Calculate the Enthropy Change of Isothermal Expansion of a Non-Ideal Gas

[Ignitia]

Homework Statement

a) Calculate the entropy change for an isothermal expansion of a Van Der Waals Gas from V1 to V2. b) Use this to calculate ΔS for 1 mole of NH3 from 2L to 20L at 298K. c) Compare this to an Ideal Gas

Relevant Equations

ΔS = ∫ (dq/T)
PV=nRT Ideal Gas
[P + a/(v/n)[SUP]2[/SUP]][v/n - b] = RT Not Ideal

Part (a)
ΔS = ∫ (dq/T)

because: dq = PdV = (nRT/V)dV

Then:
ΔS = ∫ (1/T)*(nRT/V)dV
ΔS = nR ∫(1/V) dV
ΔS =nR[ln(V2/V1)]

Part (b)
This is where I'm stuck. I know [P + a/(v/n)²][v/n - b] = RT can be solved for P and simplified to
P = [RT/(v-b)]-[a/v²] since n=1mol

But I don't know how to proceed from here, to solve for v and in turn solve ΔS. I can't use Boyle's Law and I was told there's another method besides Newton-Rapson.

Part (c)
Take ΔS =nR(ln(V2/V1) from part (a) and input: V2=20L |V1=2L

ΔS = (1mol)(8.3145 J*mol/K)(ln(10))=19.15 J/K

19.15J/K - (part b) = Final Answer

 

[Ignitia]

I think I got it:
P = [RT/(v-b)]-[a/v²]

Integrate with respect to T:

dP/dT = RT/[(v/n)-b] (d/dT) - [a/v²] (d/dT)

[a/v²] (d/dT) goes to 0 so

dP/dT = R/[(v/n)-b]

And a Maxwell states: dP/dT = dS/dV

so: dS/dV = R/[(v/n)-b]

dS = R/[(v/n)-b] dV

ΔS = ∫ R/[(v/n)-b] dV

And then just integrate from 2L to 20L? Is this correct for Part b?