How Does Surface Tension Relate to Maxwell Relations?

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SUMMARY

This discussion focuses on the relationship between surface tension and Maxwell relations in thermodynamics. The key equation presented is the differential form of the internal energy, dU = T dS - p dV + μ dN + γ dA, where γ represents surface tension and A is the area. The discussion emphasizes how Maxwell relations can be derived by recognizing the partial derivatives of internal energy with respect to entropy, volume, particle number, and area. The conclusion is that understanding these relationships is crucial for connecting surface tension to thermodynamic principles.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically internal energy and Maxwell relations.
  • Familiarity with differential calculus in the context of thermodynamics.
  • Knowledge of surface tension and its role in physical chemistry.
  • Ability to manipulate and interpret partial derivatives in thermodynamic equations.
NEXT STEPS
  • Study the derivation of Maxwell relations in detail, focusing on their applications in thermodynamics.
  • Explore the concept of surface energy and its mathematical representation in thermodynamic equations.
  • Learn about the implications of surface tension in various physical systems and its measurement techniques.
  • Investigate the relationship between thermodynamic potentials and their derivatives, particularly in the context of phase transitions.
USEFUL FOR

This discussion is beneficial for students and professionals in physics and physical chemistry, particularly those studying thermodynamics, surface phenomena, and their applications in material science.

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Homework Statement



Discuss the variation of force of surface tension with the help of maxwell relations ?


Homework Equations





The Attempt at a Solution



It is a question from previous year question paper , my exams are going so i am asking for little help as i don't know how to connect the two as the books that i have don't mention it anywhere
 
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You can do this by writing the first law in differential form

dU=T\,dS-p\,dV+\mu\,dN

and adding a term for surface energy to let you set up Maxwell relations.
 
i am still not getting it ?
 
Surface tension adds an additional energy term \gamma\,dA where \gamma is the surface energy and A is the area.

Maxwell relations arise because the equation I wrote above is really

dU=\left(\frac{\partial U}{\partial S}\right)_{V,N,A}dS+\left(\frac{\partial U}{\partial V}\right)_{S,N,A}dV+\left(\frac{\partial U}{\partial N}\right)_{S,V,A}dN+\left(\frac{\partial U}{\partial A}\right)_{S,V,N}dA

and we've assigned the variables T, -p, \mu, and \gamma to the partial derivatives. Therefore

\left(\frac{\partial T}{\partial V}\right)=\left(\frac{\partial^2 U}{\partial S\,\partial V}\right)=\left(\frac{\partial^2 U}{\partial V\,\partial S}\right)=-\left(\frac{\partial p}{\partial S}\right)

You should be able to apply the same reasoning to differentials involving \gamma.
 

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