Thermodynamics-internal energy of a surface of a liquid

[rayman123]

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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?

 

[Thaakisfox]

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.

 

[rayman123]

hm I am not quite following...could you give me an example?

 

[Thaakisfox]

$$dF=-SdT + \gamma dA$$

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

 

[rayman123]

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

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