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.

  • $\begingroup$ 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

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.

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

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