The Functional Form of the Internal Energy

CHEM 331
Physical Chemistry
Fall 2014
The Functional Form of the Internal Energy
The Internal Energy (U) is the energy due to the translational, rotational, vibrational and
electronic energies of the molecules that comprise a system as well as the relativistic rest-mass
energy of the electrons and nuclei that comprise the molecules and finally to the potential energy
of interaction between the molecules of a system. It can be calculated using molecular models of
a system employing Statistical Thermodynamics. But, it can only be measured in the manner
outlined previously. If a system transits between states A and B adiabatically, then:
U = W
If the path is not adiabatic, then U is given by the First Law of Thermodynamics:
U = Q + W
Attempts to measure U by measuring each of the above microscopically mentioned energies will
surely fail. Therefore, only changes in U are measurable:
U = UB - UA
The Internal Energy is a State Function; U depends only on 0thesystem's states A and B. It
does not depend on how the system transits between them. This means its differential dU is
exact. And this means mixed second partial derivatives of U are independent of the order of
differentiation:
=
When a system in a state specified by TA, PA, and VA transits to another state specified by TB, PB,
and VB,  is determined by:

=
(TB, PB, VB) -
(TA, PA, VA)
Therefore, is functionally dependent on the state variables T, P and V. However, since an
equation of state will constrain the state variables, is formally a function of T and V, (T,V),
or a function of T and P, (T,P). (T is not eliminated because we are considering
thermodynamics afterall.) For reasons that will become clear later, taking to be a function of
T and V will be the better choice. (From here on out, we will drop the notational difference
between the internal energy U and the molar internal energy and let the context determine
which of these we are referring to. We will specifically use only when the context may lead to
ambiguity.)
Therefore we can write:
dU =
dT +
dV
To determine U, we can integrate the above differential over an appropriate surface
representing the functional form of U.
Peter Atkins and Julio de Paula
Physical Chemistry, 8th Ed.
Each of the above partial derivatives of U,
and
, has physical meaning. To that
meaning now!
We start with the derivative with respect to temperature, which specifies the temperature
dependence of U. This derivative is the Constant Volume Heat Capacity:
Cv ≡
The heat capacity relates the amount of heat absorbed by a system to its temperature change. To
show the Cv defined in this manner is the heat capacity of the system, we write dU in both its
forms:
dU
= Q + W = Q - P dV
(P-V work only)
=
dT +
dV = Cv dT +
dV
Thus,
Cv dT +
dV =
Q - P dV
If we restrict ourselves to constant volume processes, then:
Cv dT = Qv
or,
Cv =
Hence, Cv is indeed the constant volume heat capacity.
Next to the derivative of U with respect to V, which specifies the dependence of U on the
system's volume. This derivative is referred to as the Internal Pressure, T.
T =
First, note that T has units of pressure:
~
~
~
~ [Pascal]
Second, the internal pressure is a measure of intermolecular cohesiveness. A high value of T
indicates strong intermolecular cohesion. This means T is dependent upon the attractive forces
between the molecules of a system; hence the name "internal pressure".
In the famous "free expansion" experiment of Joule, T was measured for gases and found to be
zero.
James Prescott Joule
Joule's Apparatus
In his "free expansion" experiment, Joule placed two cylinders connected by a stopcock into a
bath of Water. One cylinder was filled gas while the other was evacuated.
Joule opened the stopcock between the cylinders to allow for the free expansion of the gas from
one cylinder into the other. Measuring the temperature of the Water before and after the
expansion, Joule found no difference in the temperatures. This means,
Q
= 0
(Since no heat flows into the Water.)
W = - Pop dV
= 0
(Since Pop = 0.)
Thus,
dU
= Q + W = 0
And, since:
dU
=
dT +
= T dV
dV = Cv dT + T dV
(Since dT = 0.)
This gives us, according to Joule,
T
= 0
for gases. Now, despite the fact that Joule took great pains to construct a very precise
thermometer, the large heat capacity of Water prevented Joule from observing the very small
temperature change which would accompany the free expansion of a Real Gas, resulting in
 ≠ 0.
However, we can still draw the conclusion that for an Ideal Gas, Joule's experiment gives:
T
= 0
This is reasonable since there are no intermolecular forces between the "particles" of an Ideal
Gas and thus no molecular cohesiveness. Note that this means U(T) for an Ideal Gas.
We will show later that for a van der Waals Gas, which correctly predicts the trends of Real
Gases, that the internal pressure can be determined via:
T
=
For CO2, a = 3.59 L2 atm / mol2. At STP,
T
=
≅ 22.4 L/mol. So,
= 0.0071 atm
As Adamson notes, "[This] is quite small. In the case of liquids (or solids) the internal pressure
is usually much greater than the external pressure (usually around atmospheric). For common
liquids Pint is around 3000 atm. Liquids, in effect, are held together by their internal pressure,
that is, by their intermolecular attraction." (A Textbook of Physical Chemistry, Arthur W.
Adamson)
Some typical internal pressures for liquids:
T (atm)
3110
3486
3829
3811
15800
Liquid
H2O
CCl4
CS2
C6H6
Hg
Even though T is fairly large for liquids and solids, U is fairly small for isothermal processes
because dV is so small for these processes:
U =
~ 0
Finally, we note that if attractive forces dominate between molecules in a system, it requires
energy (dU > 0) to "pull" them apart when the volume is increased (dV > 0) and:
T =
> 0
If on the other hand repulsive forces dominate between molecules in a system, then energy is
released (dU < 0) when they are "pulled" apart as a result of a volume increase (dV > 0) and:
T =
< 0
Now we re-write the First Law in terms of a new state function. We define the Enthalpy (H) as:
H ≡ U + PV
The purpose in doing this is that this new function is important for discussions of
Thermochemistry. To see this, we write:
H + dH = U + d(U + PV) = U + dU + (P + dP) (V + dV)
= U + dU + PV + P dV + V dP + dP dV
≅ U + PV + dU + P dV + V dP
since dP dV ~ 0. Also, on the right hand side of the equation we have (U + PV), which is equal
to H. So,
dH
= dU + P dV + V dP
We also have:
dU
= Q + W = Q - P dV
dH
= Q - P dV + P dV + V dP
(for P-V work)
So,
Now we have the First Law of Thermodynamics in terms of H:
dH
= Q + V dP
If our processes occur at constant pressure (many chemical reactions occur under atmospheric
conditions, for which P is constant), then,
dH
= QP
Or, integrating, we obtain:
H = QP
For this reason, the enthalpy is frequently referred to as the "Heat Function". Its use in
thermochemical measurements, where the pressure under which a chemical reaction occurs is
constant, should be obvious.
Note also that the "natural" state variables for H are T and P;
the differential dH as:
(T,P). This means we can write
dH =
dT +
dP
And again the derivatives
and
have physical meaning.
Turning to the temperature dependence of H first, we define the constant pressure heat capacity
Cp as:
Cp
≡
Again, we must demonstrate that it makes sense to define the heat capacity in this way. We do
so by noting that at constant P:
dH
=
dT
= Qp
So,
Qp = Cp dT
And there we have it:
Cp
=
Example
For Silver metal:
p
= 23.43 + 0.00628 T [J/K mol]
What is H for the Isobaric heating of Ag from 25oC to 961oC; from Room Temp to the melting
point of Silver?

