Thermodynamics-internal energy of a surface of a liquid

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Homework Help Overview

The discussion revolves around the internal energy of a liquid's surface, specifically examining the relationship between internal energy, surface tension, and Helmholtz free energy. The original poster presents a differential equation involving internal energy, surface tension, and area, and seeks to derive a Maxwell relation and express internal energy per unit area.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the Helmholtz free energy from the given differential form and seeks assistance in showing the final expression for internal energy per unit area. Other participants suggest using the differential form to identify partial derivatives and apply Young's theorem.

Discussion Status

Participants are actively engaging with the problem, with some providing insights on how to differentiate the Helmholtz free energy and apply relevant theorems. There is a recognition of the relationships between the variables involved, but no explicit consensus has been reached on the correctness of the final expressions.

Contextual Notes

Participants are working within the constraints of thermodynamic principles and the specific relationships defined by the problem, including the assumption that internal energy and entropy are proportional to area.

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[/tex]

Homework Statement


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

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
[itex](\frac{\partial S}{\partial A})_{T}=-(\frac{\partial \gamma}{\partial T})_{A}[/itex]
The internal energy and the entropy are proportional to the area A. Show that the internal energy per unit area is
[itex]u(T)= \frac{U}{A}=\gamma-T-(\frac{\partial \gamma}{\partial T})_{A}[/itex]

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 [tex]dW=\gamma dA[/tex]
so my dF= -SdT+[tex]\gamma dA[/tex]

How to show the last expresion?
Anyone willing to help?
 
Last edited:
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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?
 
[tex]dF=-SdT + \gamma dA[/tex]

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

[tex]\frac{\partial F}{\partial T}\bigg |_A = -S[/tex]

Now differentiate wrt A keeping T constant:

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

Now do the same for the other part:

[tex]\frac{\partial F}{\partial A}\bigg |_T = \gamma[/tex]

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

is this correct

[tex]\frac{\partial^2 F}{\partial T\partial A} = (\frac{\partial \gamma}{\partial T})_A[/tex]

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

and the last part

[tex]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?[/tex]
 

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