Kinetic Theory of Gases
„
Connect microscopic properties (kinetic
energy and momentum) of molecules to
macroscopic “state” properties of a gas
(temperature and pressure).
Molecular Speeds
1
3
K = mv 2 = kT
2
2
Since
3
1
1
3
K = mv 2
and PV = NkT
∴ K = mv 2 = kT
2
2
2
Temperature is a measure of the average
kinetic energy (internal energy?) of the gas.
For constant volume, pressure increases
directly proportional to an increase in
average kinetic energy (temperature) AND
an increase in the number of molecules.
Ideal Gas Law
Using moles
PV = nRT
vp ⇒
2kT
df (v)
= 0 ⇒ vp =
m
dv
mv 2
⎛ m ⎞ 2 2 − 2 kt
f (v) = 4πN ⎜
⎟ v e
⎝ 2πkT ⎠
∞
∫0
Note
v=
f (v)dv = N
1 ∞
8kT
∫ vf (v)dv = πm
N 0
vrms =
1 ∞ 2
3kT
v f (v)dv =
∫
0
N
m
Real Gasses
Using molecules
„
PV = NkT
n = number of moles
N = number of molecules
R = universal gas constant
k = Boltzmann’s constant
= 8.315 J / (mol-K)
3kT
= vrms
m
Distribution of molecular speeds,
But
„
v2 =
The Maxwell-Boltzmann distribution
1
2 ⎛1
⎞
PV = Nmv 2 = N ⎜ mv 2 ⎟
3
3 ⎝2
⎠
„
⇒
‰
‰
„
= 1.38 x 10-23 J / K
“Real” gases do not follow the ideal gas law precisely. However, at low
pressure and temperatures “not too close” to the liquefaction point, the
ideal gas law is quite accurate and useful for “real” gases.
‰
„
Pressure not too high
Not near liquefaction point
Why the “breakdown”?
‰
This is called the Equation of State. Why “Ideal”?
•Dilute, Volume of Molecules is ~ Zero
•No Attractions Between Molecules
•Temperature must be in Absolute Units, K
Reasonable
Near liquefaction
Intermolecular forces matter
P-V Diagram
‰
‰
‰
Each line is at a constant temperature
Solid = Real, Dashed = Ideal
Point c is the critical point, curve C is critical temp.
1
Phase Diagrams
Vapor Pressure
„
Since the liquid has a distribution of
molecular speeds
‰
Some molecules will “escape”
„
„
Since the gas has a distribution of
molecular speeds
‰
Some molecules will be “recaptured”
„
Triple Point is important as a reference standard
What happens above the critical point?
„
„
‰
At equilibrium: saturated vapor
pressure (SVP)
What is the temperature on a day when the partial pressure of water is
530 Pa and the relative humidity is 40.0%?
Liquid temperature = SVP temperature
„
Boiling occurs
Law of partial pressures
‰
The total pressure of a mixture of gasses = the
sum of vapor pressure of the constituent gasses
„
„
This is temperature dependent.
Vapor pressure “explains” boiling
‰
„
Condensation
The gas will exert a “vapor” pressure
„
Vapor Pressure
Evaporation
PT = P1 + P2 + …
Relative Humidity
RH =
‰
partial pressure of H 2 O
× 100%
SVP of H 2O
Dew Point: Temperature at which unsaturated air
will become saturated.
2
Real Gasses - Better Approximations
„
„
Ideal Gas
nRT
V
First Order
‰
Clausius Equation of State (EoS)
P=
„
P=
nRT
V − nb
Second Order
‰
Van der Waals
P=
nRT
a
−
V − nb (V / n )2
Random Walk - Diffusion
„
On average, the diffusing substance
will move from a region of high
concentration to a region of low
concentration. Why?
J = DA
Mean Free Path
Molecular concentration
Assume the other molecules are not moving,
then the number of molecules in the cylinder is
NC =
N
N
Vcyl = π (2r ) 2 v Δt
V
V
Mean Free Path is average
distance between collisions
lM =
d
v Δt
1
=
=
NC N π (2r ) 2 v Δt N 4πr 2
V
V
If you account for the movement of
the other molecules
lM =
1
N
4 2πr 2
V
A student walks across a field that is 60 m x 40 m. Estimate the student’s
mean free path if there are (a) 10, (b) 100, (c) 300 other students on the
field.
C1 − C2
Δx
J = Rate of Diffusion
D = Diffusion Constant (gas dependent)
A = Cross-Sectional Area
3
„
Heat
Heat
What is heat?
„
‰
‰
„
‰
‰
It is a measure of internal energy.
It is a measure of the “closeness” of thermal equilibrium.
‰
What happens when two systems that are not in
thermal equilibrium come in contact?
‰
„
„
‰
Since heat is the “flow” of energy from one system
to another, the standard SI unit is the joule J
“Traditional” unit is the calorie (cal)
„
‰
1 cal is the heat needed to raise 1g of water from 14.5°C
to 15.5°C
„
„
Used to measure energy content of food
Conversion
„
„
4.186 J = 1 cal,
4.186 x 103 J = 1 kcal
British Thermal Unit (Btu) 1 kcal = 3.97 Btu
When the temperature of system changes, there has
been heat flow Q
Heat capacity connects heat flow to temperature change:
Q = CΔT
„
A kilocalorie (kcal) or Calorie (Cal) is 1000 cal
„
‰
„
Heat will only flow from the system with the higher
temperature to the system with the lower temperature
Heat will only flow from the system with the higher average
internal energy to the system with the lower average
internal energy
Total internal energy does not matter.
Heat Capacity
Units
‰
Heat (Q) is the “flow” or “transfer” of energy from
one system to another
Often referred to as “heat flow” or “heat transfer”
Requires that one system must be at a higher
temperature than the other
„
Energy will be transferred from the higher temperature
system to the lower energy system until thermal
equilibrium is reached – that is they have the same
temperature.
Heat
What is heat?
‰
What do we know about temperature?
‰
„
What do we mean when we say something is hot or cold?
Is “heat” the same as temperature?
„
Heat capacity C depends on the material and also on the
quantity of material present.
Eliminate quantity dependence by introducing specific
heat c and molar heat capacity c′:
Q = mcΔT
m = mass
Q = nc′ΔT
n = number of moles
4
Specific Heats and Molar Heat Capacities
Measuring Heat Capacities
„
„
Calorimeter
System is adiabatically
isolated from the environment
‰
Net Q = 0 from or into the device
Heat lost = heat gained
‰
Conservation of energy
‰
An axe head consisting of 1.8 kg of iron is left outdoors one cold winter’s night
and is brought indoors when the outside temperature is a brisk 240K. The room
into which it is brought is initially at a nice, comfortable 293K and 1.0 atm of
pressure. The volume of the room, which is well insulated, is 38 m3. Assuming
that the axe head comes to thermal equilibrium with the air in the room, by how
much is the temperature of the room lowered? (Ignore the thermal interaction
with furniture, walls, and so forth. Use cair = 0.172 cal/g-K and 28.8 g/mol for the
molecular weight of air.)
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