Recent content by tetrakis

Thank you, I've gotten the correct equation now.

for ##\frac{\partial \psi _2}{\partial N_1}## I'm taking ##\frac{\partial N _2}{\partial N_1}## and ##\frac{\partial ln(rN_2)}{\partial N_1}## as zero based on the assumption that it is similar to taking the derivative of a constant, would this be correct?

You may find this helpful: http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Atomic_Theory/Intermolecular_Forces/Polarizability It's essentially the ease in which the electron cloud of a neutral molecule can be distorted in order to produce a dipole moment.

Hey Chet, Thanks for the help, I've got the third term now. Am I even close with the other 2 terms? I've been playing around with them a bit more, but I can't quite seem to get it, especially the first term with the ln(1-ϕ2)

Can't really see your IR spectrum very clearly, but it does look similar to the reference IR spectrum for 1-bromobutane.

might be low solubility, and you're making an emulsion?

Homework Statement (Note: This isn't an assignment problem, more a curiosity about the derivation of an equation - hopefully it is still posted in the right forum..) I have working through the derivation for the partial molar Gibbs free energy of mixing from the Flory-Huggins expression for...

okay, fixed. $$ \frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cos(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t) $$ I'm sorry but I'm not sure what you mean by the argument...

making the entire thing $$ \frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cosh(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t) $$

okay, so for $$e^{-s}\frac{1}{s(s^2+6)} \rightarrow \frac{1}{6}e^{-s}\frac{1}{s} -\frac{1}{6}e^{-s}\frac{s}{s^{2}+6} \rightarrow \frac{1}{6}u(t-1) - \frac{1}{6}cosh(\sqrt{6})u(t-1)$$?

right, okay, and I can also apply this to the last term? L-1{2/(s2+6}=(2/√6)sin(6t)? how could I approach the exponential terms?

right! so I have 1/(s2(s2+6))=(1/6)(1/s2) - (1/6)(1/(s2+6) which the inverse laplace transform is (1/6)t-(√6/6)sin(√6t) ?

So, with the algebra corrected I have Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6) the inverse of the first term L-1{1/(s2(s2+6))} I can't find anything like this in my table, most of the entries have s2+k2 which I can't break 6 down into?

Homework Statement y''+6y=f(t), y(0)=0, y'(0)=-2 f(t)= t for 0≤t<1 and 0 for t≥1 Homework Equations The Attempt at a Solution L{y''}+6L{y}=L{t}-L{tμ(t-1)} where μ(t-1) is Unit Step Y(s)=L{y} sY(s)-y(0)=L{y'} and y(0)=0 s2Y(s)-sy(0)-y'(0)+6Y(s) where y(0)=0 and...