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Transcript
```Chapter 14: The Ideal Gas Law and Kinetic Theory
Fluids – Chapter 11 - cover
14.1: Molecular Mass, the Mole and Avogadro’s Number.
14.2: The Ideal Gas Law.
14.3: Kinetic Theory of Gases.
14.2 Kinetic Theory of Gases
The particles are in constant,
random motion, colliding with
each other and with the walls of
the container.
Each collision changes the
particle’s speed.
As a result, the atoms and
molecules have different
speeds.
14.2 Kinetic Theory of Gases
THE DISTRIBUTION OF MOLECULAR SPEEDS
An apparatus used to
measure atomic velocities.
J.A. Eldridge 1927 .
(Recall problem 8-15.)
14.2 Kinetic Theory of Gases
Measuring temperature
The molecules of the fluid carry momentum and exert an impulse on the
surface when they bounce from it. We can calculate the average force
on the wall exerted by the colliding molecules.
14.2 Kinetic Theory of Gases
With every collision an impulse of Δ(mv)
is imparted to the wall.
where v is the component of velocity in
the “normal” direction to the wall.
14.2 Kinetic Theory of Gases
If a single molecule repeatedly strikes
the wall with a mean time of Δt between
collisions, then the average force exerted
on the wall by this molecule is:
14.2 Kinetic Theory of Gases
If a single molecule repeatedly strikes
the wall with a mean time of Δt between
collisions, then the average force exerted
on the wall by this molecule is:
14.2 Kinetic Theory of Gases
So, for a single molecule, the average
force exerted on the wall is:
For N molecules in the container, one
third of them (on average) strike a wall
during time Δt .
The total force on a wall is thus:
The quantity
is the average value of the squared speed.
14.2 Kinetic Theory of Gases
The quantity
is the average value of the squared speed since
we are dealing with a (Maxwellian) distribution of speeds.
We can define the “root-mean-square” or “rms” speed:
14.2 Kinetic Theory of Gases
Having calculated the force on a wall,
we can now calculate the pressure, P = F / A
(where A is the wall’s area), inside the
container:
where V is the container volume.
14.2 Kinetic Theory of Gases
We have the pressure:
Notice:
We can identify:
as the average (translational) kinetic energy of
a molecule in the gas. So:
14.2 Kinetic Theory of Gases
We have (from kinetic theory):
(the ideal gas law).
This gives a connection between the
temperature of a gas and the average kinetic
energy of the constituent molecules:
14.2 Kinetic Theory of Gases
Problem 14-44: A cubical box with side of length 0.300 m
contains 1.000 moles of neon gas at room temperature (293 K).
What is the average rate (in atoms/s) at which neon atoms
collide with one side of the container? The mass of a single
neon atom is 3.35 x 10-26 kg.
14.2 Kinetic Theory of Gases
Problem 14-44: A cubical box with side of length 0.300 m
contains 1.000 moles of neon gas at room temperature (293 K).
What is the average rate (in atoms/s) at which neon atoms
collide with one side of the container? The mass of a single
neon atom is 3.35 x 10-26 kg.
Solution:
1.000 mole of neon → number of neon atoms inside is equal to
The collision rate with a single wall is:
where t is the (mean) time between collisions for any single
atom striking the wall.
14.2 Kinetic Theory of Gases
Problem 14-44: A cubical box with side of length 0.300 m
contains 1.000 moles of neon gas at room temperature (293 K).
What is the average rate (in atoms/s) at which neon atoms
collide with one side of the container? The mass of a single
neon atom is 3.35 x 10-26 kg.
Solution:
For a cubical container of side length L, the mean time between
collisions is:
14.2 Kinetic Theory of Gases
Problem 14-44: A cubical box with side of length 0.300 m
contains 1.000 moles of neon gas at room temperature (293 K).
What is the average rate (in atoms/s) at which neon atoms
collide with one side of the container? The mass of a single
neon atom is 3.35 x 10-26 kg.
Solution:
Thus, the collision rate in the cubical container (1.000 moles of
gas) is:
14.2 Kinetic Theory of Gases
Problem 14-44: A cubical box with side of length 0.300 m
contains 1.000 moles of neon gas at room temperature (293 K).
What is the average rate (in atoms/s) at which neon atoms
collide with one side of the container? The mass of a single
neon atom is 3.35 x 10-26 kg.
Solution:
Calculate the mean squared velocity using the connection
between the temperature of the gas and the average kinetic
energy of the constituent atoms:
14.2 Kinetic Theory of Gases
Problem 14-44: A cubical box with side of length 0.300 m
contains 1.000 moles of neon gas at room temperature (293 K).
What is the average rate (in atoms/s) at which neon atoms
collide with one side of the container? The mass of a single
neon atom is 3.35 x 10-26 kg.
Solution:
We now have the collision rate:
14.2 Kinetic Theory of Gases
Problem 14-44: A cubical box with side of length 0.300 m
contains 1.000 moles of neon gas at room temperature (293 K).
What is the average rate (in atoms/s) at which neon atoms
collide with one side of the container? The mass of a single
neon atom is 3.35 x 10-26 kg.
