A tabulation of more precise thermodynamic data than are presented in Appendix
D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and
\(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard
state pressure of 1 bar.
$$\begin{array}{llll}
\hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ}
\\\
& \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K
}^{-1} \\
\hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\
\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\
\hline
\end{array}$$
(a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at
\(298.15 \mathrm{K}\) for the vaporization:
\(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2}
\mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you
obtain will differ slightly from that on page 838 because here, the standard
state pressure is 1 bar, and there, it is 1 atm.
(b) Use the result of part (a) to obtain the value of \(K\) for this
vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\)
(c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to
millimeters of mercury.
(d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page
838 and calculate the vapor pressure of water at 298.15 K in a fashion similar
to that in parts (b) and
(c). In this way, demonstrate that the results obtained in a thermodynamic
calculation do not depend on the convention we choose for the standard state
pressure, as long as we use standard state thermodynamic data consistent with
that choice.