chapter 16 kinetic theory of gases. ideal gas model 2 1. large number of molecules moving in random...
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Chapter 16
Kinetic Theory of Gases
Ideal gas model
2
1. Large number of molecules moving in random directions with random speeds.
2. The average separation is much greater than the diameter of each molecule.
3. Obey the laws of classical mechanics; interact only when they collide. (ignore potential energy)
4. All collisions are perfectly elastic.
“Many elastic particles move randomly, no interaction”
Pressure in a gas (1)
3
like an umbrella in rain
Consider a gas in cube
pressure → collisions of the molecules
For one collision: 2 xp mv
.
. x
y
zl
Until next collision: 2
x
lt
vAverage force:
2
xmvp
Ft l
Pressure in a gas (2)
4
Average force:2
xmvpF
t l
Other kinds of collision?
For all molecules in the cube:2
ix
i
mvF
l
2
2( ) ix
ix
vv
N
2 x
mNv
l
2 2 2 2+ x y zv v v v
2 2 2= x y zv v v
22
3
x
vv random velocities
2
3
vmN
l.
. x
y
zl
. .
.
..
.
Pressure in a gas (3)
5
2
3
vmNF
lNet force:
Pressure on the wall:
/N N V
2
3
mvF NP
S Sl
Number density:
Average (translational) kinetic energy: 21
2kE mv
Pressure in a gas:2
3
N kP E
.
. x
y
zl
. .
.
..
.
S
Pressure in explosion
6
Example1: Gunpowder explodes in 10cm3 space. The explosion produces 0.1mol gas and 2×105J energy. Estimate the instantaneous pressure.
Solution: Explosion → extra pressure
2
3 N kP E 2
3
k
NE
V
2
3 totalE
V5
5 3
2 2 10
3 10
J
m
10 51.3 10 1.3 10 Pa atm
Molecular interpretation of temperature
7
Pressure in a gas:2
3
N kP E
Compare with the ideal gas law: PV NkT
3
2 kE kT
The average translational kinetic energy of molecules in an ideal gas is directly proportional to the absolute temperature.
Molecular interpretation of temperature:
Molecule kinetic energy
8
Example2: (a) What is the average kinetic energy of molecules in an ideal gas at 37℃? (b) If H2 and
O2 are both at 37℃, which kind of molecules
moves faster on average?Solution: (a) Average kinetic energy:
3
2kE kT 23 213
1.38 10 310 6.42 102
J
(b) Same T → same average kinetic energy
H Om m H Ov v 2 ?kE
vm
Mean speed and rms speed
9
Example3: 5 particles have the following speeds, given in m/s: 1, 2, 3, 4, 5. Calculate (a) the mean speed and (b) the root-mean-square (rms) speed.
Solution: (a) Mean speed: 3 /v m s
(b) rms speed : 2 2 2 2 2
2 1 2 3 4 5
5
rmsv v 11 3.32 / m s
! rmsv v
vrms for an ideal gas:
for an ideal gas? v
2 3 =k
rms
E kTv
m m
Distribution of molecular speeds
10
How many molecules move faster than vrms?
Distribution
Gauss distribution
Maxwell distribution of speeds:3
22( ) 4
2
mf v N v
kT
21
2
mv
kTe
▲ Ideal gas in equilibrium state at temperature T
f (v)
v
Maxwell distribution of speeds
11
3
22( ) 4
2
mf v N v
kT
21
2
mv
kTe
constants: N, m, k, T
f (v)
v
f (v) dv: number of molecules with speed between v and v+dv
dv
f (v)
v1 v2
Number of molecules with speed v1< v <v2 :2
1
( ) v
vN f v dv
Sum all molecules:0
( )
f v dv N
Shape of the curve
12
3
22( ) 4
2
mf v N v
kT
21
2
mv
kTe
Most probable speed vp
f (v)
v
( )0
df v
dv
vp
2 p
kTv
m
Distribution for different temperature:
f (v)
v300K
600K
Chemical reaction
& temperature
Average speed
13
Solution: How to calculate average values?
