Transcript Chapter 11

Chapter 11
The Behavior of Gases
Kinetic Theory
• Kinetic Theory – all molecules are in
constant motion.
– Collisions between gas molecules are perfectly
elastic.
• Diffusion – movement of molecules from
areas of high concentration to low
concentration.
• Rate of diffusion – the size and mass of the
molecule.
– Smaller, lighter molecules move faster.
Pressure
• Gas pressure – due to collisions of gas
molecules on an object.
• Atmospheric pressure – due to collisions of
air molecules on an object.
– 1 atm = 760 mm Hg = 30 in Hg = 14.7 psi
• Partial pressure – the portion of pressure
that one gas contributes to the total
pressure in a mixture of gases.
Dalton’s Law of Partial Pressure
• The total pressure of a mixture of gases is
equal to the sum of the partial pressures.
• PT = P1 + P2 + P3
• Pair = PN2 + PO2 + PCO2
Pressure vs. Moles
(at constant volume)
• Same volume containers at constant
temperature:
• If 1 mole of gas exerts 1 atm of pressure
and we add another mol of gas  twice as
many particle will have twice as many
collisions  exert twice the pressure
(2atm)
Directly proportional
2 mol.
1mol:
# moles,
P
Pressure vs. Volume
(at constant Temperature)
Start with 1 L of gas at 1 atm.
V P
½ volume  2x P
V P
2x volume  ½ P
P
1
.5
2
V
1
2
.5
Pressure vs. Volume
(at constant Temperature)
- As volume decreases, the pressure
increases proportionally.
- As volume increases, the pressure
decreases proportionally.
- As one goes up, the other goes down: P
and V are Inversely Proportional.
- P1V1 = P2V2
Boyle’s Law
• For a given mass of gas, at
constant temperature, the pressure
of the gas varies inversely with the
volume.
P1V1 = P2V2
Volume vs. Temperature
(at constant Pressure)
Start with 1 L of gas at
100 K and 1 atm.
K = oC + 273
Heat the gas the
molecules speed up
and hit the top,
pushing it to
maintain constant pressure.
200 K = 2 L
2x T  2x V
T= V
Volume vs. Temperature
(at constant Pressure)
Start with 1 L of gas at
100 K and 1 atm.
K = oC + 273
Cool the gas the
molecules slow
down, fewer
collisions w/the
top  so it falls.
50 K = ½ L
½x T ½x V
T= V
Charles’ Law
• For a given mass of gas, at
constant pressure, the volume of
the gas varies directly with its
Kelvin temperature.
V1T2 = V2T1
Pressure vs. Temperature
(at constant volume)
Start w/ 1 L at 100 K
and 1 atm.
Heat the gas  the
moles speed up and
increase the # of collisions,
which increases the
2x T = 2x P
pressure.
T = P
Pressure vs. Temperature
(at constant volume)
Start w/ 1 L at 100 K
and 1 atm.
Cool the gas  the
moles slow down and
decrease the # of
collisions,which
decreases the pressure.
½x T = ½x P
T = P
Gay-Lusaac’s Law
• For a given mass of gas, at
constant volume, the pressure of
the gas varies directly with its
Kelvin temperature.
P1T2 = P2T1
Combined Gas Law
• Combines Boyle’s, Charles’, and GayLusaac’s Laws into one equation.
•P1V1T2 = P2V2T1
• When using the combined gas law,
UNIT MUST AGREE and all
temperatures must be in Kelvin.
Moles Meets Gas Laws
• We know that the volume of a gas is
proportional to its number of particles and
the pressure of a gas is proportional to its
number of particles, which means:
• V~ # mol and P ~ # mol
or
V~n
and P ~ n
Moles Meets Gas Laws
• We also know that if the temperature of a
gas increases, its pressure increases and if
the temperature of a gas increases, its
volume increases. This means:
• T ~ P and T ~ V
• so we can write PV ~ nT
Moles Meet Gas Laws
• In order to make this proportion useful as a
mathematical expression we can derive a
constant by solving PV/nT using the values
for 1 mole of a gas at STP. This constant
will be called “R”.
• Substituting into the equation we get:
( 1 atm) ( 22.4 L)
( 1 mol) ( 273 K)
= .0821 atm L = R
mol K
Ideal Gas Law
• PV = nRT
• When using this equation, units MUST be
the same as those of the R value therefore:
– Pressure must be in ________
atm
L
– Volume must be in _____
– n must be in ________
mol
K
– Temperature must be in _____
The Ideal Gas Law applies to real and
Ideal gases under ALL conditions.
