Transcript CHAPTER 12

Unit 8: Gas Laws
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Comparison of Solids, Liquids, & Gases
• The density of gases is much less than that of
solids or liquids.
• Gases are easily compressed and they
completely fill any container in which they
occupy
– Tells us that gas molecules are far apart &
interactions among them are weak
• Gases exert pressure on their surroundings;
in turn, pressure must be exerted to confine a
gas
• Gases diffuse into one another (i.e. they are
miscible, unless they react with one another)
– Mix completely e.g. air is a mixture of gases
– Conversely, different gases in a mixture do not
separate on standing
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Pressure
• Pressure is force per unit area.
– lb/in2 commonly known as psi
• Atmospheric pressure is measured using
a barometer.
• Definitions of standard pressure
–
–
–
–
–
1 atmosphere
760 mm Hg
76 cm Hg
760 torr
101.3 kPa
Hg density = 13.6 g/mL
Mercury barometer: the air pressure is measured in
terms of the height of the mercury column i.e. the
vertical distance between the surface of the mercury
in the open dish and that inside the closed tube. The
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pressure exerted by the atmosphere is the pressure
exerted by the column of mercury
Boyle’s Law: Inverse relationship of pressure &
volume
p = 1 atm
p = 2 atm
p = 4 atm
10 L
5L
Pressure
doubled
25°
C
Volume
reduced
by half
Pressure
doubled
25°
C
Volume
reduced
by half
2.5 L
25°
C
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Boyle’s Law: The Volume-Pressure Relationship
Graphical Representation of Boyle’s Law
• At constant temp and with equal # mols, volume of a gas
varies inversely with pressure:V α 1
P

V = 1 k1
P

PV = k1
• The product of volume (V) and pressure (P) is a constant
– This allows us to equate initial and final conditions to solve practical
problems.
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Fig. 12-4, p. 405
Boyle’s Law: The Volume-Pressure
Relationship
• Used to calculate:– Volume resulting from pressure change
– Pressure resulting from volume change
• P1V1 = k1 for one sample of a gas.
• P2V2 = k2 for a second sample of a gas.
• k1 = k2 for the same sample of a gas at the same T.
• Thus we can write Boyle’s Law mathematically as:
P1V1 = P2V2
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Boyle’s Law: The Volume-Pressure
Relationship
• Example 1: At 25oC a sample of He has a volume of
4.00 x 102 mL under a pressure of 7.60 x 102 torr.
What volume would it occupy under a pressure of
2.00 atm at the same T?
P
2
P11 V
V11 PP2 2VV
P1 V1  P2 V22
P1 V1
V2  P1 V1
V2  P2
P2



760
torr
400
mL

760 torr 400 mL 


1520
torr
1520 torr
 2.00 10 2 mL
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Boyle’s Law: The Volume-Pressure
Relationship
• Notice that in Boyle’s law we can use any
pressure or volume units as long as we
consistently use the same units for both P1 and
P2 or V1 and V2.
• Use your intuition to help you decide if the
volume will go up or down as the pressure is
changed and vice versa.
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Charles’ Law: The Volume-Temperature Relationship
p = 1 atm
p = 1 atm
p = 1 atm
4L
2L
1L
273 K
Temperature (K)
doubled
Volume
doubles
Temperature (K)
doubled
546 K
Volume
doubles
1092 K
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Charles’ Law:
The Volume-Temperature Relationship
• Charles’s law states that the volume of a gas is
directly proportional to the absolute temperature
at constant pressure.
– Gas laws must use the Kelvin scale to be correct.
• Relationship between Kelvin and centigrade.
K = o C + 273
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Charles’ Law:
The Volume-Temperature Relationship
• Volume of a gas varies directly with temperature if the
pressure and # of mols of gas remain constant
V α T  V = Tk2
 V = k2
T
• Ratio of volume (V) and temperature (T) is a constant
• This allows us to equate initial and final conditions to solve
practical problems
V1 V2

T1 T2
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Charles’ Law: The Volume-Temperature Relationship
• Example 2: A sample of hydrogen, H2, occupies
1.00 x 102 mL at 25.0oC and 1.00 atm. What
volume would it occupy at 50.0oC under the same
pressure?
T1 = 25 + 273 = 298
T2 = 50 + 273 = 323
V1 V2
V1T2

