Surface Tension ?

mkbh_10

1. The problem statement, all variables and given/known data

Discuss the variation of force of surface tension with the help of maxwell relations ?


2. Relevant equations



3. The attempt at a solution

It is a question from previous year question paper , my exams are going so i am asking for little help as i don't know how to connect the two as the books that i have don't mention it anywhere

Mapes

You can do this by writing the first law in differential form

[tex]dU=T\,dS-p\,dV+\mu\,dN[/tex]

and adding a term for surface energy to let you set up Maxwell relations.

mkbh_10

i am still not getting it ?

Mapes

Surface tension adds an additional energy term [itex]\gamma\,dA[/itex] where [itex]\gamma[/itex] is the surface energy and [itex]A[/itex] is the area.

Maxwell relations arise because the equation I wrote above is really

[tex]dU=\left(\frac{\partial U}{\partial S}\right)_{V,N,A}dS+\left(\frac{\partial U}{\partial V}\right)_{S,N,A}dV+\left(\frac{\partial U}{\partial N}\right)_{S,V,A}dN+\left(\frac{\partial U}{\partial A}\right)_{S,V,N}dA[/tex]

and we've assigned the variables [itex]T[/itex], [itex]-p[/itex], [itex]\mu[/itex], and [itex]\gamma[/itex] to the partial derivatives. Therefore

[tex]\left(\frac{\partial T}{\partial V}\right)=\left(\frac{\partial^2 U}{\partial S\,\partial V}\right)=\left(\frac{\partial^2 U}{\partial V\,\partial S}\right)=-\left(\frac{\partial p}{\partial S}\right)[/tex]

You should be able to apply the same reasoning to differentials involving [itex]\gamma[/itex].