Determine Joule-Kelvin coefficient for gas given equations of state

AI Thread Summary
The discussion focuses on calculating the Joule-Kelvin coefficient (μ) for a gas using equations of state. Key calculations involve determining the thermal expansion coefficient (α) and heat capacity at constant pressure (c_P) through relationships involving volume (V), temperature (T), and entropy (S). The user successfully derives expressions for α and c_P but encounters an issue where substituting these values into the equation for μ results in μ equaling zero. There is a suggestion to express parameters in terms of pressure (P), volume (V), temperature (T), and particle number (N) to simplify the equations. The conversation highlights the challenge of correctly applying thermodynamic principles to achieve accurate results.
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Homework Statement
The equations of state of a gas are ##P=\frac{U}{V}## and ##T=3B(U²/NV)^{1/3}##. Determine ##\alpha## and ##\mu##.
Relevant Equations
##U##: internal energy; ##T##: temperature; ##\mu##: Joule-Kelvin coefficient; ##B##: positive constant; ##V##: volume; ##N##: number of moles; ##\alpha##: coefficient of thermal expansion; ##P##: pressure; ##c_P##: heat capacity at constant pressure.
Hi

##\mu=\frac{\alpha TV–V}{N c_P}##. So, firstly, I have to calculate ##\alpha## and ##c_P##. So ##\alpha=\frac{1}{V} \frac{\partial V}{\partial T}## at constant ##P##. I can write ##U=PV##, then I replace it in the equation of ##T##, solve for ##V## and then I differentiate with respect to ##T##.

Then, ##c_P=\frac{T}{N} \frac{\partial S}{\partial T}## at constant ##P##. Do I have to find the fundamental equation for ##S## using Euler and Gibbs-Duhem relations, or is there an easier way?
 
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Like Tony Stark said:
Then, ##c_P=\frac{T}{N} \frac{\partial S}{\partial T}## at constant ##P##. Do I have to find the fundamental equation for ##S## using Euler and Gibbs-Duhem relations, or is there an easier way?
Try using ##c_P = \frac 1 N \left( \frac{dQ}{dT} \right)_P## along with the first law.
 
TSny said:
Try using ##c_P = \frac 1 N \left( \frac{dQ}{dT} \right)_P## along with the first law.
Thanks! I have arrived to ##c_P=\frac{2T^2}{9B^3P}## and ##\alpha=\frac{NT^2}{9B^3P^2V}##. But when I replace this identities in the expression for ##\mu## I get ##\mu=0##
 
Like Tony Stark said:
Thanks! I have arrived to ##c_P=\frac{2T^2}{9B^3P}## and ##\alpha=\frac{NT^2}{9B^3P^2V}##.
I believe these are correct. They will simplify nicely if you use ##P=\frac{U}{V}## and ##T=3B(U^2/NV)^{1/3}## to express ##B^3## in terms of ##P##, ##V##, ##T##, and ##N##.

But when I replace this identities in the expression for ##\mu## I get ##\mu=0##
I don't get ##\mu = 0##.
 
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