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1206 - Concepts in
Physics
Wednesday, November 11th
Notes
• Midterms can be picked up today - anytime
after 11:30 am
• For those who would like to write the
midterm again, you can do so on Friday at
4:30 pm. But you need to put your name
onto the list that is passing around (unless
you have not attended the first one).
Notes
• equation inside equations - this is a NONO
• Future assignments will include a section
with short questions, so that you can
practice using “your own words” to
describe concepts in physics
• Also for future assignments, we will be
strict about units. Every time you don’t
write a unit of don’t check that you unit
calculations are correct, points will be
subtracted ....
Notes
• List of topics you are interested in for last
few lectures - via email “topics of interest”
Thermodynamics
So far we have talked about heat and work separately, but they often occur together simultaneously. In an automobile engine for example, fuel is burned at relatively high
temperature, some of its internal energy is used for doing the work of driving the
pistons up and down, and the excess heat is removed by the cooling system to
prevent overheating. Thermodynamics is built upon the fundamental laws that heat
and work obey.
As before we will talk about systems, where we mean the collection of objects.
Everything that is not part of the system is called surrounding. For example (if we
go back to the engine in an automobile), we could define the gasoline burner as our
system, which makes than the pistons, exhaust system, the radiator, and the outside air
the surroundings. The system and its surroundings are separated by walls of some
kind. Walls can permit heat to flow through the, such as those of the engine block and
they are called diathermal walls. Perfectly insulating walls that do not permit heat
to flow between the system and its surroundings are known as adiabatic walls.
The state of a system (physical condition) will play an important role in
understanding what the laws of thermodynamics say about the relationship between
heat and work.
The other important player in this topic is energy, especially conversion of one type of
The Laws of Thermodynamics
Zeroth law: Thermodynamic Equilibrium and Temperature
First law: Work, Heat and Energy
Second law: Entropy
We will discuss all of them today, but come back on Friday to get a more detailed
understanding and see how they “act” together.
Zeroth law of thermodynamics
ZEROTH LAW OF THERMODYNAMICS
Two system individually in thermal equilibrium with a third system are in thermal equilibrium
with each other.
Remember the
example we had on
Monday with the
metal skillet.
If you wait long
enough the
temperature
throughout will be
the same as the heat
source it is touching
- the two systems
are in thermal
equilibrium.
Another example that we all know from experience is that hot and cold water
mix and will be at a temperature that is in between the at the end. By the end
of the week we will be able to describe this quantitatively - or in other words
we can translate this effect into the language of mathematics.
microscopic
“picture” of mixing
water at different
temperatures
Red dots are hot water and blue dots are cold water. In the case of water we
don’t need to “shake the box” - since opening the tap will produce a turbulent
flow and cause a lot of movement.
First law of Thermodynamics
THE FIRST LAW OF THERMODYNAMICS
The internal energy of a system changes from an initial value Ui to a final value of Uf due to
heat Q and work W:
ΔU = Uf - Ui = Q - W
Q is positive when th system gains heat and negative when it loses heat. W is positive
when work is done by the system and negative when work is done on the system.
Earlier we discussed the work-energy theorem (W = Ef - Ei) which tells us that work
is change in energy. (At this point this was mechanical energy - so the sum of potential
and kinetic energy). For example, the atoms and/or molecules of a substance exert
forces on one another. As a result, they have kinetic and potential energy. These and
other kinds of molecular energy constitute the internal energy of a substance. When a
substance participates in a proces involving energy in the form of work and heat, the
internal energy of the substance can change.
Thus the first law of thermodynamics is just the conservation of energy principle
applied to heat, work and the change in the internal energy.
Example: positive and negative work
A system and its surroundings are shown in the figure below. In part (a) the system
gains 1500 J of heat from its surroundings, and 2200 J of work is done by the system of
the surroundings. In part (b), the system also gains 1500 J of heat, but 2200 J of work is
done on the system by the surroundings. In each case, determine the change in the
internal energy of the system.
