Download Set 3

Thermodynamics: Lecture 3
Chris Glosser
February 13, 2001
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OUTLINE
I. Work: Two Types
(A) Configuration Work
(B) Dissipative Work
(C) Work done in an adiabatic process
II. Heat
(A) Units and theMechanical Equivalence of Heat
(B) Heat Capacity
(C) Latent Heat
III. The First Law
IV. Mayer’s Equation
V. Enthalpy: Definition
VI. More on Enthalpy
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Work: Two Types
There are basically two types of work that you can do on a system. Either
the system can return the energy, in which you have a reversable process,
or it cannot, in which case we call the process irreverseable. Here are some
examples of reversable and irreversable processes:
Reverseable:
1. Compressing a piston adiabatically.
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System
Gas
Battery
Magnet
Intensive
P (pressure)
E(electromotive force)
B(magnetic field)
−
extensive
dW
V (volume)
P dV
V (charge)
Edq
M (Magnetization) BdM
2. Sending current through an (ideal) inductor in a circuit.
Irreverseable:
1. Stiring your coffee with a straw.
2. Sending current through a resistor.
2.1
Configuration Work
If the work done on a system is reverseable, we call it configuration work.
This is because in almost all cases, reverseable processes do something to
change the macroscopic configuration of the system. This in general can
be undone. Consider the compression of an ideal gas by a piston. Of the
system is adiabatically isolated, then by “easing up” on the piston, I allow
the gas to return the energy to the piston.
We usually think of the configuration work done on an ideal gas, but it
could take a variety of forms (see table ??). The most common use is the
work done on magnets. We shall examine this in excruciating detail toward
the end of the semister.
For the time being, we will focus on an ideal gas. Let’s calculate the
configuration work for an isothermal process, and an isobaric process (isochoric implies no work is done):
Isothermal Configuration Work - Ideal Gas
W = nRT
Z
Vf
Vi
Vf
dV
= nrT ln
V
Vt
(1)
Isobaric Configuration Work - Ideal Gas
W =P
Z
Vf
Vi
dV = P (Vf − Vi )
(2)
I will ask you to calculate the work done in the case of an inductor in a
circuit as a Homework exercise.
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2.2
Dissipative Work
Irrecoverable energy is said to be dissipated. Therefore, an Irreverseable
process does dissipative work on a system. An example is work done by
stirring a fluid. The work done on such a system by an external torgue τ is:
−
dW
= −τ dθ,
(3)
Where θ is the angle that the rod moves. It is quite clear that this work is
always negative, and therefore the work done by stirring cannot be recovered.
The second example is that of a resistor in a circuit. As you all know,
the battery “works” on the charge in order to produce a current of charge
(q). Recall the expression for the energy dissipation of a resistor over time;
−
dW
= −I 2 R dt.
(4)
We may re-express this in terms of our state variables,
−
dW
= −IR I dt = −E dq
(5)
Again, if you reverse the direction of charge flow, you do not recover any
energy. You just dissipate more energy.
As we shall soon see, all processes have an irreversable component. However, we shall assume that there are processes for which the configuration
work is much, much greater than the dissipative work, and that our arguments will hold for these as well.
2.3
Work done in an adiabatic process
An adiabatic process is a process in which no heat is added to the system.
In such a casr, the first law of thermodynamics tells us that the total work
done must be exact, since it is equivalent to the change in internal energy.
Wad =
Z
b
a
−
dW
ad
(6)
In such a case, it makes sense to write down a formula for the work done for,
say an Ideal Gas. This is one way to compare our models for the internal
energies of the system.
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Heat
What if the system is not adiabatic? Well, the work is no longer exact,
that’s for sure. Recalling the first law, and noting that the “adiabatic work”
is equivalent to the (negative) internal energy of the system, we may write
down, for any process;
−
−
−
dQ
= dW
− dW
(7)
ad.
That is:
The heat flow into a system is the work done minus the adiabatic work done.
This is, of course, equivalent to the first law, which states:
The heat supplied is the work done by the system, plus the internal energy
of the system
This is, of course, just the conservation of energy.
