Understanding Expansion, Compression and Entropy Coefficients

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SUMMARY

This discussion focuses on the derivation of the temperature change with respect to volume at constant entropy, specifically represented as (DT/DV)s. The user successfully manipulates thermodynamic equations involving the expansion coefficient (a), isothermal compression coefficient (k), heat capacity at constant pressure (cp), and heat capacity at constant volume (Cv) to arrive at the expression (DT/DV)s = a*T/(Cv*k). The discussion also includes relevant equations for entropy (S), internal energy (U), and pressure (P), confirming the relationships between these thermodynamic variables.

PREREQUISITES
  • Understanding of thermodynamic principles, particularly entropy and internal energy.
  • Familiarity with the concepts of expansion and compression coefficients.
  • Knowledge of heat capacities, specifically Cp and Cv.
  • Basic proficiency in calculus, particularly partial derivatives.
NEXT STEPS
  • Explore the derivation of Maxwell's relations in thermodynamics.
  • Study the implications of the Gibbs free energy in relation to entropy and temperature.
  • Investigate the role of the Helmholtz free energy in thermodynamic processes.
  • Learn about the applications of the Clausius-Clapeyron equation in phase transitions.
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Students and professionals in thermodynamics, physicists, and engineers seeking to deepen their understanding of thermodynamic relationships and their applications in various systems.

tsuwal
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So, until now I know:
(DV/DS)p=(DT/Dp)s=a*T/cp*(rho) (enthalpy)
(Dp/DT)v=(DS/DV)t=-a/k (helmoltz)
(DS/Dp)t=-(DV/DT)p=-Va (gibbs)

a=expansion coefficient
k=isothermal compression coefficient
cp=heat capacity at constante pressure

I want to deduce DT/DV at constant entropy=(DT/DV)s. BUT HOW?
Let me try to write S(T,V), then,
dS=Cv/T*dT-a/k*dV
putting S=0, i get,
a/k*dV=Cv/T*dT <=> (DT/DV)s=a*T/Cv*k

am I right?
 
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Is this the one you want?

[tex]\begin{array}{l}<br /> T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_V} \\ <br /> {\left( {\frac{{\partial T}}{{\partial V}}} \right)_S} = \left[ {\frac{\partial }{{\partial V}}{{\left( {\frac{{\partial U}}{{\partial S}}} \right)}_V}} \right] = \frac{{{\partial ^2}U}}{{\partial V\partial S}} \\ <br /> \end{array}[/tex]

and

[tex]\begin{array}{l}<br /> P = - {\left( {\frac{{\partial U}}{{\partial V}}} \right)_S} \\ <br /> {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V} = - \left[ {\frac{\partial }{{\partial S}}{{\left( {\frac{{\partial U}}{{\partial S}}} \right)}_V}} \right] = - \frac{{{\partial ^2}U}}{{\partial S\partial V}} \\ <br /> \end{array}[/tex]

Therefore

[tex]{\left( {\frac{{\partial T}}{{\partial V}}} \right)_S} = - {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V}[/tex]
 


Hey, thanks for worring so much, but until there I knew...
I want to evaluate that derivative further and write in terms of a,k,Cv,Cp,T,p,... as I did
(∂T/∂p)s=a*T/cp*(rho)
 

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