Chapter 5: Problem 5
A feeling for the numbers: comparing multiplicities Boltzmann's equation for the entropy (eqn 5.29 ) tells us that the entropy difference between a gas and a liquid is given by \(S_{\text {gar }}-S_{\text {liquid }}=k_{E} \ln \frac{W_{\text {Qas }}}{W_{\text {llavid }}}\) From the macroscopic definition of entropy as \(\mathrm{d} S=\mathrm{d} Q / \mathrm{T}\) we can make an estimate of the ratios of multiplicities by noting that boiling of water takes place at fixed \(T\) at 373 K. (a) Consider a cubic centimeter of water and use the result that the heat needed to boil water (the latent heat of vaporIzation) is given by \(Q_{\text {weviration }}=40.66 \mathrm{kJ} / \mathrm{mol}\) (at \(100^{\circ} \mathrm{C}\) ) to estimate the ratio of multiplicities of water and water vapor for this number of molecules. Write your result as 10 to some power. If we think of multiplicities in terms of an ideal gas at fixed \(T,\) then $$\frac{w_{1}}{W_{2}}=\left(\frac{V_{1}}{V_{2}}\right)^{N}$$ What volume change would one need to account for the liquid/vapor multiplicity ratio? Does this make sense? (b) In the chapter we discussed the Stirling approximation and the fact that our results are incredibly tolerant of error. Let us pursue that in more detail. We have found that the typical types of multiplicities for a system like a gas are of the order of \(W \approx \exp \left(10^{25}\right) .\) Now, let us say we are off by a factor of \(10^{1000}\) in our estimate of the multiplicities, namely, \(W=10^{1000} \exp \left(10^{25}\right) .\) Show that the difference in our evaluation of the entropy is utterly negligible whether we use the first or second of these results for the multiplicity. This is the error tolerance that permits us to use the Stirling approximation so casually!
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