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.