Dalton’s Law
I.
A. The total pressure of a mixture of gases equals the sum of the pressures
each gas would exert independently
1. Ptotal = P1+ P2 + …
2. Partial pressures is the pressure a gas in a mixture
would exert if it were alone in the container
3.
Particularly useful for determining the pressure a dry gas
collected over water: Ptotal = Pwet gas = Pgas + Pwater
4.
B.
P1 
Ptotal
Pwater vapor depends on the temperature, look up in table
Combining Dalton’s Law and Ideal Gases
1. We can assume each gas will behave ideally in the mixture
n RT
n 2 RT
P3  3
V
V
n RT n 2 RT n 3 RT
 RT
RT
 P1  P2  P3  ...  1


 (n 1  n 2  n 3 )
 n total 
V
V
V
V
 V
n 1RT
V
2.
P2 
It’s the total number of particles present that is important
a. The volume of the individual particle is very small
b. The forces among particles are very small



C. Problems
1. Example: 46L He and 12L O2 at 25 oC and 1atm are pumped into a 5.0L
tank. What are the partial and total pressures?
a. Calculate the number of moles of each gas from the ideal gas law
nHe 
1.0atm46L
PV

 1.9mol
RT 0.08206Latm / molK 298K 
nOx 
1.0atm12L
PV

 0.49mol
RT 0.08206Latm / molK 298K 
b. Calculate partial pressures of each gas from new conditions
PHe 
nRT 1.9mol 0.08206lat / molK 298K 

 9.3atm
5.0L
V
POx 
0.49mol 0.08206lat / molK 298K   2.4atm
5.0L
c. Add partial pressures: 9.3atm + 2.4atm = Ptot = 11.7atm
2. Mole fraction = c1 = moles of molecule 1 divided by moles total
n1
 V 

χ1 
n  P
n total
 RT 
χ1 
P1 (V/RT)
P1 (V/RT)  P2 (V/RT)  P3 (V/RT)  ....

P1
Ptotal
P1  c1PT
3. Example: Find cO2 if PO2 = 156torr in air at PT = 743torr.
a. cO2 = PO2/PT = 156torr/743torr = 0.210
b. 21% of the air molecules are oxygen
4. Example: Calculate PN2 if cN2 is 0.7808 when PT = 760torr.
PN2 = cN2 x PT = (0.7808)(760torr) = 593torr
5. Example: 0.650L of gas at 22 oC is collected over water in the
decomposition reaction of KClO3. Calculate PO2 in this gas and the
amount of KClO3 in the reaction. PH2O= 21torr at 22 oC. PT = 754 torr
a. 2KClO3(s) -------> 2KCl(s) + 3O2(g)
b. Find PO2 from Daltons Law: PO2 = PT – PH2O = 754-21 = 733torr
c. Use ideal gas law to find moles O2
n
0.964atm0.650L  0.0259mol
PV

RT 0.08206Latm / molK 295K 
d. Calculate moles KClO3 needed to make this O2.


0.0259mol O2  2mol KClO 3   0.0173mol KClO 3  122.6g   2.12 g KClO 3

3mol O 2

 mol 
II.
The Kinetic Molecular Theory of Gases
A. Empirical Laws
1. Gas Laws we have just studied
2. Describe how gases behave, but don’t explain why they behave that way
B. Theory or Model
1. Explains why gases behave as they do
2. Describing an Ideal Gas with the Kinetic Molecular Theory (KMT)
a. Gas particles very small compared to distance between them (assume
gas molecules have no volume)
b. Molecules constantly and rapidly move in a straight line until they
bump into each other or the wall (this causes pressure)
3.
C.
c.
Assume that the gas molecules’ attraction for each other is negligible
d.
Average kinetic energy is proportional to the temperature (K)
Real gas molecules do have volumes, do attract each other
Test: can the theory predict the experimental observations of PV = nRT?
1. Pressure is inversely proportional to Volume (Boyle’s Law)
a. KMT: Decrease in Volume means particles hits wall more often
b. This results in an increase in Pressure
2. Pressure is directly proportional to Temperature
a. KMT: As temperature increases, gas speed increase
b. Pressure increases as the collisions with the wall are harder
3. Volume is directly proportional to Temperature (Charles’s Law)
a. KMT: As temperature increases, gas speed increase
b. If pressure is to remain the same, the volume must increase
4. Volume is directly proportional to number of moles (Avogadro’s Law)
a. KMT: As moles increases, more collisions with the walls occur
b. If pressure is to remain the same, the volume must increase
5. Mixtures of Gases (Dalton’s Law)
a. KMT: Identity of the gas molecule doesn’t change (ideal) properties
b. Adding another gas increases pressure same as adding first gas
D. Ideal Gas Law—Derivation from KMT
1. Physics (NA = Avogadro’s number, m = mass of particle, m = velocity)


