Thermodynamics-internal energy of a surface of a liquid

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SUMMARY

The discussion focuses on deriving the Maxwell relation from the differential form of the Helmholtz free energy, specifically for a liquid's surface with surface tension γ and area A. The Helmholtz free energy is expressed as F = U - TS, leading to the differential dF = -SdT + γdA. Participants confirm that the Maxwell relation is established as (∂S/∂A)ₜ = -(∂γ/∂T)ₐ, using Young's theorem and the properties of exact differentials. The internal energy per unit area is derived as u(T) = γ - T - (∂γ/∂T)ₐ.

PREREQUISITES
  • Understanding of Helmholtz free energy (F = U - TS)
  • Knowledge of Maxwell relations in thermodynamics
  • Familiarity with exact differentials and partial derivatives
  • Concept of surface tension (γ) in liquids
NEXT STEPS
  • Study the derivation of Maxwell relations in thermodynamics
  • Learn about the implications of surface tension on thermodynamic properties
  • Explore the concept of exact differentials in calculus
  • Investigate the relationship between internal energy and entropy in thermodynamic systems
USEFUL FOR

Students and professionals in physics and engineering, particularly those specializing in thermodynamics and fluid mechanics, will benefit from this discussion.

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Homework Statement


The differential of the internal energy of a surface of a liquid with surface tension \gamma and area A may be written as
dU=TdS+\gamma dA

Write down the corresponding form of the Helmholtz free energy F= U -TS. Using the fact that these equations involve exact differentials,derive the Maxwell relation
(\frac{\partial S}{\partial A})_{T}=-(\frac{\partial \gamma}{\partial T})_{A}
The internal energy and the entropy are proportional to the area A. Show that the internal energy per unit area is
u(T)= \frac{U}{A}=\gamma-T-(\frac{\partial \gamma}{\partial T})_{A}

Homework Equations



The first part I solved in this way:

F= U-TS
dF= dU-Tds-Sdt
dF= TdS+dW-Tds-Sdt=dW-Sdt
I found on the internet that the mechanical work needed to increase a surface is dW=\gamma dA
so my dF= -SdT+\gamma dA

How to show the last expresion?
Anyone willing to help?
 
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For the maxwell relation use the differential form of the Helmholtz free energy you derived, and read off the partial derivatives, now differentiate them again and use young's theorem.
 
hm I am not quite following...could you give me an example?
 
dF=-SdT + \gamma dA

this is a differential for the function F(T,A). Hence we can see that:

\frac{\partial F}{\partial T}\bigg |_A = -S

Now differentiate wrt A keeping T constant:

\frac{\partial^2 F}{\partial A\partial T} = -\frac{\partial S}{\partial A}\bigg |_T

Now do the same for the other part:

\frac{\partial F}{\partial A}\bigg |_T = \gamma

differentiate wrt. T keeping A constant and then use Young's theorem, and you are done
 
oh that makes really sense

is this correct

\frac{\partial^2 F}{\partial T\partial A} = (\frac{\partial \gamma}{\partial T})_A

so (\frac{\partial S}{\partial A})_{T}=-(\frac{\partial \gamma}{\partial T})_{A} by Young's theorem.

and the last part

u(T)= \frac{U}{A}= \frac{TS+\gamma A}{A}=\frac{TS}{A}+\gamma= \gamma -T(\frac{\partial \gamma}{\partial T})_{A}[/itex]<br /> is this correct?
 

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