Solution:
Inserting numerical data:
14.2 Kinetic Theory of Gases
Problem 14-38: Two gas cylinders are identical. One contains
the monatomic gas argon (Ar), and the other contains an equal
mass of the monatomic gas krypton (Kr). The pressures in the
cylinders are the same, but the temperatures are different.
Determine the ratio of the average kinetic energy of a krypton
atom to the average kinetic energy of an argon atom.
14.2 Kinetic Theory of Gases
Problem 14-38: Two gas cylinders are identical. One contains
the monatomic gas argon (Ar), and the other contains an equal
mass of the monatomic gas krypton (Kr). The pressures in the
cylinders are the same, but the temperatures are different.
Determine the ratio of the average kinetic energy of a krypton
atom to the average kinetic energy of an argon atom.
Solution:
The ratio of the average kinetic energies is equal to the ratio of
the Kelvin temperatures of the gases.
14.2 Kinetic Theory of Gases
Problem 14-38: Two gas cylinders are identical. One contains
the monatomic gas argon (Ar), and the other contains an equal
mass of the monatomic gas krypton (Kr). The pressures in the
cylinders are the same, but the temperatures are different.
Determine the ratio of the average kinetic energy of a krypton
atom to the average kinetic energy of an argon atom.
Solution:
We are not given the temperatures but these are related to the
given information by the ideal gas law:
We have gas containers of same V with equal pressures P
containing the same mass, m, of gas. We will need to
determine n (ratio) for the two cylinders.
14.2 Kinetic Theory of Gases
Problem 14-38: Two gas cylinders are identical. One contains
the monatomic gas argon (Ar), and the other contains an equal
mass of the monatomic gas krypton (Kr). The pressures in the
cylinders are the same, but the temperatures are different.
Determine the ratio of the average kinetic energy of a krypton
atom to the average kinetic energy of an argon atom.
Solution:
We will need to determine n (ratio) for the two cylinders.
14.2 Kinetic Theory of Gases
Problem 14-38: Two gas cylinders are identical. One contains
the monatomic gas argon (Ar), and the other contains an equal
mass of the monatomic gas krypton (Kr). The pressures in the
cylinders are the same, but the temperatures are different.
Determine the ratio of the average kinetic energy of a krypton
atom to the average kinetic energy of an argon atom.
Solution:
The number of moles can be calculated from the mass m (in
grams) and the mass per mole M (in grams per mole):
14.2 Kinetic Theory of Gases
Problem 14-38: Two gas cylinders are identical. One contains
the monatomic gas argon (Ar), and the other contains an equal
mass of the monatomic gas krypton (Kr). The pressures in the
cylinders are the same, but the temperatures are different.
Determine the ratio of the average kinetic energy of a krypton
atom to the average kinetic energy of an argon atom.
Solution:
The kinetic energy ratio is thus given by:
14.2 Kinetic Theory of Gases
Problem 14-38: Two gas cylinders are identical. One contains
the monatomic gas argon (Ar), and the other contains an equal
mass of the monatomic gas krypton (Kr). The pressures in the
cylinders are the same, but the temperatures are different.
Determine the ratio of the average kinetic energy of a krypton
atom to the average kinetic energy of an argon atom.
Solution:
The result:
14.2 Kinetic Theory of Gases
We can define the “internal energy” of the gas
as the total translational energy of the N atoms
in the gas:
Thus, we have:
(for a monoatomic gas).
The internal energy of the gas is determined
by the Kelvin temperature.
14.2 Kinetic Theory of Gases
The internal energy of a monatomic gas is given by the total translational
energy of the N atoms in the gas:
In thermodynamics, we identify a monatomic gas as having 3 “degrees
of freedom” corresponding to the 3 available directions (x,y,z) in space
for translation. There is an internal energy of ½kT per atom per degree
of freedom.
14.2 Kinetic Theory of Gases
For a gas consisting of diatomic molecules ( air ie. N2 ) there are two
extra degrees of freedom which can “store” internal energy: rotation
about two axes. This gives 5 degrees of freedom altogether, each
providing an internal energy of ½kT per molecule. Thus, for a diatomic
molecule:
Three translational degrees of freedom plus two rotational degrees of freedom.
14.2 Kinetic Theory of Gases
Problem 14-42: Compressed air can be pumped underground
into huge caverns as a form of energy storage. The volume of a
cavern is 5.6 x 105 m3 and the pressure of the air in it is 7.7 x
106 Pa. Assume that air is a diatomic ideal gas whose internal
energy U is given by
If one home uses 30.0 kWh of energy per day, how many
homes could this internal energy serve for one day?
14.2 Kinetic Theory of Gases
Problem 14-42: Since we are treating the air as a diatomic ideal
gas (PV = nRT), it follows that
U  52 nRT  52 PV 
5
2
 7.7×106 Pa 5.6 105 m3   1.11013 J
The number of joules of energy consumed per day by one house is

3 J  h   3600 s 
8
30.0 kW  h   30.0 10

1.08

10
J


s  1 h 

The number of homes that could be served for one day by
1.1  1013 J of energy is
 1 home
1.1×10 J 
8
1.08
×10


13


5
 = 1.0 ×10 homes
J
PHYS1020 General Physics I
```