 nN A  1 mμ 2  
2

1

2


P 
(KE)avg  N A  mμ 2 

3
V
2






2  n KEavg 
PV 2
P 

 KEavg


3
V
n
3

2. KMT: average KE is directly proportional to T(K)
PV
PV
T
 RT
n
n
E. The Meaning of Temperature
1. KMT: average KE is directly proportional to T(K)
PV
2
3
2.
 RT  (KE)avg  (KE)avg  RT
n
3
2
F. Root Mean Square Velocity
2
μ rms  μ
1. Root mean square velocity = mrms
2. Deriving an expression for mrms
a.
1
 3
3RT
(KE)avg  N A  mμ 2   RT  μ 2 
 μ rms 
NAm
2
 2
b.
 particles  kg  kg
(NA )(m)  

c.
μ rms 
mol
3RT
M
3RT
NAm


 particle   mol  M  molecularmassin kg


R  8.3145 J/Kmol needed to give m  meters/second
3. Example: Calculate mrms for He at 25 oC.
μ rms 
3RT

M
2
3(8.3145 J/Kmol)(298K)
6
6 kgm
3
 1.86 x 10 J/kg  1.86 x 10

1.36
x
10
m/s
2
(4g/mol)(1kg/1000g)
kgs
4. Range of velocities of a gas sample
a. Mean free path = avg. dist. between collisions ~ 1 x 10-7 m at STP
b. Many collisions produce large range of velocities
c. 500 m/s ~ mrms at STP, but velocities are widely ranging
d. Temperature greatly effects the distribution (KMT)
G. Effusion and Diffusion
1. Effusion = movement of gas into vacuum through a small opening
a. Example: Find ratio of effusion rate for H2 and UF6.
b. Graham’s Law: Rate of effusion of gas1
M2
352.02
Rate of effusion of gas 2


M1
2.016
 13.2
2. KMT: effusion depends on average velocity of the gas particles
3RT
M1
M2
Rate of effusion of gas 1 μ rms for gas 1



Rate of effusion of gas 2 μ rms for gas 2
3RT
M1
M2
3. Diffusion = mixing of gases
a. NH3(g) + HCl(g) -------> NH4Cl(s)
b. Expected speed of mixing would allow estimation of distances:
Rate of effusion of gas NH3
Rate of effusion of gas HCl

μ rms for NH 3
μ rms for HCl

M HCl
M NH3

36.5
 1.5
17
c. Multiple collisions with air gases complicate the model of diffusion
d. The ratio of distance traveled is < 1.5; mixing time is several minutes
H. Real Gases
1. No gas is ideal, although most are close at low P and high T
2. Where does the KMT fail in describing Real Gases?
3. For an ideal gas, PV/nRT = 1 at all pressures and temperatures
203 K
4. Modifying the Ideal Gas Law
a. Real gas molecules have volume, which reduces the Volume available
b. An empirical constant b for each gas is determined
P
nRT
nRT
 P'
V
V  nb
c. Real gas molecules attract each other, making Pobs < P’
d. The higher the concentration of particles, the larger the effect
e. The number of interacting pairs depends on (concentration)2
i. N particles has N-1 partners
ii. Divide by 2 to eliminate counting each pair twice
N(N  1) N 2

2
2
f.
n
N  concentration 
V
Pobs
n
 P'a 
V
2
The correction for V and P combine in van der Waals equation
Pobs
2
2

 nRT   n 
n 
  a    Pobs  a  V - nb   nRT
 

 V - nb   V 
 V  

g. a and b are varied until the best fit of observation is found
h. Low pressure = large volume, where volume of particles is negligible
i. High temperature = fast motion, where attractions are negligible
III.
Atmospheric Chemistry
A.
B.
Components: N2 = 78%, O2 = 21%, Ar, CO2, less than 1%, H2O is variable
Smog Production in the lower atmosphere
1. Burning fossil fuels produces NOx = NO and NO2
2. NO2 + light -------> NO + O
3. O + O2 -------> O3 ------> O2 + O* (high energy O atom)
4. O* + H2O -------> 2OH radicals
5. OH + NO2 -------> HNO3 (nitric acid)
6. OH + hydrocarbons -------> photochemical smog
7. Prevalent in urban areas; harmful to respiratory system
8. Combated by public transportation, cleaner burning fuels
C.
Acid Rain
1. S(in coal) + O2 -------> SO2
2. 2SO2 + O2 -------> SO3
3. SO3 + H2O -------> H2SO4 (sulfuric acid)
4. Harmful to buildings and organisms
5. Need to remove sulfur from coal (Scrubbing)
a.
CaCO3 -------> CaO + CO2
b. CaO + SO2 -------> CaSO3 (solid calcium sulfite)