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Thermodynamic identities

Posted by peeterjoot on March 7, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Impressed with the clarity of Baez’s entropic force discussion on differential forms [1], let’s use that methodology to find all the possible identities that we can get from the thermodynamic identity (for now assuming $N$ is fixed, ignoring the chemical potential.)

This isn’t actually that much work to do, since a bit of editor regular expression magic can do most of the work.

Our starting point is the thermodynamic identity

$\begin{aligned}dU = d Q + d W = T dS - P dV,\end{aligned} \hspace{\stretch{1}}(1.0.1)$

$\begin{aligned}0 = dU - T dS + P dV.\end{aligned} \hspace{\stretch{1}}(1.0.2)$

It’s quite likely that many of the identities that can be obtained will be useful, but this should at least provide a handy reference of possible conversions.

Differentials in $P, V$

This first case illustrates the method.

$\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {P}} \right)_{V} dP +\left( \frac{\partial {U}}{\partial {V}} \right)_{P} dV- T\left( \left( \frac{\partial {S}}{\partial {P}} \right)_{V} dP + \left( \frac{\partial {S}}{\partial {V}} \right)_{P} dV \right)+ P dV \\ &= dP \left( \left( \frac{\partial {U}}{\partial {P}} \right)_{V} - T \left( \frac{\partial {S}}{\partial {P}} \right)_{V} \right)+dV \left( \left( \frac{\partial {U}}{\partial {V}} \right)_{P} - T \left( \frac{\partial {S}}{\partial {V}} \right)_{P} + P \right).\end{aligned} \hspace{\stretch{1}}(1.0.3)$

Taking wedge products with $dV$ and $dP$ respectively, we form two two forms

$\begin{aligned}0 = dP \wedge dV \left( \left( \frac{\partial {U}}{\partial {P}} \right)_{V} - T \left( \frac{\partial {S}}{\partial {P}} \right)_{V} \right)\end{aligned} \hspace{\stretch{1}}(1.0.4a)$

$\begin{aligned}0 = dV \wedge dP \left( \left( \frac{\partial {U}}{\partial {V}} \right)_{P} - T \left( \frac{\partial {S}}{\partial {V}} \right)_{P} + P \right).\end{aligned} \hspace{\stretch{1}}(1.0.4b)$

Since these must both be zero we find

$\begin{aligned}\left( \frac{\partial {U}}{\partial {P}} \right)_{V} = T \left( \frac{\partial {S}}{\partial {P}} \right)_{V}\end{aligned} \hspace{\stretch{1}}(1.0.5a)$

$\begin{aligned}P =-\left( \frac{\partial {U}}{\partial {V}} \right)_{P}- T \left( \frac{\partial {S}}{\partial {V}} \right)_{P}.\end{aligned} \hspace{\stretch{1}}(1.0.5b)$

Differentials in $P, T$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {P}} \right)_{T} dP + \left( \frac{\partial {U}}{\partial {T}} \right)_{P} dT-T \left( \left( \frac{\partial {S}}{\partial {P}} \right)_{T} dP + \left( \frac{\partial {S}}{\partial {T}} \right)_{P} dT \right)+\left( \frac{\partial {V}}{\partial {P}} \right)_{T} dP + \left( \frac{\partial {V}}{\partial {T}} \right)_{P} dT,\end{aligned} \hspace{\stretch{1}}(1.0.6)$

$\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {P}} \right)_{T} -T \left( \frac{\partial {S}}{\partial {P}} \right)_{T} + \left( \frac{\partial {V}}{\partial {P}} \right)_{T}\end{aligned} \hspace{\stretch{1}}(1.0.7a)$

$\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {T}} \right)_{P} -T \left( \frac{\partial {S}}{\partial {T}} \right)_{P} + \left( \frac{\partial {V}}{\partial {T}} \right)_{P}.\end{aligned} \hspace{\stretch{1}}(1.0.7b)$

Differentials in $P, S$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {P}} \right)_{S} dP + \left( \frac{\partial {U}}{\partial {S}} \right)_{P} dS- T dS+ P \left( \left( \frac{\partial {V}}{\partial {P}} \right)_{S} dP + \left( \frac{\partial {V}}{\partial {S}} \right)_{P} dS \right),\end{aligned} \hspace{\stretch{1}}(1.0.8)$

$\begin{aligned}\left( \frac{\partial {U}}{\partial {P}} \right)_{S} = -P \left( \frac{\partial {V}}{\partial {P}} \right)_{S}\end{aligned} \hspace{\stretch{1}}(1.0.9a)$

$\begin{aligned}T = \left( \frac{\partial {U}}{\partial {S}} \right)_{P} + P \left( \frac{\partial {V}}{\partial {S}} \right)_{P}.\end{aligned} \hspace{\stretch{1}}(1.0.9b)$

Differentials in $P, U$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= dU - T \left( \left( \frac{\partial {S}}{\partial {P}} \right)_{U} dP + \left( \frac{\partial {S}}{\partial {U}} \right)_{P} dU \right)+ P\left( \left( \frac{\partial {V}}{\partial {P}} \right)_{U} dP + \left( \frac{\partial {V}}{\partial {U}} \right)_{P} dU \right),\end{aligned} \hspace{\stretch{1}}(1.0.10)$

$\begin{aligned}0 = 1 - T \left( \frac{\partial {S}}{\partial {U}} \right)_{P} + P \left( \frac{\partial {V}}{\partial {U}} \right)_{P} \end{aligned} \hspace{\stretch{1}}(1.0.11a)$