 V2 =
T1 T2
T1
1.00 10 2 mL  323 K
V2 =
298 K
 108 mL
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Standard Temperature and Pressure
• Have seen that both temperature & pressure affect the
volume of a gas
• Often convenient to choose some “standard” temperature
and pressure as a reference point for discussing gasses.
• Standard temperature and pressure is given the symbol
STP.
– Standard P  1 atm or 101.3 kPa
– Standard T  273 K or 0 oC
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The Combined Gas Law Equation
• Boyle’s and Charles’ Laws combined into one statement
is called the combined gas law equation.
– Useful when the V, T, and P of a gas are changing.
Boyle' s Law
Boyle' s Law
P1V1  P2 V2
P1V1  P2 V2
Charles' Law
Charles'
V1 VLaw
2

T1V1 TV2 2
T1
T2
For a given sample of gas : The combined gas law is :
PV
P1 V1 P2 V2
k

T
T1
T2
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The Combined Gas Law Equation
• Example 3: A sample of nitrogen gas, N2, occupies 7.50
x 102 mL at 75.00C under a pressure of 8.10 x 102 torr.
What volume would it occupy at STP?
V1 = 750 mL V2 = ?
T1 = 348 K
= 273mL
K
V1T=2 750
V2 = ?
P1 = 810 torrT1 =P348
K torrT2
2 = 760
P1 = 810
P torr
VT
Solve for V2 =
1
1
2
SolvePfor
2 T1V2
= 273 K
P2 = 760 torr
P1 V1 T2
=
P2 T1

810 torr 750 mL 273 K 

760 torr 348 K 
 627 mL
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Avogadro’s Law
• Mathematically Avogadro’s law can be stated as: At
constant temperature & pressure, the volume, V, occupies
by a gas is directly proportional to the number of mols,
n, of gas
Vαn
or V = kn
or
V = k
n
(constant temp & pressure)
• For 2 samples of a gas at the same temperature & pressure,
the relation between volumes and number of moles can be
represented as:
V1 = V2
n1
n2
(constant T , P)
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Using Avogadro’s Law
• Example 4: If 5.50 mol of CO occupy 20.6 L, how
many liters will 16.5 mol of CO occupy at the same
temperature and pressure?
• What do we know?
– n1 = 5.50 mol
– V1 = 20.6 L
n2 = 16.5 mol
V2 = ? L
– V2 = V1n2 = (20.6 L)(16.5 mol)
n1
(5.50 mol)
= 61.8 L CO
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Avogadro’s Law and the Standard Molar Volume
• Avogadro’s Law states that at the same temperature and
pressure, equal volumes of two gases contain the same
number of molecules (or moles) of gas.
• If we set the temperature and pressure for any gas to be
STP, then one mole of that gas has a volume called the
standard molar volume.
• The standard molar volume is 22.4 L at STP.
– This is another way to measure moles.
– For gases, the volume is proportional to the number of moles.
• 11.2 L of a gas at STP = 0.500 mole
– 44.8 L = ? moles
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Summary of Gas Laws: The Ideal Gas Law
Boyle’s Law - V  1/P (at constant T & n)
Charles’ Law – V  T (at constant P & n)
Avogadro’s Law – V  n (at constant T & P)
Combine these three laws into one statement
V  nT/P
• Convert the proportionality into an equality.
V = nRT/P
• This provides the Ideal Gas Law.
PV = nRT
•
•
•
•
– R = 0.0821 L . atm / mol . K
– It is a proportionality constant called the universal gas constant.
An ideal gas is one that exactly obeys these gas laws. Many gases
show slight deviations from ideality, but at normal temperatures, &
pressures the deviations are usually small enough to ignore
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Summary of Gas Laws: The Ideal Gas Law
•
Example 5: What volume would 50.0 g of ethane, C2H6,
occupy at 1.40 x 102 oC under a pressure of 1.82 x 103
torr?
–
1.
2.
3.
To use the ideal gas law correctly, it is very important that all of
your values be in the correct units!
T = 140 + 273 = 413 K
P = 1820 torr (1 atm/760 torr) = 2.39 atm
50 g (1 mol/30 g) = 1.67 mol
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Summary of Gas Laws: The Ideal Gas Law
PV = nRT 
nRT
V=
P
nRT
V=
nP
RT
V=
L atm 