(a)
(b)
Surroundings
Surroundings
System
System
Work
Heat
Work
Heat
In figure (a) the system loses more energy
in doing work than it gains in the form of
heat, so the internal energy of the system
decreases. Thus, we expect the change in
the internal energy ΔU to be negative. In
part (b), the system gains energy in the
form of both heat and work. The internal
energy of the system increases, and we
expect ΔU to be positive.
(a) heat is positive, Q = 1500 J, since it is gained by the system. Work is positive, W = 2200
J, since it is done on the system. Therefore:
ΔU = Q - W = 1500 J - 2200 J = -700 J
(b) heat is positive, Q = 1500 J, since it is gained by the system. Work is negative, since it is
done on the system W = -2200 J. Thus:
ΔU = Q - W = 1500 J - (-2200 J) = 3700 J
Example: An ideal Gas
The temperature of three moles of a monatomic ideal gas is reduced from Ti = 540 K to
Tf = 350 K by two different methods. In the first method 5500 J of heat flows into the gas,
while in the second 1500 J of heat flows into it. In each case find (a) the change in the
internal energy and (b) the work done by the gas.
Since the internal energy of a monatomic ideal gas is U = 3/2 nRT and since the number
of moles n is fixed, only a change in temperature T can alter the internal energy. Because
the change in T is the same in both methods, the change in U is also the same. From the
given temperatures, the change ΔU in internal energy can be determined. Then, the first
law of thermodynamics can be used with ΔU and the given heat values to calculate the
work for each of the methods.
(a) Using equation for internal energy of a monatomic ideal gas, we find:
ΔU = 3/2nR(Tf-Ti) = 3/2(3.0 mol)[8.31J/(mol K)](350 K - 540 K) = -7100 J
(b) with the calculated ΔU and the given heat we can now determine the work:
1st method: W = Q - ΔU = 5500 J - (-7100 J) = 12600 J
2nd method: W = Q - ΔU = 1500 J - (-7100 J) = 8600 J
The gas does more work in the first method.
Example for first law of thermodynamics:
Elevated Tank Problem
A 50.0 m 3 closed tank is initially filled with air at 20.0 0C and 100 kPa. A pump fills the tank 3/4
full of water in such a way that the temperature of the air and water in the tank remain constant at
20.0 0C. Assume that the height of the center of mass of the water when the tank is 3/4 full is
40.0 m above the level of a lake from which the tank is filled.
(A)
What is the final pressure of the air in the tank when the tank is 3/4 full ?
YOUR TURN
(A) What is the final air pressure in the tank?
We can apply the ideal gas law (discussed on Monday)
When the tank is filled to 3/4, we have 1/4 of air left - compared to completely filled
with air in the beginning. The temperatures are constant, so we can eliminate them.
The pressure of the air is given.
ISOBARIC Pressure is Constant (ΔP = 0)
Example:
Gas Heated in a Cylinder fitted with a movable frictionless piston.
The pressure the atmosphere and the pressure due to the weight of the piston
remains constant as the gas heats up and expands.
First Law Implications: ΔU = Q - W
Unlike some of the other processes below neither the heat Q , the work W, or the change in
internal energy ΔU are necessarily zero in a constant pressure process.
For an ideal gas, constant pressure work is easily determined, W = ∫ PdV = PΔV
Part of the heat that flows into the system causes the temperature to rise,, the rest goes into
work.
Example: Isobaric expansion of water
One gram of water is placed in a cylinder, and the pressure is maintained at 2.0 x 105 Pa.
The temperature is raised by 31 C. In one case, the water is in the liquid phase and
expands by the small amount of 1.0 x 10-8 m3. In another case, the water is in the gas
phase and expands by the much greater amount of 7.1 x 10-5 m3. For the water in each
case, find (a) the work done and (b) the change in the internal energy.
In both cases the material expands, so work is done by the system (water) on the
surroundings and is, therefore, positive. The work is done at a constant pressure, so it
can be found form W = PΔV. Once the work is known, the first law of
thermodynamics, ΔU = Q - W, can be used to find the change in internal energy,
provided a value fot the heat Q can be found. The heat needed to raise the temperature
can be obtained from Q = cmΔT, where c = 4186 J/(kg C) is the specific heat capacity
of liquid water and cp = 2020 J/(kg C) is the specific heat capacity of water vapor at
constant pressure.
