2
$\begingroup$

Lets say I have a function:$$\nu=\frac{RT}{P}+B_{p}(T)RT$$ and I am trying to find $\left(\frac{\partial \nu}{\partial T}\right)_{P}$. I understanding that the partial derivative of the first term is just $\frac{R}{P}$ but the second term has two terms that depend on T. Do I use the product rule like regular derivatives? If so, I think the answer would look something like this:

$$\left(\frac{\partial \nu}{\partial T}\right)_{P}=\frac{R}{P}+RT \frac{\partial B_{P}(T)}{\partial T}+RB_{P}(T)$$

Am I correct in saying this or does the product rule not apply to partial derivatives like I was thinking.

|cite|improve this question
$\endgroup$
  • $\begingroup$ This is the right way to do this, yes. I'm not sure about your notation. Why $\left(\frac{\partial v}{\partial T}\right)_P$ and not just $\frac{\partial v}{\partial T}$ in the left hand side ? $\endgroup$ – Pixel Dec 8 '13 at 8:15
  • $\begingroup$ Happy to see some one here playing with the mathematics in equations of state ! $\endgroup$ – Claude Leibovici Dec 8 '13 at 8:46
  • $\begingroup$ @pbs, the notation is saying the change in $\nu$ with $T$ while keeping $P$ constant. That is the notation my professor has given us and also how it is shown in the Thermodynamics book. I was just trying to be consistent $\endgroup$ – Greg Harrington Dec 8 '13 at 19:18
  • $\begingroup$ @GregHarrington I see. In that case wouldn't $\left(\frac{\partial v}{\partial T}\right)_{P,R}$ be more appropriate... $\endgroup$ – Pixel Dec 9 '13 at 12:57
1
$\begingroup$

I agree with pbs, you did it right, and I assume in your field or work you explicitly label the variable that you consider a constant (P in this case). It many branches of math and physics that notations is not used because you assume it is is so.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Yes that is the reason. Thank you $\endgroup$ – Greg Harrington Dec 8 '13 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.