1.67 mol 0.0821 L atm413 K 
mol K 413 K 



1
.
67
mol
0
.
0821


mol K 

2
.
39
atm

P
2.39 atm
 23.6 L
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Dalton’s Law of Partial Pressures
An illustration of Dalton’s law. When the 2 gases A and B
are mixed in the same container at the same temperature,
they exert a total pressure equal to the sum of their partial
pressures.
Dalton’s Law of Partial Pressures
• Dalton’s law states that the pressure exerted by
a mixture of gases is the sum of the partial
pressures of the individual gases.
Ptotal = PA + PB + PC + ....
– Where:
PA = nA RT
V
PB = nBRT
V
PC = nCRT
V
etc
– The pressure that each gas exerts in a mixture is called its
partial pressure. No way has been devised to measure the
pressure of an individual gas in a mixture; it must be calculated
from other quantities.
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Dalton’s Law of Partial Pressures
• Example 6: If 100 mL of hydrogen, measured at 25.0 oC
and 3.00 atm pressure, and 100 mL of oxygen,
measured at 25.0 oC and 2.00 atm pressure, were forced
into one of the containers at 25.0 oC, what would be the
pressure of the mixture of gases?
PTotal  PH 2  PO 2
 3.00 atm + 2.00 atm
= 5.00 atm
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Dalton’s Law of Partial Pressures
• Example 7: A 10.0 L flask contains 0.200 mol of
methane, 0.300 mol of hydrogen and 0.400 mol of
nitrogen at 250C
– (a) What is the pressure (in atm) inside the flask?
– (b) What is the partial pressure of each component of
the mixture of gases?
(a) Solution:
• Given # mols of each component. Can use ideal gas
equation to find total pressure from # mols:
ntotal = 0.200 + 0.300 + 0.400 = 0.900 mols
V= 10.0L T = 25 + 273 = 298K
Ptotal = nRT = (0.900mol)(0.0821L.atm / mol.K) (298K) = 2.20atm
V
10.0L
Dalton’s Law of Partial Pressures
(b) Solution:
• Now we find partial pressures for each component:
PCH4 = (nCH4)RT = (0.200)(0.0821)(298) = 0.489 atm
V
10.0
Similar calculations for N2 and H2 give:
PH2 = 0.734 atm
PN2 = 0.979 atm
• As a check, the sum of all the partial pressures should
be equal to the total pressure (Dalton’s Law):
0.489 + 0.743 + 0.979 = 2.20 atm
• Problem solving tip: Sometimes the amount of gas is expressed in
other units that can be converted to mols. E.g. molar mass. Can
then convert mass  mols
Mole Fraction and Partial Pressure
• Can describe the composition of any gas mixture
in terms of the mole fraction of each component
.
• The mole fraction, XA, of component A in a
mixture is defined as:
XA =
no. mol A
total no. mol of all components
• Like any other fraction, mole fraction is
dimensionless.
• Can relate the mole fraction of each component
to its partial pressure
Mole Fraction and Partial Pressure
XA = PA
Ptotal
similarly,
XB = PB
Ptotal
and so on
• We can rearrange these equations to give another
statement of Dalton’s Law of Partial Pressures:
PA = XA Ptotal
PB = XB Ptotal
and so on
The partial pressure of each gas is equal to the
mole fraction in the gas mixture times the toal
pressure of the mixture
Mole Fraction and Partial Pressure
• Example 14: Find the mole fractions of the gases in
example 13.
– A 10.0 L flask contains 0.200 mol of methane, 0.300 mol of
hydrogen and 0.400 mol of nitrogen at 250C
• Solution, using the mols given:
– X methane = nmethane / ntotal
– X hydrogen = nhydrogen / ntotal
– X nitrogen = nnitrogen / ntotal
= 0.200 /0.900 = 0.222
= 0.300/ 0.900 = 0.333
= 0.400/ 0.900 = 0.444
• Alternatively, could use partial & total pressure in
Example 13 (b):
– X methane = Pmethane / P total
– X hydrogen = Phydrogen / P total
– X nitrogen = Pnitrogen / P total
= 0.489 atm / 2.20 atm = 0.222
= 0.734 atm / 2.20atm = 0.334
= 0.979 atm / 2.20 atm = 0.445
The Kinetic-Molecular Theory
The basic assumptions of the kinetic-molecular
theory for an ideal gas (one that obeys the gas
laws):1.Gases consist of discrete molecules which are
small and very far apart relative to their own size,
and occupy no volume (they can be considered as
points)
–
The observation that gases can be easily
compressed supports this.
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The Kinetic-Molecular Theory
2. The gas molecules are in continuous random,
straight-line motion with varying speeds.
3. The collisions between gas molecules and with
the walls of the container are elastic, (that is, no
energy is gained or lost during the collision).
–
At any given instant only a small fraction of the gas
molecules are involved in collisions.
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The Kinetic-Molecular Theory
4. There are no attractive or repulsive forces
between molecules.
–
–
NOTE: at HIGH pressures and LOW temperatures
(conditions under which a gas liquefies attractions
and repulsions between gas molecules become
significant and the gas behaves non-ideally.
A real gas is one that does not behave as an ideal
gas due to interactions between gas molecules.
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