A) A sample calculation at 100°C follows:
The data in the steam tables was converted from m³/kg into cm³/mol for consistency with the given units of a and b. Thus, from steam tables, at P=1 bar \widehat{V} = 1.6959 m³/kg.
\left(1.6959 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}\right)\left(\frac{100 \mathrm{~cm}}{\mathrm{~m}}\right)^{3}\left(\frac{1 \mathrm{~kg}}{1000 \mathrm{~g}}\right)\left(\frac{18.02 \mathrm{~g}}{\mathrm{~mol}}\right)=30,560 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}
This process can be repeated for all of the data points in the steam tables. This same volume is used for the sample calculation in part B.
B) The Van der Waals EOS is:
P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}^{2}}
Thus when V = 30,560 cm³/mol:
P=\frac{\left(83.14 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}\right)(373.15 \mathrm{~K})}{30,560-30.48 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}}-\frac{5.535 \times 10^{6} \frac{\mathrm{bar~cm}^{6}}{\mathrm{~mol}^{2}}}{\left(30,560 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}\right)^{2}}=1.025 \,\mathrm{bar}
Using a spreadsheet, a full plot of P vs. V from the VDW equation can be constructed and plotted, like the following:
As explained in Example 2-6, VDW is a cubic equation and there can be as many as three distinct values of V that produce the same P. In such a case, we might interpret the largest V as the vapor value and the smallest V as the liquid value. This is discussed further in Section 7.2.4.
C) The most straightforward way to do a comparison is to construct one or more plots that contain both the data from the steam tables and the VDW calculations from part B.
However care must be taken to construct the plots in a way that enables fair comparisons.
For example a plot like this one would suggest the agreement is outstanding:
And the agreement is indeed very good for the vapor phase at this temperature. However the liquid phase predictions are not at all good as illustrated by this plot:
This plot shows that the VDW equation predicts liquid molar volumes in the range of 38-39 cm³/mol for pressures of 5-500 bar. If you insert the actual liquid molar volumes (~18 cm³/mol) into the VDW equation, you get a negative pressure, because V<b.
The above perhaps suggests that the value of “b” given in the problem statement must be wrong. Since the whole intent of the parameter “b” (as described in section 2.3.4) is to represent the “minimum possible molar volume” of the compound, it makes no sense that the experimentally measured V would be smaller than b. But the values of a and b given in this problem statement are indeed the “correct” ones for water, according to a technique that will be illustrated in Section 7.2.5. The problem is not the values of a and b themselves, the problem is that the van der Waals EOS isn’t robust enough to model both liquid and vapor molar volumes.
At T=200°C the vapor pressure of water is over 15 bar, and at 300°C it is over 85 bar, so much more vapor-phase data is available. The agreement is again good for the vapor phase (though departures become more noticeable as pressure increases) but very poor for the liquid phase.
400 and 500°C are above the critical point so the complication of two phases is no longer present in the data at these temperatures. Also, the van der Waals equation P vs. V plot no longer has maxima or minima at these temperatures; the volume simply decreases with increasing pressure. This, too, is explained further in Section 7.2.5.
