Transcript Chapter 21: Molecules in motion

Chapter 21: Molecules in motion
• Diffusion: the migration of matter down a concentration gradient.
• Thermal conduction: the migration of energy down a temperature
gradient.
• Electric conduction: the migration of electric charge along an
electrical potential gradient.
• Viscosity: the migration of linear momentum down a velocity
gradient.
• Effusion: the emergence of a gas from a container through a small
hole.
Molecular motion in gases
•
•
•
PV = 1/3 nMc2
(21.1)
where M = mNA, the molar mass of the molecules,
c is the root mean square speed of the molecules:
c = < v2 >1/2 (21.2)
c=
 3RT 


 M 
1/ 2
(21.3)
The root mean square speed of the gas molecules is proportional to the
square root of the temperature and inversely proportional to the square
root of the molar mass.
Important concepts for gases
1/ 2
 8 RT 
c  

 M 

•
Mean speed
•
The most probable speed
1/ 2
•
 2 RT 
c*  

 M 
Relative mean speed:
1/ 2
 8kT 

c rel  21 / 2 c  
  
_
_

•
Collision frequency
m A mB
m A  mB
(reduced mass)

_
Z   c rel  
 crel P
kT
21.2 Collisions with walls and
surfaces
• The collision flux, Zw, the number of collisions with the area in a
given time interval divided by the area and the duration of the
interval.
Zw 
p
(2mkT )1 / 2
• Collision frequency can be obtained by multiplication of the collision
flux by the area of interest.
21.3 The rate of effusion
• Graham’s law of effusion: the rate of effusion is inversely
proportional to the square root of the molar mass.
rate of effusion  Z w A0 
pA0
(2mkT )1 / 2

pA0 N A
(2MRT )1 / 2
• Vapor pressures of liquids and solids can be measured based on the
above equation.
Example 21.2 Cesium (m.p. 29oC, b.p. 686 oC) was introduced into a
container and heated to 500 oC. When a hole of diameter 0.500mm
was opened in the container for 100s, a mass loss of 385 mg was
measured. Calculate the vapor pressure of liquid cesium at 500K.
Solution: Despite the effusion, the vapor pressure is constant inside
the container because the hot liquid metal replenishes the vapor.
Consequently, the effusion rate is constant!
The mass loss Δm in an interval Δt is related to the collision flux by:
Δm = ZwA0mΔt
atom.
Where A0 is the area of the hole and m is the mass of one
m
A0 mt
Zw =
p
(2mkT )1 / 2

m
A0 mt
plug in numbers, one gets p = 11k Pa
• Self-test 21.2: How long would it take 1.0 g of
Cs atoms to effuse out of the over under the
same conditions as listed in example 21.2?
(260 s)
• Self-test: There is 1.0 g of Cs solid in the
effusion oven, how long does it take to effuse
out of the oven?
•
Step 1: Derive an expression that shows how the pressure of a gas inside an
effusion oven varies with time if the oven is not replenished as the gas escapes.
•
step 2: calculate the derive of P:
dP kT dN

dt
V dt
step 3: the rate of change of the number of molecules is equal to the collision
frequency with the hole multiplied by the area of the hole:
dN
pA0
  Z w A0  
( 2mKT )1 / 2
dt
so
dP
 kT 
 

dt
 2m 
1/ 2
PA0
V
step 4: integrate the above equation
P = P0e-t/τ
1/ 2
 2m 
 

kT


V
A0
(Remark: how does the temperature (or the size of the hole) affect the decrease of
pressure?)
21.4 Transport properties of a
perfect gas
• Experimental observations on transport properties shows that the
flux of a property is proportional to the first derivative of other related
properties.
• The flux of matter is proportional to the first derivative of the
concentration (Fick’s first law of diffusion): J(matter)  dN
dz
• The rate of thermal conduction is proportional to the temperature
gradient: J(energy)  dT
dz
dN
• J(matter) = - D dz
diffusion coefficient (m2s-1);
•
J(energy) = - k dT/dz
(J K-1 m-1 s-1)
D is called the
k is called the coefficient of thermal conductivity
•
J(x-component of momentum) =

dv x
dz
, η is the coefficient of viscosity.
Table 21.3
Diffusion
1 _
D c
3
As represented by the above Figure, on average the molecules passing through the area
A at z = 0 have traveled about one free path.
The average number of molecules travels through the imaginary window A from Left to
Right during an interval Δt is
ZwA Δt
(L→R)


1
1
Because Zw = N (   ) c So
(L→R)
N (   ) c A Δt
4
4
The average number of molecules travels through the imaginary window A from Right to
Left during an interval Δt is

1
(R →L)
N (  ) c A Δt
4
The net number of molecules passing through the window A along the z direction is:

1
A
N (  ) c Δt
4
By definition the flux of molecules along z direction can be calculate as
J(z) = ( 1 N c A Δt - 1 N c A Δt )/(A Δt )
( )

1
N (   ) c A Δt 4
4
J(z) =
( )
4


1
1
N (  ) c  N ( ) c
4
4
The number density N(-λ) and N(λ) can be represented by number density N(0) at z =0
dN
dN
(
)0
N(-λ) = N(0) - λ ( ) 0 N(λ) = N(0) + λ
dz
Therefore:
then we get D =
J(z) =
1 
c
2
dz
1
dN
 c (
)0
2
dz
(different from what we expected)
A factor of 2/3
needs to be
introduced.
So we get
D=
1 
c
3
Thermal conduction
J (energy )  k
dT
dz
1 _
k =  c CV ,m [ A]
3
where CV,m is the molar heat capacity at constant volume.
Because λ is inversely proportional to the molar concentration of the
gas, the thermal conductivity is independent of the concentration
of gas, and hence independent of the gas pressure.
One exception: at very low pressure, where the mean free path is
larger than the size of the container.
•
J(x-component of momentum) =

dv x
dz
, η is the coefficient of viscosity.
Viscosity
• The viscosity is independent of
the pressure.
• Proportional to T1/2

_
1
M c[ A]
3
Measuring the viscosity
• Poiseuille’s formula:
dV ( p12  p22 )r 4

dt
16lp0
Calculations with Poiseuille’s
formula
•
Example: In a poiseuille flow experiment to measure the viscosity of air at
298K, the sample was allowed to flow through a tube of length 100cm and
internal diameter 1.00mm. The high-pressure end was at 765 Torr and the
low-pressure end was at 760Torr. The volume was measured at the latter
pressure. In 100s, 90.2cm3 of air passed through the tube.
•
Solution: Reorganize Poseuille’s equation:
( p12  p22 )r 4

16lp0 (dV / dt )
 
{( 765  133.3 Pa ) 2  (760  133.3 Pa ) 2 } (5.00  10 4 m ) 4
9.02  10 5 m 3
16  (1.00  10 m )  (760  133.3 Pa )  (
)
100 s
1
  1.82  10 4 kgm 1 s 1
(1 Pa  1kgm 1 s 2 )