Thermodynamics: determining potentials

In summary, you discussed how to determine the Helmholtz free energy F(T, V), the Gibbs free energy G(T, p), and the enthalpy H(S, p) for an imaginary substance characterized by thermal energy U = (NS^2)/V^2. You explained the concept of a Legendre transform and provided a step-by-step guide for finding each of these energies by replacing variables and using differentiation. You also pointed out a sign error in the solution provided by the student and helped them correct it.
  • #1
Chiborino
21
0

Homework Statement


Consider an imaginary substance which is characterized by thermal energy
U=[itex]\frac{NS^2}{V^2}[/itex]

(a) Determine the Helmholtz free energy F(T, V).
(b) Determine the Gibbs free energy G(T, p).
(c) Determine the enthalpy H(S, p)

Homework Equations


F=U-TS (maybe dF = dU - sdT = -pdV - sdT?)
G=U-TS+pV = F+pV (dG = -sdT + Vdp
H=U+pV (dH = TdS + Vdp)

The Attempt at a Solution


I'm really lost when it comes to this. My professor hasn't done any examples with this sort of problem, the book doesn't have anything like this in it, and I can't find a problem similar online anywhere. So my best guess is to just throw the N(S/V)^2 into the potential formula for each problem and circle it. But I get the feeling that F(T,V) =N(S/V)^2 -TS isn't a valid answer. A quick explanation of how to proceed would be extremely helpful.
 
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  • #2
The energy U(S,V) is a function of S and V while the Helmholtz free energy F(T,V) is a function of T and V. To go from U to F, you're replacing the variable S with its conjugate, T. This is what's called a Legendre transform.

If you differentiate U, you get
$$dU = \frac{\partial U}{\partial S} dS + \frac{\partial U}{\partial V} dV.$$ Comparing this to the first law of thermodynamics, ##dU = T\,dS-p\,dV##, you can see that ##T = \frac{\partial U}{\partial S}## and ##p = -\frac{\partial U}{\partial V}##. The first one is the one you want because you want to replace S with T.

For part a, start by differentiating U with respect to S to find T in terms of S and V. Invert that to find S in terms of T and V. You should get ##S = \frac{TV^2}{2N}##. Then use this to eliminate S from the expression
$$F = U - TS = \frac{NS^2}{V^2} - TS.$$ In the end, you should have ##F(V,T)=-\frac{V^2T^2}{4N}##.

To find Gibbs free energy G, note that you're starting with F and replacing V by p. Similarly, to find the enthalpy H, you're starting with G and replacing T by S. You follow the same basic procedure each time.
 
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  • #3
Thank you!
That's such an easy problem, I don't know why professor blew over the concept in class.
I don't know if you feel like checking me over, but for G I got [itex]\frac{3Np^2}{T^2}[/itex] and the algebra's giving me trouble for H. I got p= -(dU/dV)= [itex]\frac{2NS^2}{V^3}[/itex], then [itex]V=(\frac{2NS^2}{p})^{\frac{1}{3}}[/itex]
 
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  • #4
I got ##G = \frac{Np^2}{T^2}##. I think you made a sign error somewhere.

I got ##H = 3\left(\frac{N S^2 p^2}{4}\right)^{1/3}##. Your expressions for p and V match what I found.
 
  • #5
I got two separate cube root expressions for H, but I see how I can factor out a 2 to combine them into what you have. And yes, I spotted the sign error. Thank you again for the help. :)
 

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the study of energy and its transformations in different systems. It focuses on the relationship between heat, work, and energy, and how these factors affect the behavior and properties of matter.

2. What are potentials in thermodynamics?

In thermodynamics, potentials refer to the various forms of energy that are associated with a system. These include internal energy, enthalpy, entropy, and free energy. Potentials are used to describe the state of a system and how it changes under different conditions.

3. How are potentials determined in thermodynamics?

Potentials in thermodynamics can be determined through various mathematical equations and formulas. For example, internal energy can be calculated using the first law of thermodynamics, while enthalpy can be determined through the relationship between internal energy, pressure, and volume.

4. Why are potentials important in thermodynamics?

Potentials are important in thermodynamics because they allow us to understand and predict the behavior of a system. They provide valuable information about the state of a system and how it will respond to changes in temperature, pressure, or other variables.

5. How are potentials used in real-world applications?

Potentials are used in a variety of real-world applications, including in the design and operation of engines, refrigeration systems, and power plants. They also play a crucial role in fields such as chemistry, biology, and materials science, where they are used to understand the properties and behavior of different substances and systems.

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