= 23.43 (T2 - T1) + ½ (0.00628) (T 22 - T12)
=
= 23.43 (1234 - 298) + ½ (0.00628) (12342 - 2982) = 26430 J/mol
Now to the pressure dependence of H. We wish to know:
= ?
To come at this derivative, we employ the Cyclic Rule of calculus:
= -1
Rearranging, we have:
= -
= -
Cp
The derivative of T with respect to P at constant H is referred to as the Joule-Thomson
Coefficient, JT.
JT =
This was measured for gases in the famous throttle experiment of James Prescott Joule and
William Thomson (Lord Kelvin).
Lord Kelvin
Joule-Thomson Apparatus
In this experiment, a gas is passed through a throttle in an adiabatically wrapped tube; at a high
pressure before it passes through the throttle to a low pressure after. The differential in pressure
(P) and temperature (T) of the gas as it passes through the throttle are then measured. Thus,
we have:
≅
for this process. The question is, has this process occurred at constant enthalpy (H)? To show
this we note the work of passing a given volume (V1) of gas through the throttle is:
W1 = -
= P1V1
The work expended after the gas passes through the throttle is:
W2 = -
= - P2V2
The total work for these processes is:
W
= W1 + W2 = P1V1 - P2V2
However, since the tube is adiabatically wrapped:
Q
= 0
meaning:
U = U2 - U1
= W = P1V1 - P2V2
Rearranging, we obtain the desired result:
U2 + P2V2 = U1 + P1V1
or:
H2 = H1
Hence, P and T have been measured for a constant enthalpic process and thus JT has been
determined for the gas.
Results for various gases, as a function of temperature, are plotted below:
Note that at Room Temperature, JT is positive for most gases; the cryogenic gases H2 and He
being the exception. This means that most gases will cool when throttled at Room Temperature.
The so called Inversion Temperature is the temperature at which JT inverts from positive to
negative.
The Joule-Thomson effect can be used as the basis for a refrigerating device. The cooled gas on
the low-pressure side is passed back over the high-pressure line to reduce the temperature of the
gas before it is expanded; repetition of this can reduce the temperature on the high-pressure side to
quite low values. If the temperature is low enough, then on expansion the temperature falls below
the boiling point of the substance and drops of liquid are produced. This procedure is used in the
Linde method for producing liquid air. The ordinary household refrigerator has a high- and a lowpressure side separated by an expansion valve, but the cooling results from the evaporation of a
liquid-refrigerant on the low-pressure side; the liquid is liquefied by compression on the highpressure side.
Gilbert W. Castellan
Physical Chemistry, 3rd Ed.
If attractive forces dominate between the molecules of the gas, then as a gas expands, energy
must be supplied to increase the separation between the molecules and hence increase their
potential energy. This energy comes at the expense of kinetic energy, therefore the molecules
slow and the temperature decreases. Since both the temperature and pressure are decreasing as
the gas expands, JT is positive.
On the other hand, if repulsive forces dominate, as the gas expands potential energy is converted
to kinetic energy due to the fact that there are fewer collisions between the molecules and the gas
warms. Since as the pressure of the gas drops, its temperature increases, JT is negative.
The Joule Effect can be used to liquefy H2 and He. In each case you pre-cool the gas such that
its temperature is below its inversion temperature; -80oC for H2 and -276oC for He. Then you
employ the Joule Effect to cool the gas to its boiling point and liquefy it. In 1898 James Dewar
succeeded in liquefying Hydrogen and in 1908 Heike Kramerlingh Onnes liquefied Helium using
this methodology.
James Dewar
Heike Kramerlingh Onnes
Now, all of these relationships derive from our taking (T,V) and (T,P). This is not to say that
we cannot take (T,P) and (T,V), or use any other state variables that are appropriate. It is
simply to say that taking (T,V) and (T,P) is useful experimentally.
So, for the Internal Energy and the Enthalpy we have:
dU = Q + W =
dT +
dV = Cv dT + T dV
and
dH =
dT +
dP = Cp dT - JT Cp dP