# 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

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.