Example4: Determine the average speed of molecules in an ideal gas at temperature T.
v
( )v f v dv
f (v) dv: dN with speed between v and v+dv
Sum of speeds of dN molecules:
0( )
v f v dv Sum of speeds of all molecules:
Average speed:0
( )
v f v dv
vN
8
kT
m
Three statistical speeds
14
Most probable speed vp :
2p
kTv
m1.41
kT
m
8
kTv
m1.60
kT
m
Average speed :v
Root-mean-square speed vrms :
3rms
kTv
m1.73
kT
m
f (v)
vvp
v
vrms
p rmsv v v
Using f (v)
15
Solution: (a) Most probable speed:
Example5: 1mol H2 gas at 300K. (a) What is vp? (b)
How many molecules have speed vp < v < vp +20m/s?
(b) Speed between vp and vp +dv : dN=f (vp) dv
2p
kTv
m
2
mol
RT
m31.58 10 / m s
231
22 24
2
mv
kTmN N v e v
kT
216.33 10 ( 1%)
2vp < v < 2vp +20m/s: / 0.2% N N
3vp < v < 3vp +20m/s: / 0.003% N N
Homework
16
If the distribution of speeds in a N-particles system
is (instead of Maxwell distribution)
00
0
sin( ), 0 constant
0.
vC v v
vf v
v v
(a) C=? (b) number of particles with f(v) > C /2.
v00 v
f(v)
Real gases
17
High T & low P → ideal gas law PV=nRT
Real gas behavior? Shown in a PV diagram:
Higher pressure
molecules be closer
Ep (attractive force) can’t be ignored
molecules get even closer
Changes of phase
18
Curve D → liquefaction
Curve C → critical point c
Critical temperature
vapor & gas
PT diagram: phase diagram
boiling / freezing / sublimation
Using phase diagram
19
Solution: (a) P > 1atm → liquid; P < 1atm → gas
Example6: Describe the phase of water in different
pressure at (a) 100℃; (b) 0℃.
100℃P = 1atm → boiling point
(b) P > 1atm → liquid;
P < P0 → gas;
P0 < P < 1atm → solid
0℃
P0 .P = 1atm → freezing point
P = P0 → sublimation point
Mean free path
20
Collisions between molecules
Mean free path: average distance
traveling between collisions
P396: Molecules →
hard spheres of radius r
2
1
4 2M
N
lr
Collision frequency
21
Solution: (a) Number density at STP:
Example7: Estimate (a) the mean free path of O2
molecules at STP and (b) the average collision frequency. (r≈1.5×10-10m)
2325 3
3
6.02 102.7 10
22.4 10
N
Nm
V
Mean free path:2
1
4 2M
N
lr
89 10 600 m r
(b) Average speed 8 / 420 / v kT m m s
average collision frequency 9 1/ 4.7 10 Mf v l s
*Diffusion
22
Diffusion: substance moves from high concentrated region to
low concentrated region
Random motion of molecules:
1 2N N more molecules moves from 1 to 2 than from 2 to 1
concentrations become equal everywhere
Diffusion equation (Fick’s law): dC
J Ddx
1 2 3
*Distribution of energy
23
All molecules of atmosphere diffuse to the space?
Boltzmann’s distribution of potential energy:
0 exp( ) pN N
E
kT
number density at position Ep =0 0 :N
pE mghIn an gravity field:
NP kT 0 exp( ) mgh
P PkTconstant T
*Height of aircraft
24
Solution: By using the equation of pressure:
Example8: Air pressure gauge on an aircraft reads 0.3atm, 0 outside. What is the height? ℃
0 exp( ) mgh
P PkT
0 0ln ln mol
P PkT RTh
mg P m g P
3
8.31 273 10ln
29 10 9.8 3
9600 m