Problem $2(20\%)$

The values of $H$ and $S$ for steam at $400C$ are listed below. Use these data to calculate the fugacity of $20\%$ steam at $400C$ and $15$MPa, assume steam is an ideal solution.

$\begin{array}{cc}\text{S}\text{t}\text{a}\text{t}\text{e}1(Low\text{p}\text{r}\text{e}\text{s}\text{s}\text{u}\text{r}\text{e})& \text{S}\text{t}\text{a}\text{t}\text{e}2\\ T_1& T_2\\ P_1& P_2=15\text{M}\text{P}\text{a}\\ H_1=3279\mathrm{.}9k\frac{J}{k}g& H_2=2975\mathrm{.}9k\frac{J}{k}g\\ S_1=9\mathrm{.}609k\frac{J}{k}{g}^{**}K& S_2=5\mathrm{.}882k\frac{J}{k}{g}^{**}K\end{array}=5\mathrm{.}882k\frac{J}{k}{g}^{**}K=9\mathrm{.}609k\frac{J}{k}{g}^{**}K=2975\mathrm{.}9k\frac{J}{k}g=3279\mathrm{.}9k\frac{J}{k}g=15$MPa$=0\mathrm{.}01$MPa$=5\mathrm{.}882k\frac{J}{k}{g}^{**}K=9\mathrm{.}609k\frac{J}{k}{g}^{**}K=2975\mathrm{.}9k\frac{J}{k}g=3279\mathrm{.}9k\frac{J}{k}g=15$MPa$=0\mathrm{.}01$MPa$=400C=5\mathrm{.}882k\frac{J}{k}{g}^{**}K=9\mathrm{.}609k\frac{J}{k}{g}^{**}K=2975\mathrm{.}9k\frac{J}{k}g=3279\mathrm{.}9k\frac{J}{k}g=15$MPa$=0\mathrm{.}01$MPa$=400C=400C$