Pressure Conversions
1 atm = 760 mm Hg = 30 in Hg
= 14.7 psi = 101.3 kPa
Problem 1
• .05 moles of a gas at a temperature of 20oC
is contained in a 150 mL vessel. What is
the pressure of this gas inside the vessel?
P=
V = 150 mL = .150 L
n = .05 mol
R = .0821 atmL/mol K
T = 20oC + 273 = 293 K
PV = nRT
Problem 1: Answer
P=
V=
n=
R=
T=
PV = nRT
.150 L
.05 mol
.0821 P(.150) = (.05)(.0821)(293)
293 K
P = 8.02 atm
Problem 2
• How many grams of bromine gas at
– 10oC and 1277 mm Hg would be
contained in a 3000 mL vessel?
P = 1277 mm x (1 atm/760 mm) = 1.68 atm
V = 3000 mL = 3 L
n=
R = .0821 atmL/mol K
T = -10oC + 273 = 263 K
Problem 2: Answer
P=
V=
n=
R=
T=
PV = nRT
1.68 atm
3L
(1.68)(3) = n(.0821)(264)
.0821
263 K
.23 mol x
5.04 = 21.59n
n = .23 mol Br2
160 g
1 mol =
36.8 g
Br2 = 2(80) = 160 g
Problem 3
• 110 g of carbon monoxide at a pressure of
35.4 in Hg and a volume of 782 mL would
be at what temperature? Express your
answer in degrees Celsius.
P = 35.4 in x (1 atm/30 in) = 1.18 atm
V = 782 mL = .782 L
C
1
x
12
=
12
n = 3.93 mol
1
mol
O
1
x
16
= 16
110
g
CO
x
---------R = .0821 atmL/mol K
28 g= 28
T=
Problem 3: Answer
P=
V=
n=
R=
T=
1.18 atm
.782 L
3.93 mol
.0821
PV = nRT
(1.18)(.782) = (3.93)(.0821)T
.92 = .32T
T = 2.86 K
oC
= 2.86 K – 273 = -270.14 oC
Real vs. Ideal Gases
• Ideal Gas
• Real Gas
• Follows the gas laws
at all conditions of
temp. and pressure.
• Particles are infinitely
small (have no vol.)
• Particles are not
attracted to one
another.
• DO NOT EXIST!
• Do not follow gas
laws at all conditions
of temp. and pressure.
• Particles have
volume.
• Particles may attract
one another when
very close.
Real Gases
• Conditions at which real gases do
NOT behave as ideal gases and
therefore do not obey the gas laws:
1. At extremely high pressures  do not
obey Boyle’s Law.
2. At extremely low temperatures  do
not obey Charles’ Law.
Reasons:
• This occurs because under these two
conditions the gas molecules are close
enough together that they begin to exert
forces on one another and behave
similarly to a liquid.
• Gas Law equations are still extremely
useful because under common conditions
the behavior of a real gas is the same as
the behavior of an ideal gas.
Density and Molecular Weight
of Gases
• Density (D) = mass/volume = m/V = g/L
• Molecular Weigh (MW) = gram/mol
• For gases we know that at STP :
– 1 mol = gfm = 22.4 L = 6.02x1023 molecules
1 atm and _________.
– STP is defined as _________
0 oC
Problem 1
• What is the density of a gas with a mass of
28 g and a volume 31 L? What is its MW?
D= M =
V
MW =
28 g
31 L
g = 28 g
mol
31 L
= .9 g/L
22.4 L
x
1 mol
= 20.23 g/mol
Problem 2
• Calculate the molecular weight of a gas
with a mass of 45 g and a volume of 6.8 L.
45 g
MW = g
=
6.8 L
mol
22.4 L
x
1 mol
= 148.24 g/mol
Problem 3
• What is the density of oxygen at STP?
D= M
V
32
g
D=
22.4 L
= 1.43 g/L
O2: mass = gfm
mass = 2(16)
mass = 32 g
volume = 22.4 L
Problem 4
• What is the density of sulfur trioxide at
STP?
D= M
V
80
g
D=
22.4 L
= 3.57 g/L
SO3: mass = gfm
S 1 x 32 = 32
O 3 x 16 = 48
80
volume = 22.4 L