Can The Joule Coefficient be Derived Using the First Law and Maxwell Relation?

In summary, the conversation discusses the Joule coefficient, μJ, which is defined as μ = (∂T/∂V)U. It is shown that μJ CV = p – αT/κ, and the relevant equations used include the specific heat at constant volume (Cv), the isothermal compressibility (κT), and the thermal expansion coefficient (α). The first law and a Maxwell relation are also utilized in the discussion.
  • #1
lee403
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The Joule coefficient, μJ, is defined as μ = (∂T/∂V)U. Show that μJ CV = p – αT/κ

Relevant equations
Cv=(∂U/∂T)v
κT=-1/V*(∂V/∂p)T
α=1/V*(∂V/∂T)p
dV(T,p)=(∂V/∂T)pdt+(∂V/∂p)TdpWhat I have attempted was to make μ = (∂T/∂V)U=(∂T/∂U)V(∂U/∂V)T=(∂U/∂V)T*Cv-1.

Then α/κT=(∂V/∂T)p/(∂V/∂p)T because the 1/V cancels.

Since (∂V/∂T)p=(∂V/∂p)T *(∂p/∂T)V
p – αT/κ becomes p-T(∂p/∂T)V

so μJ*Cv = (∂U/∂V)T (because Cv and Cv-1 cancel)
(∂U/∂V)T= p-T(∂p/∂T)V

I try to make the assumption that it is an ideal gas so (∂p/∂T)V becomes nR/V
so then p-T(nR/V)= p-p= 0. But that would mean (∂U/∂V)T = 0.

I don't know that this statement is true or if it is safe to assume an ideal gas. I am afraid that maybe I missed a negative or that I used properties of partial differentials incorrectly. Can anyone provide assistance?
 
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  • #2
Hello. Welcome to PF.

You don't want to assume an ideal gas.

You've gotten to μ = (∂U/∂V)T*Cv-1. So, you need to deal with the factor (∂U/∂V)T. The first law will be helpful. Then, maybe use a Maxwell relation.
 
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1. What is the Joule Coefficient Derivation?

The Joule Coefficient Derivation is a mathematical equation used to calculate the change in internal energy of a substance due to a change in its temperature.

2. How is the Joule Coefficient Derivation derived?

The Joule Coefficient Derivation is derived from the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

3. What are the units of the Joule Coefficient?

The Joule Coefficient has units of energy per temperature (J/K). This represents the amount of energy required to raise the temperature of a substance by one degree Kelvin.

4. What is the significance of the Joule Coefficient in thermodynamics?

The Joule Coefficient is significant in thermodynamics because it relates the change in internal energy of a substance to its temperature change. This allows us to understand and predict the behavior of substances under different thermal conditions.

5. How is the Joule Coefficient used in practical applications?

The Joule Coefficient is used in various practical applications, such as in the design of heating and cooling systems, as well as in the study of phase transitions in materials. It is also used in the development of thermometers and other temperature measuring devices.

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