(a) process is isobaric, so the work is given by:
Wliquid = PΔV = (2.0 x 105 Pa) (1.0 x 10-8 m3) = 0.0020 J
Wgas = PΔV = (2.0 x 105 Pa) (1.0 x 10-8 m3) = 14 J
(b) Using ΔU = Q - W and Q = cmΔT, we find ΔU = cmΔT - W
ΔUliquid = [4186 J/(kg C)](0.0010 kg)(31 C) - 0.0020 J = 130 J - 0.0020 J = 130 J
ΔUgas = [2020 J/(kg C)](0.0010 kg)(31 C) - 14 J = 63 J - 14 J = 49 J
ISOTHERMAL Temperature is Constant (ΔT = 0)
Example1:
Boiling of water in the open air.
In general most isobaric phase changes are isothermal.
In this example the system does work as the steam-produced pushes against
the atmosphere as it expands. Neither the heat Q , the work W, or the change in internal
energy ΔU are zero. The water change phase (from liquid to gas)
Example 2:
In general for an Ideal gas U is only a function of the temperature so that ΔU is always equal
to zero for an isothermal process. Since ΔU = 0 then W = Q from the First Law.
What has to happen for this process to be isothermal is that the gas in a cylinder is
compressed slowly enough that heat flows out of the gas at the same rate at which is being
done on the gas.
The ideal gas law can be used to determine the work done W = PV ln(Vf/Vo) which is also
the equation for Q.
Note that P1V1 = P2V2 = nRT, the ideal gas law for an isothernal process.
ln denotes the natural
logarithm to the base
e = 2.71828. One can
relate is back to the
common logarithm to
the base ten by ln(Vf/
Vi) = 2.303 log(Vf/Vi)
Example: Isothermal Expansion of an Ideal Gas
Two moles of the monatomic gas argon expand isothermally at 298 K, from an initial
volume of Vi = 0.025 m3 to a final volume of Vf = 0.050 m3. Assuming that argon is an
ideal gas, find (a) the work done by the gas, (b) the change in the internal energy of the
gas, and (c) the heat supplied to the gas.
(a) The work done by the gas can be found from:
W = nRT ln(Vf/Vi) = (2 mol) [8.31 J/(mol K)] (298 K) ln(0.050 m3/0.025 m3) = 3400 J
(b) The internal energy of a monatomic ideal gas i U = 3/2nRT and does not change
when the temperature is constant, therefore ΔU = 0 J
(c) The heat Q supplied can be determined from the first law of thermodynamics:
Q = ΔU + W = 0 J + 3400 J = 3400 J
ADIABATIC No heat flows into or out of the system (Q = 0)
Example
Compression of a Gas in an Insulated Cylinder.
In this case any change in the internal energy of the gas is due to work done on it or by it,
ΔU = -W.
Normally if ΔU changes the temperature of a system will change. Any temperature rise or fall
is due to the work done or by the gas alone and not due to heat flowing into or out of the
system since Q = 0.
If a process is carried out fast enough the heat flow will be small and the process can be
approximate as being adiabatic. This happen because heat flow is in general a slow process.
Observe that we did not say that Q is constant because it not a state variable. Q represent an
energy transfer not the heat energy of the system.
For an ideal gas the work W = 3/2nR(Ti - Tf)
More details on Friday ...
Second law of Thermodynamics
THE SECOND LAW OF THERMODYNAMICS
Heat flows spontaneously from a substance at a higher temperature to a substance at
a lower temperature and does not flow spontaneously in the reverse direction.
THE SECOND LAW OF THERMODYNAMICS
The total entropy of the universe does not change when a reversible process occurs
and does not increase when an irreversible process occurs
The first version is easier to understand, after Friday’s lecture the second version will
obvious as well. Basically this is telling you that you cannot create energy, as we have
seen for mechanical energy already (perpetuum mobile)
We will look at entropy on the next slide (and in a mathematical way on Friday),
but in lay language entropy means that a system left alone (without outside
influence) will always go from a state of more order to a state of less order.
Entropy:
Disorder is more
probable than
order.
It is impossible to
go from disorder
to order for an
isolated system.