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Lecture 37
Last Lecture:
Equations of state. The ideal gas.Kinetic model of a gas. (Chap. 18 Sec. 1-4)
This Lecture:Maxwell-Boltzmann distribution.
Van der Waals Equation. (Chap. 18 Sec. 5)
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Last time we connected the Macroscopic
and Microscopic pictures of an ideal gas
2total translational kinetic energy of molecules
3pV
23 1
2 2kT m v Average kinetic energy of one molecule
(for monoatomic ideal gas)
23
A
1.38 10 J/KR
kN
Boltzmanns constantwhere
when combined with the ideal gas law: PV = nRT
we found:
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Two separate containers of gas are in thermal equilibrium with each other.One contains He and the other contains Ar. Which of the following
statements is correct?
1. .
2. .
3. .
2 2
He Ar
2 2
He Ar
2 2
He Ar
v v
v v
v v
3 They have the same average kinetic energy.
2K E kT
21 3 The heavier mass must have slower speeds.2 2
m v kT
Clicker Question: Molecular speeds
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Root-mean-square velocities
The key quantity is mean of squared velocities
2
rmsv v
Take the square-root back to the units of velocity
2
rms
1 3
2 2mv kT
Example: 3 molecules at 400,500,600 ms-1
2 2 2
400 500 600average 500 m/s
3
400 500 600rms 507 m/s
3
Not the same!
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v
Mean free path
Assume molecules are rigid spheres of radius r. Only one moves, with speed v.
It will collide with another molecule when their centers are at distance 2r.
2r
vdt
ie, in dtit collides if there are other molecules with centers inside this cylinder:
2
2 N
dN r vdt V
Collision per unit time:24
dN Nr v
dt V
If all molecules are moving, a similar calculation gives24 2
dN Nr v
dt V
mean 2
1
4 2
Vt
dN r vN
dt
Average time between collisions:
Average displacement between collisions
(mean free path)mean
24 2
Vvt
r N
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Molecular speeds
Not all molecules have the same speed.
If we have Nmolecules, the number of molecules with speeds
between vand v + dv is:
( )dN Nf v dv
( )f v = distribution function (probability density)
= probability of finding a molecule with speed between v
and v + dv
( )f v dv
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Maxwell-Boltzmann distribution
2
3/ 2
2 /(2 )( ) 42
mv kT mf v v ekT
Maxwell-Boltzmann
distribution
higher T
higher speeds aremore probable
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Distribution = probability density
= probability of finding a molecule with speed between vand v + dv( )f v dv
Normalization:
0( ) 1f v dv
2
11 2
( ) probability of finding molecule with speeds between andv
vf v dv v v
= area under the curve
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Most probable speed, average speed, rms speed
2mp
kTv
m
Most probable speed (where
f(v) is maximum)
00
0
( ) 8( ) ...
( )
vf v dv kTv vf v dv
mf v dv
Average speed
2 20
3( ) ... kTv v f v dv m
Average squared speed
2
rms
3kTv v
m rms speed
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Beyond an ideal gas
We used this picture to derive the
Ideal Gas Equations
But real gases are more complicated
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The Van der Waals state equation
The ideal gas equation is the first approximation to the behavior of gases.Now lets start to correct the idealization of a gas.
Reduce the assumptions by including:
molecule size reduces volume to move
molecular interactions (attraction) reduces pressure
2
2
np a V nb nRT
V
a, b are determined empirically.
Different for each gas
This model is used for more extreme conditions. If the gas is diluted
(ie, n/Vis very small), the corrections are negligible.
2
21n np a V b nRT
VV
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pVdiagrams
Expansion at constant
pressure
(isobaric process)
Convenient tool to represent states and transitions from one state toanother (line represents a series of thermal states)
V
p
B
VA VB
A
states
process
If we treat the helium in the
balloon as an ideal gas, we can
predict T for each state:
A/BA/B
pVT
nR
Example: helium in balloon expanding in the room and warming up
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ThispVdiagram can describe:
V
p
B
pA
pB
A
A. A tightly closed container cooling
down.
B. A pump slowly creates a vacuuminside a closed container.
C. Either of the two processes
In either case, volume is constantand pressure is decreasing.
In case A, becauseT decreases.
In case B, because n decreases.
(isochoric process)
Clicker Question: Constant volume
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Isothermal curves
For an ideal gas,nRT
pV
(For constant n, a hyperbola for each T;Boyles Law )
1 2 3 4
T T T T Each point in apVdiagram
is a possible state (p, V, T )
Isothermal curve = all states with the same T
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A container is divided in two by a thin wall. One side contains an ideal
gas, the other has vacuum. The thin wall is punctured and disintegrates.
Which of the following is the correctpVdiagram for this process?
Initial state
Final state
2
Initial state
Final state
Initial state
Final state
3
Initial state
Final state
4
1A B
C D
Clicker Question: Free expansion
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Final state has larger V, lowerp
During the rapid expansion, the gas does NOT uniformly fill Vat a uniformp
hence it is not in a thermal state.
hence no states during process
hence this process is not represented by line
Initial state
Final state
2
Initial state
Final state
Initial state
Final state
3
Initial state
Final state
4
1A B
C D
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Further beyond the ideal gas
When a real gas is compressed, it eventually becomes a liquid
Decrease volume at constant
temperatureT2:
At point a, vapor begins to condense
into liquid.
Between a and b: Pressure and Tremain
constant as volume decreases, more of
vapor converted into liquid. At point b, all is liquid. A further
decrease in volume will require large
increase inp.
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The critical temperature
For T >> Tc, ideal gas.
critical temperature = highest temperature where a
phase transition happens.
T
p
solidliquid
gas
Triple
point
Critical point
Supercritical fluid
Critical point for water: 647K and 218 atm
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pVT diagram: Ideal gas
States are points on
this surface.
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pVT diagram: Water
Phase transitions
appear as angles.