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In this answer one obtains an expression of entropy of a Van der val gas:

$$S(T,V)=S_0+\int_{T_0}^T\frac{C_v(T)}{T}dT+nR \rm{ln} \left( \frac {V-nb}{V_0-nb} \right) $$

where $S$ is the entropy, $C_v$ is the heat capcity, $T$ is the temperature, $n$ is the number of molecules and $R$, $V_0$ , $S_0$, $b$ are constants.

How does one determine the constants $S_0$ and $V_0$

I suspect $S_0$ is the same as in the ideal gas case with Sackur-Tetrode equation? How does one find the constant $V_0 $?

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  • $\begingroup$ Because of the logarithm, there is no possible way to fix the exact value at, say, infinity, or at small sizes. So that this expression can only be fixed at an arbitrary state, where you get to arbitrarily pick its value of entropy $S_0$. The volume of the arbitrarily chosen state is $V_0$. i.e. you get to pick both. $\endgroup$
    – naturallyInconsistent
    May 13 at 5:46
  • $\begingroup$ @naturallyInconsistent see this: en.wikipedia.org/wiki/Sackur%E2%80%93Tetrode_equation#Formula $\endgroup$
    – More Anonymous
    May 13 at 5:48
  • $\begingroup$ I know about the Sackur-Tetrode. My above comment is already written in a way that should read as bringing the known stuff from Sackur-Tetrode to your case. $\endgroup$
    – naturallyInconsistent
    May 13 at 5:49
  • $\begingroup$ Also, this is a known problem with classical descriptions of gases. Their entropy formula does not have the correct behaviour that nerst postulate says they should $\endgroup$
    – naturallyInconsistent
    May 13 at 5:51
  • $\begingroup$ Ah I see. Feel free to write an answer with references? $\endgroup$
    – More Anonymous
    May 13 at 5:52

1 Answer 1

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For practical calculations we set the entropy to an arbitrary value, usually zero, at a chosen temperature and pressure: $$ S_0 = S(T_0,P_0) = 0 $$ $V_0$ then must be obtained by solving the van der Waals equaition: $$ P_0 = \frac{RT_0}{V_0-b} - \frac{a}{V_0^2} $$

To complete the calculation of thermodynamic properties we also need to set the enthalpy (or the internal energy, but not both) to zero, usually at the same state.

This is known as the reference state and fixes the absolute values of $S$ and $H$ (or $U$) from which all other values can be obtained. The state we pick is chosen arbitrarily. For example, the reference state for the steam tables engineers use is set for the liquid at the triple point of water, 0.01°C, 0.006117 bar.

The arbitrariness of the reference state causes consternation and confusion among students, but it is no different than setting the reference state for the gravitational potential energy at "sea level". The second law fixes the absolute entropy at $T=0$ K and we can use statistical mechanics to obtain self-consistent values, as the Sackur-Tetrode equation for point mass molecules. We need the absolute value of entropy when we are dealing with reactions, because we must have a common basis to compare the entropy of different molecules. For all other problems the arbitrary reference state suffices because we are interesed in calculations of $\Delta S$ between states, which are independent of the reference state.

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