Now that we have defined all independent energy quantities in terms of p, V, T, and S, we summarize them in terms of their natural variables:
dU5T dS2p dV (4.14)
dH5T dS1V dp (4.15)
dA 5 2S dT2p dV (4.16)
dG5 2S dT1V dp (4.17)
These equations are important because when the behaviors of these energies on their natural variables are known, all thermodynamic properties of the system can be determined.
For example, consider the internal energy, U. Its natural variables are S and V;
that is, the internal energy is a function of S and V:
U5U(S, V)
As discussed in the last chapter, the overall change in U, dU, can be separated into a component that varies with S and a component that varies with V. The variation of U with respect to S only (that is, V is kept constant) is represented as ('U/'S)V, the
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partial derivative of U with respect to S at constant V. This is simply the slope of the graph of U plotted against the entropy, S. Similarly, the variation of U as V changes but S remains constant is represented by ('U/'V)S, the partial derivative of U with respect to V at constant S. This is the slope of the graph of U plotted versus V. The overall change in U, dU, is therefore
dU5 a'U 'Sb
V dS 1 a'U 'Vb
S dV But from the natural variable equation, we know that
dU5T dS2p dV
If we compare these two equations, the terms multiplying the dS must be equal, as must the terms multiplying the dV. That is,
a'U 'Sb
V dS5T dS a'U
'Vb
S dV5 2p dV We therefore have the following expressions:
a'U 'Sb
V5T (4.18)
a'U 'Vb
S
5 2p (4.19)
Equation 4.18 states that the change in internal energy as the entropy changes at constant volume equals the temperature of the system. Equation 4.19 shows that the change in internal energy as the volume changes at constant entropy equals the negative of the pressure. What fascinating relationships! It means that we do not have to actually measure the change in internal energy versus volume at constant entropy—if we know the pressure of the system, the negative value of it equals that change. Because these changes represent slopes of plots of internal energy versus entropy or volume, we know what those slopes are for our system. So, if we know how U varies with S and V, we also know T and p for our system.
Furthermore, many such partial derivatives can be constructed that cannot be determined experimentally. (Example: Can you construct an experiment in which the entropy remains constant? That can sometimes be extremely diffi- cult to guarantee.) Equations like 4.18 and 4.19 eliminate the need to do that:
They tell us mathematically that the change in internal energy with respect to volume at constant entropy equals the negative of the pressure, for example.
There is no need to measure internal energy versus volume. All we need to measure is the pressure.
Finally, in many derivations, partial derivatives like these will show up.
Equations like 4.18 and 4.19 allow us to substitute simple state variables for more complicated partial derivatives. This will be extremely useful in our fur- ther development of thermodynamics and accounts partially for its real power.
ExamplE 4.5
Show that the expression on the left-hand side of equation 4.18 yields units of temperature.
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Other relationships can be derived from the other natural variable equations.
From dH:
a'H 'Sb
p5T (4.20)
a'H 'pb
S5V (4.21)
From dA:
a'A 'Tb
V
5 2S (4.22)
a'A 'Vb
T5 2p (4.23)
and from dG:
a'G 'Tb
p5 2S (4.24)
a'G 'pb
T
5V (4.25)
If we know that G is a function of p and T, and we know how G varies with p and T, we also know S and V. Also, knowing G and how it varies with p and T, we can de- termine the other state functions. Because
H5U1pV and
G5H2TS we can combine the two equations to get
U5G1TS2pV
Substituting from the partial derivatives in terms of G (that is, equations 4.24 and 4.25), we see that
U5G2Ta'G 'Tb
p2pa'G 'pb
T (4.26)
Expressions for the other energy state functions can also be determined. The point is, if we know the values for the proper changes in one energy state function, we can use all of the equations of thermodynamics to determine the other changes in energy state functions.
Solution
The units of U are J/mol, and the units of entropy are J/mol#K. Changes in U and S are also described using those units. Therefore, the units on the derivative (which is a change in U divided by a change in S) are
J/mol
J/mol#K5 l/K1 5K which is a unit of temperature.
ExamplE 4.5 (continued)
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