Im trying to do my chem homework and I'm stumped. 


  Determine the solubility of $\ce{N2}$ in water exposed to air at 25°C, if the
  atmospheric pressure is 1.2 atm, assume that the mole fraction of
  nitrogen is 0.78 in air and the Henry's Law constant for nitrogen in
  water at this temperature is $6.1 × 10^{-4} M/atm$.



I have the answer, but i want to know how to solve for it. I've been reading through my notes and the book trying to find a reference through which to approach this problem and i have found none. If anyone can guide me in the right direction that would suffice



Question: Series expansion of large expression
Answer: To solve this problem under the assumption that the stated quantities are small, you can insert a perturbation expansion for the unknown phi and solve for its expansion coefficients in the smallness parameter:

First I'm copying the definitions from the question to be self-contained:

Ecp = (s1*s3*Abs[J]/DE^2/2 - 
         1/DE*s2*s3*KroneckerDelta[1, s1])*(t1^2 - t2^2) /. 
      KroneckerDelta[1, s1] -&gt; (1 + s1)/2 /. t1 -&gt; tt1*DE /. 
    t2 -&gt; tt2*DE /. xi -&gt; xii*DE /. J -&gt; Jj*DE
Ec2p = s2/DE^2*KroneckerDelta[1, s1] (t1^2 - t2^2)^2/Abs[J] /. 
      KroneckerDelta[1, s1] -&gt; (1 + s1)/2 /. t1 -&gt; tt1*DE /. 
    t2 -&gt; tt2*DE /. xi -&gt; xii*DE /. J -&gt; Jj*DE
Esp = s2*s3/DE^2*t1*t2*xi*
        Abs[J]/(Abs[xi] KroneckerDelta[-1, s1] + 
           Abs[J] KroneckerDelta[1, s1]) /. 
       KroneckerDelta[1, s1] -&gt; (1 + s1)/2 /. 
      KroneckerDelta[-1, s1] -&gt; (1 - s1)/2 /. t1 -&gt; tt1*DE /. 
    t2 -&gt; tt2*DE /. xi -&gt; xii*DE /. J -&gt; Jj*DE


Now I define the derivative that you're trying to set to zero:

deriv = 
 D[E00 + Cos[phi]*Ecp + Cos[phi]^2*Ec2p + Sin[phi]*Esp, phi]

(*
==&gt; (DE s2 s3 tt1 tt2 xii Abs[DE Jj] Cos[phi])/(
 1/2 (1 + s1) Abs[DE Jj] + 
  1/2 (1 - s1) Abs[DE xii]) - (DE^2 tt1^2 - 
    DE^2 tt2^2) (-(((1 + s1) s2 s3)/(2 DE)) + (s1 s3 Abs[DE Jj])/(
    2 DE^2)) Sin[phi] - ((1 + s1) s2 (DE^2 tt1^2 - DE^2 tt2^2)^2 Cos[
   phi] Sin[phi])/(DE^2 Abs[DE Jj])
*)


Now I make the assumption that tt1, tt2, Jj, xii are all proportional to the small parameter ϵ and that I can then also write phi = ϕ0 + ϵ ϕ1 to first order in that small parameter. The expansion of deriv in powers of ϵ has two nonvanishing terms if we go to third order. Setting these two non-vanishing terms to zero allows us to solve for the two unknowns ϕ0 and ϕ1. This solves the problem of finding phi to first order in the small parameter. 

Solve[
 Thread[CoefficientList[
    Normal@Series[
      Simplify[
       deriv /. 
         Thread[{tt1, tt2, Jj, 
            xii} -&gt; ϵ {tt1, tt2, Jj, xii}] /. 
        phi -&gt; (ϕ0 + ϵ ϕ1), 
       DE Jj ϵ &gt; 0 &amp;&amp; DE xii ϵ &gt; 0], {ϵ, 0,
        3}], ϵ] == 0], {ϕ0, ϕ1}]

(*
==&gt; {{ϕ0 -&gt; 
   0, ϕ1 -&gt; -((
    4 Jj tt1 tt2 xii)/((1 + s1) (tt1^2 - tt2^2) (Jj + Jj s1 + xii - 
       s1 xii)))}, {ϕ0 -&gt; -Pi, ϕ1 -&gt; -((
    4 Jj tt1 tt2 xii)/((1 + s1) (tt1^2 - tt2^2) (Jj + Jj s1 + xii - 
       s1 xii)))}, {ϕ0 -&gt; Pi, ϕ1 -&gt; -((
    4 Jj tt1 tt2 xii)/((1 + s1) (tt1^2 - tt2^2) (Jj + Jj s1 + xii - 
       s1 xii)))}}
*)


In order to get useful results, I had to add some assumptions about certain products of your parameters being positive. You may have to check whether those assumptions are appropriate for your data.