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Compressing a rod:
The equation of state for a rod is:
     T
l  l0
F
  is the stress;  
is the strain.
A
l0
Again, if the compression is done quasi-statically (e.g.,
with a vise), then
From the eq. of state:
So:
Isothermal and adiabatic processes:
Compressing the rod may generate some internal energy,
and the rod may heat up. But if the process is carried out
VEEEEEEERYYYYYYYY SLOOOOOOWLYYYYY then the
heat will dissipate, and the rod will not change its T.
We can then calculate the isothermal work:
f

V
V 
2
2
Wsys       d  
 f  0
2kT
 kT   0

However, if the process is done very quickly, the
Heat has no time to escape, so Q=0.
Then the First Law becomes: 0 = ΔU + W
Such a process is called adiabatic.
More about quasi-static work:
We have already calculated the work done by
expanding gas:
In this apparatus, described on Page 20,
the piston is loaded with sand. We can
remove the sand grain after grain and
thus slowly expand the gas.
At any moment we can reverse
the process – i.e., start putting
sand grains back on the pile,
and we can return to the initial
state going through exactly the
same intermediate states as
when the gas was expanded – or
going back along the same path
In the p-V space.
HEAT CAPACITY
It is an extremely important quantity in thermodynamics
because it can be relatively easily measured (e.g.,
measurments the heat capacity of metals like Cu or Al
are often done in high-school physics classes).
Adding or removing thermal energy from a system (in
the form o heat) results in a temperature change.
But remember – this is not always true! You can keep
adding heat to boiling water, and its temperature will
be 212 F and will not change as long as there is water
in the cauldron!!! But it is so because in the process
of boiling water changes its phase – in other words, it
undergoes a phase transition (a.k.a. “phase transformation”). So, the above applies only to systems
that do not undergo a phase transition.
However, for now let’s forget about phase transitions.
So, an infinitesimal reversible heat transfer
causes a temperature change dT, and:
The coefficient Cα is called the heat capacity.
Why the subscript α ? Well, heat may be transferred to the
system in different ways. We may keep one of the system
Parameters constant, and α refers to that parameter – e.g.:
for a process in which V = const.
for a process in which p = const.
Heat capacities can be expressed in terms of derivatives
of state parameters. This is a very important part of the
heat capacity theory!!!
Let’s recall the mathematical formula expressing the First
Law:
For a gas,
expanded or
compressed
But we can also take advantage of the fact that U is a
function of two parameters in this particular process –
namely, U = U(T,V). So:
Combining the two equations, we get:
If the volume is kept constant, dV = 0, so:
Meaning that:
ENTHALPY:
In thermodynamics, in addition to the “natural” state
parameters, we introduce a number of state functions
that help to organize and to simplify the calculations.
The most important of these are:
• The “Helmholtz free energy”, traditionally denoted as F ;
• Enthalpy, traditionally denoted as H ;
• The “Gibbs function”, traditionally denoted as G.
Mathematically, these functions are Legendre Transformations of the internal energy U, which can be though of
as a function of entropy S, volume V, and the number of
particles N. But we will not discuss the theory of the
Legendre transformation now.
In literature, the F, H, and G functions are often called
“thermodynamic potentials”.
The enthalpy is defined simply as:
Let’s take the differential of H:
But we also know that (First Law):
By combining, we get:
Because H is a state parameter, we can write H = H(T,p)
and take the differential:
Then, we combine the last two equations:
We get:
At constant pressure, dp = 0, so that:
for p = const.
From which we immediately obtain:
EXAMPLES:
A wire is kept at zero tension between two firmly planted
vertical posts:
Initial temperature is TH and the tension τ = 0.
Then the wire is cooled down to temperature TL
which causes that the wire gets tensioned because
It “wants” to get shorter, but cannot, because the
length L does not change!
Suppose that we know the Young modulus of the wire
material, and its thermal expansion coefficient.
We want to find the expression for the tension in a cooled
wire.
Let’s take the length L as a function of T and the tension:
Then:
But dL = 0 , so that:
Now, we can use some coefficient that have been
introduced in order to characterize the “response” of
materials to external influences.
The isothermal Young modulus ET is a commonly
Used “material characteristic” – you can find the values of
ET listed in many books – it is defined as:
1  L 
1
, where A - cross section area of a wire or rod
  
L   T ET A
The coefficient of linear thermal
expansion is defined as:
1  L 
L   
L  T 
(in other words, it’s the elongation of a wire of unit length
heated up by a unit of temperature).
So, we can rewrite:
If the temperature change is small enough that αL and ET do
not change significantly, we get: