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We start from a jacobian matrix: $$J=\begin{pmatrix}\left.\dfrac{∂ρ}{∂T}\right|_p&\left.\dfrac{∂ρ}{∂p}\right|_T\\\left.\dfrac{∂U}{∂T}\right|_p&\left.\dfrac{∂U}{∂p}\right|_T\end{pmatrix},$$ with $\rho(p,T)$, $U(p,T)$, $p(\rho,U)$ and $T(\rho, U)$. We know that it is possible to get the derivative of the inverse map using

$$\begin{pmatrix}\left.\dfrac{∂T}{∂ρ}\right|_U & \left.\dfrac{∂T}{∂U}\right|_ρ\\ \left.\dfrac{∂p}{∂ρ}\right|_U & \left.\dfrac{∂p}{∂U}\right|_ρ\end{pmatrix} = J^{-1}.$$

The question is: is it possible, with no more information, to obtain analytically $\left.\dfrac{∂U}{∂ρ}\right|_T$ or $\left.\dfrac{∂p}{∂ρ}\right|_T$, and how?

Context: I am trying to compute the speed of sound in a gas

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If $U(T,p) = U(T(\rho,U),p(\rho,U))$, the chain rule gives the relation $$ \frac{\partial U}{\partial \rho}(T(\rho,U),p(\rho,U)) = \frac{\partial U}{\partial T}(T(\rho,U),p(\rho,U))\frac{\partial T}{\partial \rho}(\rho,U) + \frac{\partial U}{\partial p}(T(\rho,U),p(\rho,U))\frac{\partial p}{\partial \rho}(\rho,U). $$

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  • $\begingroup$ So, If I want to get $\partial U / \partial \rho$ at constant temperature, can I only take the second term? It seems to me that what you propose is not at constant temperature, and something would be be missing by dropping the first term. $\endgroup$
    – seb007
    Oct 3 '18 at 9:28
  • $\begingroup$ Yes, of course if you want the expression at constant temperature you should drop the first term, and substitute $T$ instead of $T(\rho,U)$ in the second. Similar expressions may be found for the density $\rho$ at constant temperature. $\endgroup$
    – Hugo
    Oct 3 '18 at 9:31
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It seems simpler than I thought.

Using the chain rule, we can write

$$\left.\frac{\partial U}{\partial \rho}\right|_T = \left.\frac{\partial U}{\partial p}\right|_T \left.\frac{\partial p}{\partial \rho}\right|_T$$

and then, simply

$$\left.\frac{\partial p}{\partial \rho}\right|_T = \frac{1}{\left.\frac{\partial \rho}{\partial p}\right|_T},$$

because $T$ is left constant. Then we have all the required derivatives.

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