Rearranging Equations - Solving for v

[eric99gt]

Alright I'm going absolutely nuts here. It's a simple matter of rearranging an equation and for the life of me I can't do it. I've been working at it for like an hour now to no avail. Here's the equation.

P=[RT/(v-b)]-[a/v(v+b)T^0.5]

I'm trying to solve for v. I've manipulated this equation every way I can think of and still can't get it. I'm sure it's something simple but I'm stuck. Please help.
Thanks

 

[MalleusScientiarum]

you'll get a quadratic in V. You are aware of this, right?

 

[eric99gt]

Nope. Completely over my head. How in the world would you get that in the form of ax^2+bx+c=0?

 

[eric99gt]

Well here's the real problem. I'm trying to get the partial of (dv/dT) with constant pressure from that equation. What I was trying to do was to find v from the equation and take the partial of that side with respect to v and the partial of the other side with respect to T. Is this what I should be doing?

 

[HallsofIvy]

Actually, you don't get a quadratic, you get a cubic! The least common denominator of the fractions, if I am reading this properly,
$$P= \frac{RT}{v-b}-\frac{a}{v(v+b)T^{0.5}}$$
(Is that what you intended? Your fractions are ambiguous.)
is v(v-b)(v+b)T^0.5, a cubic in v.
Cubics in general are hard to solve. Since you say you really want the derivative of y wrt T, P being constant, I would recommend "implicit differentiation". Writing the equation as
$$P= RT(v-b)^{-1}- av^{-1}(v-b)^{-1}T^{-1/2}$$
differentiate both sides: using product and chain rules, wrt T:
$$0= R(v-b)^{-1}- RT(v-b)^{-2}v'+av^{-2}(v-b)^{-1}T^{-1/2}v'+ av^{-1}{v-b}^{-2}T^{-1/2}v'+ (a/2)v^{-1}(v-b)^{-1}T^{-3/2}$$
and solve for v', the derivative of v with respect to T.

 

[eric99gt]

I think my biggest problem is I'm not really seeing how to solve partials. Anyone have any good literature about the subject. Thanks.