$\begin{aligned}T \left( \frac{\partial {S}}{\partial {P}} \right)_{U} = P \left( \frac{\partial {V}}{\partial {P}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.11b)$

Differentials in $V, T$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {V}} \right)_{T} dV + \left( \frac{\partial {U}}{\partial {T}} \right)_{V} dT - T \left( \left( \frac{\partial {S}}{\partial {V}} \right)_{T} dV + \left( \frac{\partial {S}}{\partial {T}} \right)_{V} dT \right)+ P dV,\end{aligned} \hspace{\stretch{1}}(1.0.12)$

$\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {V}} \right)_{T} - T \left( \frac{\partial {S}}{\partial {V}} \right)_{T} + P \end{aligned} \hspace{\stretch{1}}(1.0.13a)$

$\begin{aligned}\left( \frac{\partial {U}}{\partial {T}} \right)_{V} = T \left( \frac{\partial {S}}{\partial {T}} \right)_{V}.\end{aligned} \hspace{\stretch{1}}(1.0.13b)$

Differentials in $V, S$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {V}} \right)_{S} dV + \left( \frac{\partial {U}}{\partial {S}} \right)_{V} dS - T dS+ P dV,\end{aligned} \hspace{\stretch{1}}(1.0.14)$

$\begin{aligned}P = -\left( \frac{\partial {U}}{\partial {V}} \right)_{S}\end{aligned} \hspace{\stretch{1}}(1.0.15a)$

$\begin{aligned}T = \left( \frac{\partial {U}}{\partial {S}} \right)_{V} .\end{aligned} \hspace{\stretch{1}}(1.0.15b)$

Differentials in $V, U$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= dU- T \left( \left( \frac{\partial {S}}{\partial {V}} \right)_{U} dV + \left( \frac{\partial {S}}{\partial {U}} \right)_{V} dU \right)+ P \left( \left( \frac{\partial {V}}{\partial {V}} \right)_{U} dV + \left( \frac{\partial {V}}{\partial {U}} \right)_{V} dU \right)\end{aligned} \hspace{\stretch{1}}(1.0.16)$

$\begin{aligned}0 = 1 - T \left( \frac{\partial {S}}{\partial {U}} \right)_{V} + P \left( \frac{\partial {V}}{\partial {U}} \right)_{V} \end{aligned} \hspace{\stretch{1}}(1.0.17a)$

$\begin{aligned}T \left( \frac{\partial {S}}{\partial {V}} \right)_{U} = P \left( \frac{\partial {V}}{\partial {V}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.17b)$

Differentials in $S, T$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \left( \frac{\partial {U}}{\partial {S}} \right)_{T} dS + \left( \frac{\partial {U}}{\partial {T}} \right)_{S} dT \right)- T dS+ P \left( \left( \frac{\partial {V}}{\partial {S}} \right)_{T} dS + \left( \frac{\partial {V}}{\partial {T}} \right)_{S} dT \right),\end{aligned} \hspace{\stretch{1}}(1.0.18)$

$\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {S}} \right)_{T} - T + P \left( \frac{\partial {V}}{\partial {S}} \right)_{T} \end{aligned} \hspace{\stretch{1}}(1.0.19a)$

$\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {T}} \right)_{S} + P \left( \frac{\partial {V}}{\partial {T}} \right)_{S}.\end{aligned} \hspace{\stretch{1}}(1.0.19b)$

Differentials in $S, U$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= dU - T dS+ P \left( \left( \frac{\partial {V}}{\partial {S}} \right)_{U} dS + \left( \frac{\partial {V}}{\partial {U}} \right)_{S} dU \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)$

$\begin{aligned}\frac{1}{{P}} = - \left( \frac{\partial {V}}{\partial {U}} \right)_{S} \end{aligned} \hspace{\stretch{1}}(1.0.21a)$

$\begin{aligned}T = P \left( \frac{\partial {V}}{\partial {S}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.21b)$

Differentials in $T, U$

$\begin{aligned}0 &= dU - T dS + P dV \\ &= dU - T \left( \left( \frac{\partial {S}}{\partial {T}} \right)_{U} dT + \left( \frac{\partial {S}}{\partial {U}} \right)_{T} dU \right)+ P\left( \left( \frac{\partial {V}}{\partial {T}} \right)_{U} dT + \left( \frac{\partial {V}}{\partial {U}} \right)_{T} dU \right),\end{aligned} \hspace{\stretch{1}}(1.0.22)$

$\begin{aligned}0 = 1 - T \left( \frac{\partial {S}}{\partial {U}} \right)_{T} + P \left( \frac{\partial {V}}{\partial {U}} \right)_{T} \end{aligned} \hspace{\stretch{1}}(1.0.23a)$

$\begin{aligned}T \left( \frac{\partial {S}}{\partial {T}} \right)_{U} = P \left( \frac{\partial {V}}{\partial {T}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.23b)$

References

[1] John Baez. Entropic forces, 2012. URL http://johncarlosbaez.wordpress.com/2012/02/01/entropic-forces/. [Online; accessed 07-March-2013].

This entry was posted on March 7, 2013 at 11:09 pm and is filed under Math and Physics Learning.. Tagged: differential form, energy, entropy, one form, partial deriviative, PHY452H1S, pressure, statistical mechanics, temperature, thermodynamic identity, two form, volume. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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