseous
molecule. Notice that this speed is inversely
related to the square root of the molecular
weight.
It should be noted, however, that the urms is not
equal to the diffusion rate of a gas through air.
Why?
Diffusion and Effusion of Gases
• What is the urms of Xe and He? This would
be the distance covered by one molecule in
one second. If no collisions occurred, this
would be the distance traveled by the
gaseous atom from a certain starting point.
Since collisions do occur on this planet, the
distance traveled away form a certain
starting point is much less.
Deviations from the Ideal Gas
Behavior
• Under ordinary conditions of pressure and
temperature most gases behave ideally.
Nonideal gas behavior becomes significant
at high pressures and/or low temperatures.
– Nonideal behavior is especially evident near the
conditions under which the gas turns to a liquid
• Johanes van der Waals studied the nonideal
behavior of gases and adjusted the ideal gas
equation.
Factors Contributing to Nonideal
Gas Behavior
• Upon increasing the pressure, the volume of
the gas molecules becomes important to the
overall/available volume
– Videally available = Vmeasured – nb
• nb corrects for the volume occupied by the molecules
• The value of ‘b’ increases with the size of the molecule
• n is equal the number of moles
How will Videally available vary with He and C3H8?
Factors Contributing to Nonideal
Gas Behavior
• As the pressure increases and the temperature
decreases, the gaseous molecules move more slowly
due to attractive forces between molecules. This will
lower the predicted pressure due to the reduced
number of collisions.
n 2a
Pideally exerted  Pmeasured  2
Vmeasured
– Large values of ‘a’ indicate strong attractive forces. More
molecules and smaller volumes also produces a large
correction term.
The Van der Waals Equation

n 2a 
 P  2 V  nb   nRT
V 

• P, V, T, and n represent the measured values just like
the ideal gas equation.
• The values, a and b, are experimentally determined
and will be provided. They are different for each
gas.
– a accounts for attractive forces between gas molecules
– b accounts for the volume of the gas molecule
The Van der Waals equation reduces to the ideal gas law at
high temperatures and low pressures. Why?
Some Problems with the Van der
Waals Equation
• Calculate the pressure exerted by 84.0 g of
ammonia, NH3, in a 5.00 L at 2000C using
the ideal gas law. Solve Example 12-17
using the van der Waal’s equation.
– What is the difference?
• Another problem.