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rayman123

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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]

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?

## 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: