1
$\begingroup$

I thought it would be fun to get an idea of how enormously large $W$ in $S = k\ln W$ is. I'd specifically like to estimate this for a mole of water at $\pu{1 atm}$ and $\pu{298 K}$. To do this, I need an estimate of the absolute molar entropy of water at those conditions, enabling me to calculate W using: $$W= e^{S/k} $$

Are there tables of absolute (as opposed to standard) molar entropies of water at various temperatures? I searched, and was unable to find one.

Alternately, is there an expression (or series of expressions) that provides an estimate for $C_p(T)$ of ice at $\pu{1 atm}$, between $\sim\pu{0 K}$ and $T_\mathrm{fus}$? If there were, I could use this to estimate the absolute entropy of ice at $\pu{1 atm}$ and $\pu{273.15 K}$ by substituting it into the following:

$$ S_T-S_0=\int_0^T \frac{C_p(T)}{T}\mathrm{d}T$$

It would then be an easy matter to add the difference in entropy between ice at $T_\mathrm{fus}$ and water at $\pu{298 K}$.

$\endgroup$
  • $\begingroup$ Look for Debye Model and Einstein Model of Specific heats. $\endgroup$ – Soumik Das Feb 18 at 5:08
3
$\begingroup$

To evaluate $S=Nk\ln(\Omega)$ it is necessary to use statistical mechanics to calculate $\Omega$.

In quantum mechanics, for any independent system in a potential there is generally a set of permissible energy levels $\epsilon_1,\,\epsilon_2,\cdots$. The energy of the whole assembly of $N$ similar systems is $E$ and the volume $V$. The number of microstates or configurations (complexions) is $\Omega$ and this depends on $E,\,V,\, N$. It is assumed that all of the different microstates with the same values of $E,\,V,\,N$ are equally probable.

There are $n_1$ systems with energy $\epsilon_1$, $n_2$ systems with energy $\epsilon_2$ and so on. The total number of different states of the assembly is the number of ways of dividing $N$ things into groups with $n_1, n_2\cdots$ in each which is $N!/(n_1!n_2!\cdots)$. The total number of different states is then the total of all of these and is

$$\sum_{\text{all possible sets }n_i} \frac{N!}{\prod_i n_i}$$

As the total energy $E$ is fixed as is $N$, then

$$\Omega = \sum \frac{N!}{\prod_i n_i} \quad \text{with constraints} \quad \sum_i n_i =N, \sum_i \epsilon_i n_i=E$$

This is a difficult expression to evaluate but it happens that the largest term makes such an overwhelming contribution that only this term need be considered. The mathematical method to find the maximum is an interesting use of Lagrange's method of undetermined multipliers and is given in texts on stat mech but covers several pages so is too long to include here.

The result is that

$$S=k\left(N\ln\left(\sum_i e^{-\epsilon_i /(kT)} \right) +\frac{E}{kT}\right) $$

where $\displaystyle Z=\sum_i e^{-\epsilon_i /(kT)}$ is the partition function. The entropy thus comes down to calculating the partition function. In fact all other thermodynamic properties can be found once $Z$ is known.

In a molecular system the $\epsilon_i$ are the energies of translation, vibration and rotation and electronic excited states (if any). Knowing the energy levels, say from spectroscopy, then allows each of these terms to be calculated. The translational partition function is calculated using the Sakur-Tetrode equation and a value $\ln(Z)\approx 70 $ is typical. For rotational motion $\ln(Z)\approx 10$ and for vibrations $\ln(Z)\approx 1.5$ for each vibrational mode. The vibrational contribution is small because at room temperature $kT\approx 210$ wavenumbers and vibrations are typically several hundred to a few thousand wavenumbers. In the case of liquid water you have a particular problem which may be mitigated somewhat by adding H bonds to a molecule as an extra vibration.

$\endgroup$
  • $\begingroup$ This is not my post, but your answer makes me want to ask: Is it appropriate to apply the Sakur-Tetrode equation to molecules in solution? Are not translational motions in solution (particularly a highly associated liquid) better described as collective (correlated, like vibrational modes)? In general, how generally applicable is the equation for S? Assuming MD results are exact, are results of the S equation you present close to exact? $\endgroup$ – Buck Thorn Feb 18 at 18:45
  • $\begingroup$ I suggested adding extra terms only as a crude approximation. Clearly it is going to be far more complicated in a liquid and I'm not sure if this is a fully solved problem. Have a look at chapter 3 in 'Properties of Liquids and Solutions' by Murrell & Jenkins and also at the Percus-Yervick theory. There are also many computer methods, see 'Computer Simulation of Liquids', by Allen & Tildesley. $\endgroup$ – porphyrin Feb 19 at 12:51
  • $\begingroup$ Actually, given that there are literature values available for absolute molar entropies it seems that, if one only wishes to get an idea of the value of W, calculating from scratch is not necessary. One can simply plug the value of the molar entropy into $S = k \ln W$, and solve for W, as I've done. If, however, one wished to understand why S has a particular value for a particular compound, then it could be useful to go through a calculation such as what you propose. $\endgroup$ – theorist Feb 22 at 3:10
0
$\begingroup$

I did find this paper from 2005 in which Zielkiewicz estimates the molar entropy of water at $\pu{298 K}$ [1].

It's behind a paywall, but the subsequent correction from 2006 is not [2].

