What is the solution for finding F in terms of T and V without integrating?

In summary: I was trying to find F as a function of T and V and I was stuck with the S term in the expression of U. I was also trying to use the definition F=U-TS but I got stuck with the S term. Is there a simpler way to solve for F? Thanks!In order to solve for F you need to integrate U-TS. U-TS is not a function, so you need to integrate it using a different method.
  • #1
wololo
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Homework Statement



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Homework Equations


Maxwell relations

The Attempt at a Solution


I have an attempt at a solution, but I am not sure if I can replace the integral of dT in the helmholtz equation by the T I found using the internal energy. Does this make sense? Thanks

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  • #2
The problem statement asks you to express F as a function only of T and V. This is not the form of your final answer.

Chet
 
  • #3
I still haven't figured out how to get rid of the entropy terms in the expression of Helmholtz free energy. Can I simply replace the S in the equation I found by heat capacity times temperature? (given dS=cdT). I don't know how to proceed after, any advice is appreciated. Am I on the correct path or did I start solving the problem wrong from the very beginning? Thanks!
 
  • #4
In your integrations you treated S as constant which is not a valid assumption.

You have an expression for T in terms of S and V. Try using that along with the definition of F in tems of U, S, and T.
 
  • #5
If I rewrite the expression for T and isolate S i get S=(27T^3V)/(64A). I can then replace the integral of -SdT to find F by the integral of -(27T^3V)/(64A)dT which equals -(27T^4V)/(256A).
It seems to make sense since I get rid of the S terms in the first integral. However I don't see how I can get rid of the S term in the expression of pressure P, so I can't integrate PdV.

I also considered using the definition F=U-TS but there again I get stuck with the S term in the expression of U. And if I use dF=dU-TdS-SdT and integrate I still have to do something with dU. I know I am probably missing some simple relation but I really can't think of anything that would rid me of S...
 
  • #6
wololo said:
If I rewrite the expression for T and isolate S i get S=(27T^3V)/(64A). I can then replace the integral of -SdT to find F by the integral of -(27T^3V)/(64A)dT which equals -(27T^4V)/(256A).
It seems to make sense since I get rid of the S terms in the first integral. However I don't see how I can get rid of the S term in the expression of pressure P, so I can't integrate PdV.

I also considered using the definition F=U-TS but there again I get stuck with the S term in the expression of U. And if I use dF=dU-TdS-SdT and integrate I still have to do something with dU. I know I am probably missing some simple relation but I really can't think of anything that would rid me of S...
Why are you integrating at all? You solved for S and got:
$$S=\frac{27}{64}\frac{VT^3}{A^3}$$
Why don't you just substitute that into the equation for U to get the internal energy in terms of T and V? Then why don't you just multiply S by T to get TS in terms of T and V? Then, F = U - TS.

Chet
 
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Related to What is the solution for finding F in terms of T and V without integrating?

1. What is the Helmholtz free energy problem?

The Helmholtz free energy problem, also known as the Helmholtz free energy minimization problem, is a mathematical problem that involves finding the equilibrium state of a thermodynamic system at a constant temperature and volume. It is named after the German physicist Hermann von Helmholtz, who first proposed the concept in the 19th century.

2. Why is the Helmholtz free energy problem important?

The Helmholtz free energy problem is important because it allows us to predict the behavior of thermodynamic systems and determine their equilibrium states. This is crucial in many fields of science and engineering, such as chemistry, physics, and materials science, where understanding the behavior of systems at equilibrium is essential for designing and optimizing processes and products.

3. How is the Helmholtz free energy problem solved?

The Helmholtz free energy problem is typically solved using mathematical techniques such as calculus and differential equations. The goal is to find the minimum value of the Helmholtz free energy function, which is a function of the system's temperature, volume, and other thermodynamic variables. This minimum value represents the equilibrium state of the system.

4. What is the relationship between the Helmholtz free energy and other thermodynamic potentials?

The Helmholtz free energy is one of the four fundamental thermodynamic potentials, along with the internal energy, enthalpy, and Gibbs free energy. It is related to these other potentials through mathematical equations, and each one has its own specific use in different thermodynamic processes. The Helmholtz free energy is particularly useful for systems at constant temperature and volume, while the Gibbs free energy is more relevant for systems at constant temperature and pressure.

5. Can the Helmholtz free energy problem be applied to real-world systems?

Yes, the Helmholtz free energy problem can be applied to real-world systems that follow the laws of thermodynamics. This includes systems such as chemical reactions, phase transitions, and even biological processes. However, in some cases, simplifications or assumptions may need to be made in order to solve the problem and apply it to a specific system.

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