The Energy $U$ and the entropy $S$ are extensive state variables, so it makes sense to talk about the energy content and entropy content of a given amount of substance at standard conditions. Texbooks however rarely talk about content but only about changes in content.

How would one proceed in order to calculate the energy and entropy content of 1 liter of water at standard conditions? For the ideal gas, we have equations of state like the ideal gas law and the Sackur-Tetrode equation which allow us to perform similar calculations. Does this mean that we need the equations of state of liquid water? Is there such a thing? Or is there an easier way?

Also for the entropy content, is there an easy relationship to the heat capacity?


1 Answer 1


If you're searching for the entropy $S(T,V,N)$ as a function of volume $V$, temperature $T$ and number of particles $N$, then you can find an easy relation with heat capacity:

$$ C_V = T \left( \frac{\partial S}{\partial T} \right)_{V,N} $$

You can find this observation for example in the book of Landau (fifth volume §13), but almost in each book of thermodynamics. It derives from the definition of $C_V$ and thermodynamic potentials:

$$ S = - \left( \frac{\partial F}{\partial T} \right)_{V,N} = - \left( \frac{\partial G}{\partial T} \right)_{p,N} = - \left( \frac{\partial \Omega}{\partial T} \right)_{V,\mu} $$

For the explicit calculus of entropy you to find over of those potentials express in terms of their own thermodynamic variables: $$ F = F(V,T,N)$$ $$ \Omega = \Omega(V, T, \mu)$$ $$ G = G(p, T, N)$$

where $\mu$ is the chemical potential, $p$ there pressure.

  • 1
    $\begingroup$ Thanks for pointing out the relationship between the heat capacity and a change in entropy. Do you also know the answer to my main question? How much entropy does 1 liter of water contain? $\endgroup$
    – Marc
    Commented Oct 2, 2017 at 9:12

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