Solve First Law of Thermo Homework: (DT/DV)s = -(DP/DS)v

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SUMMARY

The discussion focuses on demonstrating the relationship (DT/DV)s = -(DP/DS)v using the first law of thermodynamics, expressed as dU = TdS - PdV. The internal energy U is defined in terms of temperature T, pressure P, volume V, and entropy S. The solution involves applying Maxwell's relations, specifically the equality of mixed partial derivatives, to derive the desired relationship between the derivatives.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with Maxwell's relations
  • Knowledge of partial derivatives and their notation
  • Basic concepts of thermodynamic variables (T, P, V, S)
NEXT STEPS
  • Study the derivation of Maxwell's relations in thermodynamics
  • Learn about the implications of the first law of thermodynamics
  • Explore the concept of partial derivatives in multivariable calculus
  • Investigate applications of thermodynamic identities in real-world systems
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Students of thermodynamics, physics majors, and anyone studying the principles of energy and entropy in physical systems will benefit from this discussion.

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Homework Statement



First law can be written dU = TdS - PdV where the internal energy U may be written in terms of any two of T,P,V,S.

I have to show that (DT/DV)s = -(DP/DS)v

where D is partial d, and the subscripts s and v mean hold those constant..

Homework Equations





The Attempt at a Solution



Not really sure how to proceed at all?
 
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This is one of the Maxwell relations.
If you have a function
f(x,y)=\left( \frac{\partial f}{\partial x} \right)_y dx + \left( \frac{\partial f}{\partial y} \right)_x dy = A dx + B dy
then, according to
\frac{\partial}{y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right)
you have
\left( \frac{\partial A}{\partial y} \right)_x = \left( \frac{\partial B}{\partial x} \right)_y
 

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