This gives an estimate of $S_\mathrm{m, \ce{H2O}, l, \pu{298 K}} \approx \pu{60 J mol-1 K-1}$, from which I calculate that $\ln W \approx\pu{4e24}$ for 1 mole of water at $\pu{298 K}$:

$$W \approx e^\pu{4e24} \approx 10^\pu{2e24} \approx (10^{23})!$$

EDIT: It turns out the standard molar entropies available in tables actually are absolute entropies!: https://en.wikipedia.org/wiki/Standard_molar_entropy Thus one can simply look up the standard molar entropy of water at 298K, which is 75.28 J/(mol K) (https://en.wikipedia.org/wiki/Water_(data_page))

This gives about the same result: $$W \approx e^\pu{5.5e24} \approx 10^\pu{2.4e24} \approx (10^{23})!$$

Formally, this assumes all states have the same probability, which would be the case for a microcanonical ensemble, since there all states have the same energy. Nevertheless, the holds in all ensembles. For instance, in a canonical ensemble, we hold the temperature constant, and allow the energy to fluctuate. However, overwhelmingly, most states in a canonical ensemble have an energy that is no more than infinitesimally displaced from the average energy. Thus their probabilities are likewise the same, and hence $S = k \ln W$ holds generally for any thermodynamic ensemble.

References

  1. Zielkiewicz, J. Structural Properties of Water: Comparison of the SPC, SPCE, TIP4P, and TIP5P Models of Water. The Journal of Chemical Physics 2005, 123 (10), 104501. https://doi.org/10.1063/1.2018637.
  2. Zielkiewicz, J. Erratum: “Structural Properties of Water: Comparison of the SPC, SPCE, TIP4P, and TIP5P Models of Water” [J. Chem. Phys. 123, 104501 (2005)]. The Journal of Chemical Physics 2006, 124 (10), 109901. https://doi.org/10.1063/1.2178809.
$\endgroup$
  • $\begingroup$ I suspect you could also obtain a comparable result solely from estimating the number of microstates and not involving Boltzmann's entropy formula. However, this is an elegant way of solving the problem too. $\endgroup$ – andselisk Feb 18 at 9:01
  • $\begingroup$ Please check because your value for log(W) is vast, using $S=R\ln(W)$ using molar units gives a more reasonable value. $\endgroup$ – porphyrin Feb 18 at 9:07
  • $\begingroup$ @andselisk I don't know how to estimate that for a collection of interacting molecules like water. $\endgroup$ – theorist Feb 18 at 10:20
  • $\begingroup$ @porphyrin I think it has to be k, not R. More specifically, $S = k \ln W$ applies when W is the number of microstates of the system as a whole, while $S = N k \ln W = n R \ln W$ applies when W is the number of microstates per particle, and the particles are non-interacting, such that you can determine W for the particles individually, and then multiply by the number of particles. $\endgroup$ – theorist Feb 18 at 10:37
  • 1
    $\begingroup$ In addition, $W \approx N_A!$ for $N_A$ molecules seems in the right ballpark to me. Please feel free to check the calculation yourself, but if I instead use $S= n R \ln W$, with n= 1 and S = 75.28 J/(mol K) I get W = 8,500 (not ln W = 8,500, but W = 8,500), which can't be correct for a mole of molecules at room temperature. $\endgroup$ – theorist Feb 18 at 10:38
0
$\begingroup$

Before attempting to address your question, a comment on how to check whether your ginormous estimate of W makes sense (even if admittedly you are interested in orders of magnitude here): work "backwards", using Stirling's approximation:

$$\ln (N!) \approx N\ln N - N$$

For instance, if $W \approx 10^{23}!$ then

$$S = k_\mathrm{B}\ln W \approx \pu{72 J K-1 mol-1}$$

(per mole, I stand corrected).

You could also have done this for instance, if $S_\mathrm{m} = \pu{60 J mol-1 K-1}$:

$$W = \exp\left(\frac{S}{R}\right) = \exp\left(\frac{\pu{60 J mol-1 K-1}}{\pu{8.3145 J mol-1 K-1}}\right) = \pu{1.36e3}$$

This is the number of configurations for one molecule.

For one mole:

$$W_\mathrm{m} = W^{N_\mathrm{A}}$$

Note W is a much more manageable number than $W_m = W^{N_A} \approx 10^{3 \times 6 \times 10^{23}}$.

In any case, from $W_m$ it follows that

$$S_\mathrm{m} = k_\mathrm{B}\ln(W_\mathrm{m}) = k_\mathrm{B}\ln\left(W^{N_\mathrm{A}}\right) = k_\mathrm{B}N_\mathrm{A}\ln W = R\ln W$$


One way to estimate $S$ might be to estimate the gas phase entropy of steam at the temperature of interest using theory (such as Sackur-Tetrode etc), then use the Clausius-Clapeyron relation with $P-T$ data to estimate $\Delta H_\mathrm{vap}$ and from there $\Delta S_\mathrm{vap}$, or to look it up from steam enthalpy charts.

$\endgroup$
  • 1
    $\begingroup$ Night Writer/Try Hard: My calculations are correct. You've gotten yourself confused because you associate $k$ (i.e., $k_B$) with a single particle, and $R$ with a mole of particles, and thus you (mistakenly) think $S = k \ln W$ gives W per molecule, even though I've used a molar value for S. That is incorrect! $S = k \ln W$ relates S for the system with W for the system. Hence if we choose to calculate W using the molar entropy of water, $S\approx72J/K$, we will obtain W per mole of water, not W per molecule of water. $\endgroup$ – theorist Feb 21 at 23:10
  • 1
    $\begingroup$ Further, my finding that $W \approx 10^{2.4 x 10^{24}} \approx 10^{23}!$ for a mole of water is in line with how large one expects W to be for a mole of particles. For instance, see the text following eq 2.25 on p 44 of Terrell Hill's An Introduction to Statistical Thermodynamics, Dover, 1986. $\endgroup$ – theorist Feb 21 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.