Rearranging Equations

1. Oct 18, 2005

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

2. Oct 18, 2005

MalleusScientiarum

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

3. Oct 18, 2005

eric99gt

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

4. Oct 18, 2005

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?

5. Oct 19, 2005

HallsofIvy

Staff Emeritus
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)T0.5, 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.

6. Oct 19, 2005

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.