chem101/3, wi2010 po 17‐1 chem101/3, wi2010 po 17‐2 Basic Assumptions
SM III
1.) Molecules have “point mass”.
(= occupy “no space”)
Kinetic Molecular Theory (KMT)
2.) Molecules are in constant, straight- line motion,
• Fundamentals
colliding with each other and the wall of the
container.
• Applications
3.) Collisions are “elastic”.
Ref
5: 6
Prob
FUP: 5: 13
(KE is exchanged but no deformation ...)
4.) No attractive/repulsive forces exist
E of C: 5: 69 - 74
see Pet. Fig. 6.16
Adv Rdg 5: 7
chem101/3, wi2010 Pet. Fig. 6.16 Molecular Motion acc. to KMT
po 17‐3 chem101/3, wi2010 General Results of KMT
At “normal conditions”, say ~ r.t., ~ 1 atm
1.) molecules move at high speed,
~ 1 km/s
2.) frequent collisions:
~ 1010 per sec.
3.) distance between collisions small,
~ 10 –7 = 100 nm;
still large when compared w/ atomic
dimensions, (~ 1000 atomic diameters)
po 17‐4 chem101/3, wi2010 po 17‐5 Detailed Results
chem101/3, wi2010 po 17‐6 Various “Speed Expressions”
1.) Molecular Speeds
Using the “basic assumptions” & simple physics,
c = urms
1
P V = 3 n MM c2
=
(root mean square)
3.) um = most probable
Illustration
2
u
u1 = 2,
u2 = 3,
1
2
3 n MM c = n R T
∴ urms
3RT
MM ; or
(kind of avg speed) ↑
u4 = 5
2+3+3+5
= 3.25
4
u =
3RT
c2 = MM
c = urms =
u3 = 3,
um = 3
P V = n R T (IGL)
∴
(= c from previous page)
2.) uavg = u
pressure and speed can be related:
1 n x MM 2
P=3
c
V
u2
1.) urms =
c∝
urms =
T
MM
see also Pet. Fig. 6.17
po 17‐7 Pet. Fig. 6.17 Distribution of Speeds for H2 at 273 K
4+9+9+25
= 3.42
4
N.B. slight difference,
as T ↑ and MM ↓
chem101/3, wi2010 22+32+32+52
=
4
chem101/3, wi2010 po 17‐8 Detailed Results ...
2.) Kinetic Energy
1
generally, KE = 2 m u2
3RT
u2 = MM
• for 1 molecule
1
1
KE = 2 mu2 = 2 m u2
MM
m= N
A
1
3 RT
1 MM 3 RT
= 2 m MM = 2 N
MM
A
3 RT
= 2 N
A
• for 1 mol
3
= ( 2 kB T)
(multiply by NA)
3
KE = 2 RT
Note: um, uav, urms , all three are ∝
T
MM
N.B. KE of a gas depends on T only !!
chem101/3, wi2010 po 17‐9 chem101/3, wi2010 po 17‐10 HT Fig. 17.1 Plots of “Maxwell - Boltzmann” Curves
Distribution of Speeds
• wide variety of speeds exists
• described by Maxwell - Boltzmann equation
ΔN
N
f(u) Δu
=
fraction of molecules
Δ is small
MM 3/2 2 –
where f(u) = 4π (
) u e
2πRT
MM u2
2RT
looks horrendous; no need to memorize;
but realize that the only variables are T & MM
N.B. Distribution curve depends on MM and T;
for illustration see HT Fig. 17.1 &
Pet.Fig. 6.18
chem101/3, wi2010 Pet. Fig. 6.18 Distribution of molecular speeds
po 17‐11 chem101/3, wi2010 Understanding Distribution Curves
po 17‐12 chem101/3, wi2010 po 17‐13 po 17‐14 Pet. Fig. 6.21 Diffusion & Effusion
Application of KMT
Effusion
chem101/3, wi2010 Diffusion
mvmt of molecules
into abs. vacuum
into existing gas
(for illustration, see Pet. Fig. 6.21)
free
obstructed by other
molecules
~ time to reach other side
~ 0.001 sec.
“mins, hours”
chem101/3, wi2010 (a) Diffusion (b) Effusion
po 17‐15 Application of Effusion
Separation of Gases w/ different MM
(e.g., 238UF6 and 235UF6)
# of molecules crossing barrier
Eff. Rate (ER) =
unit time
• faster molecules reach pinhole more often
• cross faster = ER ↑
1
MM
1
MM or ER = const.
ER1
same as : ER =
2
ER2 =
po 17‐16 Application ...
∴
Effusion Experiments useful for
• separation of gases
• determination of MM’s
• see textbook problems , Pet. 6 73 -76,
for practice
• ∴ ER ∝ u , recall u ∝
∴ER ∝
chem101/3, wi2010 MM2
MM1 or
MM1
MM2 ER1
1
MM
chem101/3, wi2010 po 17‐17 Practice Problem, Pet. 6.74
N2 effuses in 38 sec; an unknown X effuses in
64 sec. What is the molar mass (MM) of X ?
Presumably, Pet. means the “same number of
moles” effuse in both cases.
Let’s call N2 = “1” and X = “2”; so n1 = n2 .
ER1 = n1/38 sec;
From theory,
ER2 = n2/64 sec
ER2
ER1 =
MM1
MM2
ER1
MM2 = MM1 (ER )2
2
n1/38
= MM1 ( n /64 )2
2
g 64
= 28 mol ( 38 )2 , since n1 = n2
g
= 79 mol