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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Effusion Which gas will effuse faster? How to determine this?
Week 3
CHEM 1310 - Sections L and M
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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
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He Effuses Faster Than O2
Week 3
CHEM 1310 - Sections L and M
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Final Reminders Exam Study Notes online Practice Exams – Recitation today – Online via WebAssign
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