The class is Thermodynamics and Kinetics of Materials, the section is on surface energy using "Phase Transfromations in Metals and Alloys" by Porter, Easterling, and Sharif.The first image shows the problem. My understanding from Eq 3.1 is that the free energy of the 2 dimensional rectangle (with no bulk and two surfaces) would be G=2AY (Y being gamma [free energy per unit area] in the text). The second image is the pertinant section. The problem gives two length variables, each with an assigned free energy (Y) and says the area is constant.The solution claims the Gibbs Energy is G=2(l1Y1+l2Y2) which I don't understand if the area should be the product of the lengths, not the sum. 3.3 If a two-dimensional rectangular crystal is bounded by sidesof lengths 1 and ₂ show by differentiation that the equilibriumshape is given by1₁_12=I2 71where ₁ and 2 are the energies of the sides and 2 respectively.(The area of the crystal 2 is constant.) 3.1 Interfacial Free EnergyIt is common practice to talk of interfacial energy. In reality, however, what isusually meant and measured by experiment is the interfacial free energy, y.The free energy of a system containing an interface of area A and free energyy per unit area is given byG=G₁ + Ay(3.1)where Go is the free energy of the system assuming that all material in thesystem has the properties of the bulk-y is therefore the excess free energyarising from the fact that some material lies in or close to the interface. It isalso the work that must be done at constant T and P to create unit area ofinterface.Consider for simplicity a wire frame suspending a liquid film, Fig. 3.1. Ifone bar of the frame is movable it is found that a force F per unit length mustbe applied to maintain the bar in position. If this force moves a small dis-tance so that the total area of the film is increased by dA the work done bythe force is FdA. This work is used to increase the free energy of the systemby dG. From Equation 3.1Equating this with FdA givesdG = ydA+ AdydydAF=y+A=(3.2)In the case of a liquid film the surface energy is independent of the area ofthe interface and dy/dA = 0. This leads to the well-known resultF = Y(3.3)i.e. a surface with a free energy y J m² exerts a surface tension of y N m-¹.

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The class is Thermodynamics and Kinetics of Materials, the section is on surface energy using "Phase Transfromations in Metals and Alloys" by Porter, Easterling, and Sharif.

The first image shows the problem. My understanding from Eq 3.1 is that the free energy of the 2 dimensional rectangle (with no bulk and two surfaces) would be G=2AY (Y being gamma [free energy per unit area] in the text). The second image is the pertinant section. The problem gives two length variables, each with an assigned free energy (Y) and says the area is constant.

The solution claims the Gibbs Energy is G=2(l1Y1+l2Y2) which I don't understand if the area should be the product of the lengths, not the sum.

3.3 If a two-dimensional rectangular crystal is bounded by sides
of lengths 1 and ₂ show by differentiation that the equilibrium
shape is given by
1₁_12
=
I2 71
where ₁ and 2 are the energies of the sides and 2 respectively.
(The area of the crystal 2 is constant.)

3.1 Interfacial Free Energy
It is common practice to talk of interfacial energy. In reality, however, what is
usually meant and measured by experiment is the interfacial free energy, y.
The free energy of a system containing an interface of area A and free energy
y per unit area is given by
G=G₁ + Ay
(3.1)
where Go is the free energy of the system assuming that all material in the
system has the properties of the bulk-y is therefore the excess free energy
arising from the fact that some material lies in or close to the interface. It is
also the work that must be done at constant T and P to create unit area of
interface.
Consider for simplicity a wire frame suspending a liquid film, Fig. 3.1. If
one bar of the frame is movable it is found that a force F per unit length must
be applied to maintain the bar in position. If this force moves a small dis-
tance so that the total area of the film is increased by dA the work done by
the force is FdA. This work is used to increase the free energy of the system
by dG. From Equation 3.1
Equating this with FdA gives
dG = ydA+ Ady
dy
dA
F=y+A=
(3.2)
In the case of a liquid film the surface energy is independent of the area of
the interface and dy/dA = 0. This leads to the well-known result
F = Y
(3.3)
i.e. a surface with a free energy y J m² exerts a surface tension of y N m-¹.

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