Taking the reciproque of a partial derivative (as seen in thermodynamics)?

[nonequilibrium]

Hello, I'm a first year physics student and in thermodynamics we always use [tex] \frac{1}{ \frac{dX}{dY} } = \frac{dY}{dX} [/tex] and I was wondering 'how true' this is, i.e. what are the conditions for this to be true? For example, if I have the equation of state of a Vanderwaals gas: [tex]\left( P + \frac{an^2^}{V^2^}\right) \left(V + bn \right) = nRT[/tex] and I want to find [tex]\left( \frac{dV}{dT} \right)_P[/tex], can I just calculate [tex]\left( \frac{dT}{dV} \right)_P[/tex] (which is much easier) and then take the reciproque?

(If not, is there another way?)

Thank you.

 

[Gerenuk]

Yes, you can do that. But note that you won't simplify things as your derivative will be a function of V so it's not necessarily easier to rearrange it for a function of T (like the first derivative would be).

 

[nonequilibrium]

Oh, it is always true? That's nice :) I thought there could be a problem if your function isn't injective.

And I don't get what you mean by "not easier"? To differentiate V with respect to T, you have to isolate V, which seems pretty much impossible, yet on the other hand, if you can differentiate T with respect to V, T is already isolated and then you just differentiate with the product rule and you're done. (and then take the reciproque)

 

[Gerenuk]

I guess there are some conditions on local properties. But as differentiating is a local operation, you don't have to worry about global behaviour. In physics things are well behaved ;)

I meant if you are asked dV/dT as a function of T, then your second method would rather yields 1/ (dT/dV) being a function of V. Then you'd still have to substitutde for V if you wanted to make it a function of T.

 

[nonequilibrium]

Well in thermodynamics we don't see state variables too much like functions, but more like numbers or relations (so there's no need to isolate, unless you want to take the partial derivative)

 

[Gerenuk]

Btw, it's an insight from thermodynamics that you can use to get more intuition for partial derivatives.

To find [tex]\left(\frac{\partial y}{\partial x}\right)_z[/tex] you just calculate with the more intuitive total differentials and write
[tex]\left(\frac{\partial y}{\partial x}\right)_z=\frac{dy}{dx}\qquad(dz=0)[/tex]
where the RHS is a normal division and not meant to imply differentiation.