Summary of Kinetic Theory

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Summary of Kinetic Theory
Physical meaning of the absolute temperature is a
measure of the average kinetic energy of a
molecule.
From this we can express the pressure of an
ideal gas as
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Phase Changes
We start with 1 gm of ice at -30 C, supply
energy at a constant rate and monitor its
temperature
Straight line with constant slopes warming up
of ice, water and steam. The slopes are inversely
proportional to their specific heats.
Horizontal lines no temperature change as you
add energy. Phase transitions melting and
vaporization.
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It takes some amount of energy to melt and
vaporize substances. Those energies per unit
mass of the materials are called heat of fusion
Lf and heat of vaporization, Lv.
Since no temperature change occurs during the
phase transitions, those energies are sometimes
called latent heats of fusion/vaporization.
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Positive energy is required to melt/vaporize. The
same amount of energy, however, is released when
the substance solidifies/condenses. The process
is perfectly reversible.
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Hot compress.
Paraffin compress I hope you never had one
applied to you!
Typical melting temperature is from 49 to 71 deg.
C (120 to 160 deg. F).
So, if you melt it, wrap it in a cloth, and apply
it to yourself (or your kid) it is going to stay
at its melting temperature, around 140 ºF till
its heat of fusion is transferred to your body
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How does it compare?
So, if you melt it, wrap it in a cloth, and apply
it to yourself (or your kid) it is going to stay
at its melting temperature, around 60 ºC (140 ºF)
till its heat of fusion is transferred to your
body
For water cooling down
To give (reject) the same amount of heat 1 kg of
water has to cool down by 35 ºC, from 60 ºC to
25 ºC (140 to 80 ºF).
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Nuclear Meltdown
A nuclear reactor in meltdown still generates
heat, 120MW. The core is 2.5x105 kg of uranium.
How much energy is required to melt the core
after it reaches it melting point?
How long will it take?
This is less than 3 minutes! Nuclear reactors
contain failsafe emergency cooling systems
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Ice Meltdown
How much energy is required to melt 10kg of
initially at -10oC?
The specific heat of ice (at this temperature) is
2050 J/kgK. The latent of fusion for water is 334
kJ/kg.
If this meltdown takes place inside of a 500W
microwave, how long does it take?
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Phase Diagram for Water
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Summary Thermal Behavior of Matter
Ideal gas law Can be derived from kinetic theory!
Kinetic theory
Phase Changes Latent heats of fusion and
vaporization
Thermal expansion, linear, cross sectional area,
and volume.
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First law of thermodynamics
Kinetic energy
Potential energy (V)
Conservation of energy in mechanics - mechanical
energy of a system is conserved
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External forces and work they do.
What if there is work done on the system or if
the system does work?
If the system is not totally isolated. Then
external forces can do work on it or it can do
work on external objects.
work of the external forces
Wsys is the work done by the system. In this case
work against friction. If Wsys gt 0 the energy
decreases.
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1st law of thermodynamics.
In thermodynamics we introduce a new type of
energy internal energy of the system, U, which
is a sum of all the internal energies of the
system (usually microscopic). Then
We further unify Wext and Wsys by introducing the
net work W. By convention this is assumed to be
the work done by the system. Hence it lowers the
internal of the system. However, if there is work
done on the system by external forces, then W is
negative.
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1st law of thermodynamics.
Example of converting work into internal energy.
500g of water in an uninsulated container is
shaken until its temperature rises 3oC. The
mechanical work to do this is 9 kJ.
(a) How much heat is transferred to the water
during the shaking?
(b) How much heat is lost to the environment?
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1st law of thermodynamics.
How do we convert work into internal energy? This
is easy!
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Potential energy of the weights is converted into
work (external work, negative W) and to increase
in the internal energy of water (positive DU)
External work done by the hands (negative W) is
converted into positive internal energy (positive
DU)
Are not we forgetting something, though?
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In order to increase internal energy of a system,
we do not necessarily have to have some work done
on it. There is a simple and common alternative
transferring heat, Q, from some hot object
(heater)