3.1
Units and the Mechanical Equivalence of Heat
From the first law, it is evident that heat is a form of energy. Therefore, the
fundamental unit of heat, the calorie, should be related to the fundamental
unit of (mechanical) energy, the Joule. This is know as the Mechanical
Equivalence of Heat
3.2
Heat Capacity
Now, let’s turn our discussion to some of the more familiar concepts of heat
transport. The heat capacity is defined to be the heat added over a change
in temperature. That is;
−
dQ
C=
(8)
dT
clearly, if this expression is to be concrete we must specify something else
about the system. Namely, we will “fix” one of the state variables. If we
fix P , then refer to the heat capacity at constant pressure CP . If we fix V ,
then this is the heat capacity at constant volume CV . The heat capacity is
extensive, and we can make it intensive by dividing by n and calling the
resultant quantity the sppecific heat capacity c:
c=
−
1 dQ
n dT
as usual, “specific” quantities are denoted with lower case letters.
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(9)
3.3
Latent Heat
During a phase transition, when the temperature doesn’t change when heat
is added or taken away, The definition of the specific heat doesn’t really
make any sense. In such cases, we define the latent heat of transformation,
L:
Q
(10)
L=
m
We generally insist that no work is done on the system while the transformation is taking place (hence; dU = Q = Lm, etc.). Hopefully, you remember
how to use this from elementary physics classes. When we discuss enthalpy
later, I will come up with a more precise definition of the latent heat.
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The First Law
Now, If you haven’t already figured it out, The first law of thermodynamics
is eqivalent to the following statements:
• Heat is a form of energy.
• Energy is conserved.
The equation form of this statement is:
−
−
dU = dQ
+ dW
(11)
−
Where I think of dW
as the work done by the system.
It is obvious that this is true for reversable processes. For irreversable
processes, we will have to wait until I introduce entropy later in teh semister.
4.1
Mayer’s Equation
Anyway, why don’t we combine our ideas of heat capacity with the first
law, and see where we get. We shall assume that the process in question is
reversable. Recall the (intensive) equation of state for an ideal gas:
P v = RT
(12)
Now, assuming that the number of particles is constant, we may write the
first law as;
−
du = dq
− P dv
(13)
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. Since the equation of state allows to eliminate one of the variables, we
may also write the energy differential as:
du =
∂u
∂v
dv +
T
∂u
∂T
dT.
(14)
v
We may eliminate the energy differential and solve for the specific heat:
−
dq =
∂u
∂v
+ P dv +
T
∂u
∂T
dT.
(15)
v
If we divide this by dT , we get the specific heat capacity;
c=
∂u
∂v
+P
T
dv
+
dT
∂u
∂T
.
(16)
v
We therefore obtain an alternate expression for the specific heat capacity at
constant V ;
∂u
cv =
.
(17)
∂T v
In the specific case that we have an Ideal Gas, you can differentiate the
equation of state to get;
P dv + v dP = R dT.
(18)
Now we can write Equation 16 in terms of the specific heat capacity at
constant volume, and use the previous equation for the ideal gas to eliminate
dv in favor of dP ;
−
dq
= (cv + R)dT − vdP
(19)
at constant pressure, then, we must have that the coefficient dT is the heat
capacity at constant pressure:
cP = cv + R
(20)
This is Meyer’s Equation. Be careful, as this relationship only applies for
an ideal gas.
4.2
Enthalpy: Definition
Enthalpy is basically the energy required to create a system out of nothing
in a constant pressure environment. Therefore, it is to be defined as:
Enthalpy (H): H = P V + U .
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Now let’s examine how enthalpy behaves in an environment of constant
pressure:
−
dH = V dP + P dV + dU = dQ
+ V dP
(21)
Of course, for constant pressure, we just have that the change of enthalpy
equals the heat flow. The enthalpy is related to the heat capacity at constant
pressure by the relation;
∂H
(22)
CP =
∂T P
The derivation of this proceeds along the same lines as the derivation of
Meyer’s equation, and is left as an exercise.
4.3
More on Enthalpy
With our definition, we have a way of expressing the Latent heat in a more
concrete manner. It is just the difference in enthalpy between the two states.
That is,
L = HGas − HLiquid
(23)
Is the the latent heat of evaporation. This idea is very natural if we consider
what enthalpy is. We completely destroy a system, then set up another
system in it’s place, with the same state variables, P, V, T . The difference in
Enthalpy gives tells us how much energy we must expend to do this, and is
therefore the logical equivalent to the latent heat.
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