pv = nrt - georgia institute of...
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Week 3 CHEM 1310 - Sections L and M 1
Wed | Sep 12, 2007
Chapter 5: Gases– 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 2
PV = nRT
THE GASEOUS STATE
Pressure atmVolume litersn molesR L atm mol-1K-1
Temperature Kelvin
Ideal Gas Law
Earlier… used the Ideal Gas Law to determine mass.
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Week 3 CHEM 1310 - Sections L and M 3
PRS Question #1
What mass of argon is contained in an 18.6L containerat 20°C if the pressure is 2.35 atm?
(1) 21.9 g
(2) 72.6 g
(3) 322 g
(4) 1.82 kg
PV =mass
(MW)x RT
Week 3 CHEM 1310 - Sections L and M 4
What mass of argon is contained in an 18.6L containerat 20°C if the pressure is 2.35 atm?
(1) 21.9 g
(2) 72.6 g
(3) 322 g
(4) 1.82 kg
PRS Question #1- Solution
Mass = P x V x MW
R x T
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?
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Week 3 CHEM 1310 - Sections L and M 5
PV = nRT
Gas Density
Ideal Gas Law
PV = RTmass
(MW)
massV
= P (MW)RT
= density
Week 3 CHEM 1310 - Sections L and M 6
PRS Question #2
What is the density of carbon tetrafluorideat 1.00 atm and 50 ºC?
1) 0.0377 g/L2) 0.244 g/L3) 3.32 g/L4) 21.4 g/L
PV = nRT What do we need to do tosolve this problem?
(1) Know chemical formula
(2) Convert Ideal Gas Law intodensity equation
(3) Be mindful of units
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Week 3 CHEM 1310 - Sections L and M 7
Gas Density Calculation
What is the density of carbon tetrafluorideat 1.00 atm and 50 ºC?
Chemical Formula forcarbon 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 8
Gas Density Calculation
What is the density of carbon tetrafluorideat 1.00 atm and 50 ºC?
Density = 3.32 g/L
(1.00 atm) (88 g/mol)
(0.08206 L atm mol-1K-1) (323.15K)
Density = [P x (MW)]/RT
1) 0.0377 g/L2) 0.244 g/L3) 3.32 g/L4) 21.4 g/L
PV = nRT
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Week 3 CHEM 1310 - Sections L and M 9
PV = nRT
Molar Mass
Ideal Gas Law
PV = RTmass
(MW)
mass x= RTPV
MW
Week 3 CHEM 1310 - Sections L and M 10
Mixtures of Gases
Dalton’s Law of Partial PressuresThe total pressure of a mixture of gasesequals the sum of the partial pressures
of the individual gases.
Ptotal = PA + PB
PAV = nART
PBV = nBRT
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Week 3 CHEM 1310 - Sections L and M 11
Partial Pressures in Gas Mixtures
Ptotal = PA + PB
PA = nARTV
PB = nBRTV
Ptotal = PA + PB = ntotal RT
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Week 3 CHEM 1310 - Sections L and M 12
Mole Fractions
nA
ntotal
nB
ntotal
ntotal = nA + nB
The mole fraction of a component in a mixture isdefined as the # of moles of the components that are in
the mixture divided by the total # of moles present.
XA = XB =
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Week 3 CHEM 1310 - Sections L and M 13
Mole Percents
nA
ntotal
Mole fractions must range from 0 – 1.Multiply mole fractions by 100 for mole percents.
XA = 0.5
Mole % = 50%
x 100XA =
Week 3 CHEM 1310 - Sections L and M 14
For Ideal Gases…
PA = nARTV
Ptotal = ntotalRTV
PA nARTVPtotal ntotal RTV
= =nA
ntotal
Therefore…
PA = XA Ptotal
= XA
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Week 3 CHEM 1310 - Sections L and M 15
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)?
PSO2 = XSO2
Ptotal
Calculate
Week 3 CHEM 1310 - Sections L and M 16
Example Problem
The gaseous mixture produced contains23.2 g O2 + 53.1 g SO2 only.
# mol O2 = 23.2 g x1 mol O2
31.98 g O2
= 0.725 molO2
# mol SO2 = 53.1 g x1 mol SO2
64.06 g SO2
= 0.829 molSO2
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Week 3 CHEM 1310 - Sections L and M 17
Example Problem
What is the partial pressure of SO2(g)?
X 0.829 mol0.725 mol + 0.829 mol
= 0.533SO2=
PSO2= XSO2
Ptotal = 0.533 x 2.13 atm
= 1.14 atm
Week 3 CHEM 1310 - Sections L and M 18
Kinetic Theory of Gases
Separation by large distances compared to size
Constant movement in random directions with adistribution of speeds.
No force exerted except during collisions
Direction = straight line except between collisions
Collisions are elastic; no energy lost duringcollisions
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Week 3 CHEM 1310 - Sections L and M 19
Molecular Collisions in Gases
Greater impulseon container
walls when themass of the gas
is greater
P ∝ mass
Week 3 CHEM 1310 - Sections L and M 20
Molecular Collisions in Gases
Greater impulseon container
walls when thedensity increases
P ∝ N
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Week 3 CHEM 1310 - Sections L and M 21
Molecular Collisions in Gases
Greater impulseon container
walls when theaverage speed
increases
P ∝ (speed)2
Week 3 CHEM 1310 - Sections L and M 22
Molecular Speeds
PV = nRT PV = (1/3) Nmū2
Recall: N = nN0 and m = M/N0
nRT = (1/3) (nN0) (M/N0) ū2
RT = (1/3) Mū2 ū2 = (3RT)/M
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Week 3 CHEM 1310 - Sections L and M 23
Molecular Speeds
M
3RTuu2
rms==
NOTE: Use SI units here…R = 8.31447 J mol-1K-1, where J = kg m2 s-2
T = KM = g/mol, where you would convert to kg/mol
Week 3 CHEM 1310 - Sections L and M 24
Molecular Speeds
�
uavg =8RT
pM
NOTE: Use SI units here…R = 8.31447 J mol-1K-1, where J = kg m2 s-2
T = KM = g/mol, where you would convert to kg/mol
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Week 3 CHEM 1310 - Sections L and M 25
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 26
Molecular Motion
A gas molecule at ordinaryconditions follows a
straight path only for ashort time before colliding
with another molecule.The overall path is a
zig-zag.
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Week 3 CHEM 1310 - Sections L and M 27
Diffusion and Effusion
DIFFUSIONthe spontaneous molecular mixing of materials
(usually liquids or gases) without chemicalcombination
EFFUSIONthe spontaneous movement of the molecules of
a gas through a hole whose size is smallcompared to their mean free path
Week 3 CHEM 1310 - Sections L and M 28
Effusion
Which gas will effuse faster? How to determine this?
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Week 3 CHEM 1310 - Sections L and M 29
Comparing Effusion Rates
Molecular weight of He = 4.0025 g/molūHelium is proportional to √(1/4) = 0.5
Molecular weight of O2 = 32 g/molūOxygen 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 30
He Effuses Faster Than O2
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Week 3 CHEM 1310 - Sections L and M 31
Final Reminders
Exam Study Notes online Practice Exams– Recitation today– Online via WebAssign
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