The heat Q is positive if the energy is
transferred to the system.
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First law of thermodynamics
work done by the system
The change in the internal energy of a system
depends only on the net heat transferred to the
system and the net work done by the system, and
is independent of the particular processes
involved.
The equation is deceptively simple One of the
forms of the general law of conservation of
energy. BUT! Be careful about the sign
conventions. Positive Q is heat transferred to
the system. Positive W is work done by the system.
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work done by the system
The heat, Q is considered positive, when it is
transferred TO the system. The system gains
energy.
The work, W, is considered positive if it is done
BY the system. The system loses energy.
WHY? The history heavily influenced by the heat
engines.
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work done by the system
Implication both heat, Q, and mechanical work,
W, are means of energy exchange between the
system and the outside world. The both are
kinds of energy in transit. Thats why they are
on the same side in the equation. Nevertheless,
the heat, Q, specifically relates to the energy
transferred due to temperature difference alone.
While, W, incorporates all other sorts of energy
transfer, most commonly mechanical work.
Statement 1 First Law Conservation of Energy
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work done by the system
Statement 2 Internal energy, U, is a function
of internal state of the system, a thermodynamic
state variable a quantity, whose value does not
depend on how a system got to a particular
state. In principle, you can measure internal
energy of the system -, by measuring the sum of
energies of all molecules, - the same way as you
can measure its temperature and pressure both
the variables, which are well defined in
thermodynamics equilibrium.
Usual ambiguity with setting the zero level of
energy, though What is energy of Uranium at
absolute zero temperature?
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Rate of change in internal energy
Continuous processes we differentiate with
respect to time to define rates of energy flow,
measured in Watts.
Gasoline burning in an automobile engine releases
energy at a rate of 160 kJ per second. Heat is
exhausted through the cars radiator at a rate of
51 kJ per second and out of the exhaust at 50 kJ
per second. An additional 23 kJ per second goes
to frictional heating within the machinery of the
car. What fraction of the fuel energy is
available for propelling the car?
E (160 124) / 160 22.5
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work done by the system
Example 1 A 40W heat source heats a gas for
25sec. During this time the gas expands and does
750J of work. What is the change in the internal
energy of the gas?
Example 2 If in an automobile engine 17 of
the total energy released in burning gasoline
ends up as work, what is the engines output if
the waste heat produced is 68 kW?
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Thermodynamic processes
We are interested in changes in internal energy,
the heat transferred to or from the system and
the work done by the system - ideally the gas
under piston
In principle, all we need to know are the ideal
gas law and the 1st law of thermodynamics
BUT There are many, processes, conditions
(idealistic or realistic), and applications
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Thermodynamic processes
Usually presented as P-V diagrams
A diagram suggests that both pressure and volume
are well defined in every point.

• The gas is in a thermodynamic equilibrium in
  every point, every moment of time
• We call that a quasi-static process.

Equilibrium with what?..
At least with itself! That is, different parts
of the gas are in equilibrium with each other.
Can be implemented if we change things slowly.
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Quasi-static processes.
Heating and boiling water on a stove.
Quasi-static? Why?
The least we can say is that the water is in
contact with burning gas from below and cool air
from above. So, its temperature is not likely
to be the same at the top and bottom, which
precludes thermodynamic equilibrium and
quasi-static process.
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Quasi-static processes.
Is this one any better?
Practically, how slowly you should go to be
quasi-static?
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Quasi-static processes are in principle
reversible
You can go back and forth along the same line of
well defined equilibrium states.
Can there be multiple paths to get from 1 to 2?
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Work done and heat transferred our major
concerns!
Work done by the gas
P pressure of the gas Dx displacement of
the piston, A area of the piston
The work is positive when the gas expands!
Differential form
Integral form also good for varying pressure
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Work done and heat transferred our major
concerns!
Work done by the gas
P (varying) pressure of the gas dV
differential volume change
P
V
Area under curve on a P-V diagram