Thermodynamics-internal energy of a surface of a liquid

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rayman123
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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?
 
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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]