Wed | Sep 12, 2007
 Chapter 5: Gases
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Gas Stoichiometry
Partial Pressure
Kinetic Theory
Effusion and Diffusion
 Exam #1 - Friday, Sep 14
– Attendance is mandatory!
– Practice exam today in recitation
Week 3
CHEM 1310 - Sections L and M
1
THE GASEOUS STATE
Ideal Gas Law
Pressure
Volume
n
R
Temperature
PV = nRT





atm
liters
moles
L atm mol-1K-1
Kelvin
Earlier… used the Ideal Gas Law to determine mass.
Week 3
CHEM 1310 - Sections L and M
2
1
PRS Question #1
What mass of argon is contained in an 18.6L container
at 20°C if the pressure is 2.35 atm?
(1) 21.9 g
(2) 72.6 g
PV =
(3) 322 g
mass
(MW)
x RT
(4) 1.82 kg
Week 3
CHEM 1310 - Sections L and M
3
PRS Question #1- Solution
What mass of argon is contained in an 18.6L container
at 20°C if the pressure is 2.35 atm?
(1) 21.9 g
Mass =
P x V x MW
RxT
(2) 72.6 g
(3) 322 g
(4) 1.82 kg
Mass =
(2.35 atm) x (18.6L) x (39.948 g/mol)
(0.08206 L atm mol-1 K-1) x (293.15K)
Mass = 72.6 g
What else can be determined using the Ideal Gas Law?
Week 3
CHEM 1310 - Sections L and M
4
2
Gas Density
Ideal Gas Law
PV =
PV = nRT
mass
(MW)
RT
mass
= P (MW) = density
V
RT
Week 3
CHEM 1310 - Sections L and M
5
PRS Question #2
What is the density of carbon tetrafluoride
at 1.00 atm and 50 ºC?
PV = nRT
1) 0.0377 g/L
2) 0.244 g/L
What do we need to do to
solve this problem?
(1) Know chemical formula
3) 3.32 g/L
(2) Convert Ideal Gas Law into
density equation
4) 21.4 g/L
(3) Be mindful of units
Week 3
CHEM 1310 - Sections L and M
6
3
Gas Density Calculation
What is the density of carbon tetrafluoride
at 1.00 atm and 50 ºC?
Chemical Formula for
carbon tetrafluoride
CF4
Density = [P x (MW)]/RT
P = 1.00 atm; MW = 88 g/mol;
R = 0.08206 L atm mol-1K-1; T = 50 + 273.15 = 323.15K
Week 3
CHEM 1310 - Sections L and M
7
Gas Density Calculation
What is the density of carbon tetrafluoride
at 1.00 atm and 50 ºC?
PV = nRT
1) 0.0377 g/L
2) 0.244 g/L
Density = [P x (MW)]/RT
(1.00 atm) (88 g/mol)
(0.08206 L atm mol-1K-1) (323.15K)
3) 3.32 g/L
4) 21.4 g/L
Week 3
Density = 3.32 g/L
CHEM 1310 - Sections L and M
8
4
Molar Mass
Ideal Gas Law
PV =
PV = nRT
mass
(MW)
RT
RT
MW = mass x
PV
Week 3
CHEM 1310 - Sections L and M
9
Mixtures of Gases
Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases
equals the sum of the partial pressures
of the individual gases.
Ptotal = PA + PB
PAV = nART
PBV = nBRT
Week 3
CHEM 1310 - Sections L and M
10
5
Partial Pressures in Gas Mixtures
Ptotal = PA + PB
PA = nART
PB = nBRT
V
V
Ptotal = PA + PB = ntotal RT
V
Week 3
CHEM 1310 - Sections L and M
11
Mole Fractions
XA =
nA
ntotal
XB =
nB
ntotal
ntotal = nA + nB
The mole fraction of a component in a mixture is
defined as the # of moles of the components that are in
the mixture divided by the total # of moles present.
Week 3
CHEM 1310 - Sections L and M
12
6
Mole Percents
nA
XA =
XA = 0.5
x 100
ntotal
Mole % = 50%
Mole fractions must range from 0 – 1.
Multiply mole fractions by 100 for mole percents.
Week 3
CHEM 1310 - Sections L and M
13
For Ideal Gases…
Ptotal =
PA = nART
ntotalRT
V
V
PA
Ptotal
=
nARTV
ntotal RTV
=
nA
ntotal
=
XA
Therefore…
PA = XA Ptotal
Week 3
CHEM 1310 - Sections L and M
14
7
Example Problem
Some sulfur is burned in excess oxygen.
The gaseous mixture produced contains
23.2 g O2 + 53.1 g SO2 only.
Its total pressure is 2.13 atm.
What is the partial pressure of SO2(g)?
PSO = XSO Ptotal
2
2
Calculate
Week 3
CHEM 1310 - Sections L and M
15
Example Problem
The gaseous mixture produced contains
23.2 g O2 + 53.1 g SO2 only.
# mol O2 = 23.2 g x
# mol SO2 = 53.1 g x
Week 3
1 mol O2
31.98 g O2
= 0.725 mol
O2
1 mol SO2
= 0.829 mol
64.06 g SO2
SO2
CHEM 1310 - Sections L and M
16
8
Example Problem
What is the partial pressure of SO2(g)?
X SO
PSO
2
2
=
=
0.829 mol
0.725 mol + 0.829 mol
X SO Ptotal
= 0.533
= 0.533 x 2.13 atm
2
= 1.14 atm
Week 3
CHEM 1310 - Sections L and M
17
Kinetic Theory of Gases
 Separation by large distances compared to size
 Constant movement in random directions with a
distribution of speeds.
 No force exerted except during collisions
 Direction = straight line except between collisions
 Collisions are elastic; no energy lost during
collisions
Week 3
CHEM 1310 - Sections L and M
18
9
Molecular Collisions in Gases
Greater impulse
on container
walls when the
mass of the gas
is greater
P
Week 3
mass
CHEM 1310 - Sections L and M
19
Molecular Collisions in Gases
Greater impulse
on container
walls when the
density increases
P
Week 3
CHEM 1310 - Sections L and M
N
20
10
Molecular Collisions in Gases
Greater impulse
on container
walls when the
average speed
increases
P
Week 3
(speed)2
CHEM 1310 - Sections L and M
21
Molecular Speeds
PV = (1/3) Nmū2
PV = nRT
Recall:
N = nN0
and m = M/N0
nRT = (1/3) (nN0) (M/N0) ū2
RT = (1/3) Mū2
Week 3
ū2 = (3RT)/M
CHEM 1310 - Sections L and M
22
11
Molecular Speeds
u rms =
u2 =
3RT
M
NOTE: Use SI units here…
R = 8.31447 J mol-1K-1, where J = kg m2 s-2
T=K
M = g/mol, where you would convert to kg/mol
Week 3
CHEM 1310 - Sections L and M
23
Molecular Speeds
uavg =
8RT
pM
NOTE: Use SI units here…
R = 8.31447 J mol-1K-1, where J = kg m2 s-2
T=K
M = g/mol, where you would convert to kg/mol
Week 3
CHEM 1310 - Sections L and M
24
12
Molecular Speed Distribution
Temp is a measure of the average
kinetic energy of molecules when
their speeds exhibit the
Maxwell-Boltzmann
distribution.
Week 3
CHEM 1310 - Sections L and M
25
Molecular Motion
A gas molecule at ordinary
conditions follows a
straight path only for a
short time before colliding
with another molecule.
The overall path is a
zig-zag.
Week 3
CHEM 1310 - Sections L and M
26
13
Diffusion and Effusion
DIFFUSION
the spontaneous molecular mixing of materials
(usually liquids or gases) without chemical
combination
EFFUSION
the spontaneous movement of the molecules of
a gas through a hole whose size is small
compared to their mean free path
Week 3
CHEM 1310 - Sections L and M
27
Effusion
Which gas will effuse faster? How to determine this?
Week 3
CHEM 1310 - Sections L and M
28
14
Comparing Effusion Rates
Molecular weight of He = 4.0025 g/mol
ūHelium is proportional to √(1/4) = 0.5
Molecular weight of O2 = 32 g/mol
ūOxygen is proportional to √(1/32) = 0.176
Helium gas has a faster avg speed than O2 gas,
therefore He will effuse faster than O2.
Week 3
CHEM 1310 - Sections L and M
29
He Effuses Faster Than O2
Week 3
CHEM 1310 - Sections L and M
30
15
Final Reminders
 Exam Study Notes online
 Practice Exams
– Recitation today
– Online via WebAssign
 Homework
– 5-7% students forget to submit their WebAssign
homework!
– 40% students have NOT entered their 9-digit GT ID #
into Eduspace profile
Week 3
CHEM 1310 - Sections L and M
31
16