Answer
a)$P=2319.7kPa$
b)$x=0.027$
c)$\rho=287.19\frac{kg}{m^3}$
Work Step by Step
Using table A-4
a) In this case we are working with a saturated mixture then:
$P=P_{sat}=2319.7kPa$
b)$m_{f}=\frac{V_{f}}{\upsilon_{f}}=\frac{\frac{1}{3}*1.8m^3}{0.001190\frac{m^3}{kg}}=504.20kg$
$m_{g}=\frac{V_{g}}{\upsilon_{g}}=\frac{\frac{2}{3}*1.8m^3}{0.086094\frac{m^3}{kg}}=13.94kg$
$m_{t}=m_{f}+m_{g}=504.20kg+13.94kg=518.14kg$
$x=\frac{m_{g}}{m_{t}}=\frac{13.94kg}{518.14kg}=0.027$
c)$\upsilon=\upsilon_{f}+x(\upsilon_{g}-\upsilon_{f})=0.001190\frac{m^3}{kg}+0.027*(0.086094\frac{m^3}{kg}-0.001190\frac{m^3}{kg})=0.003482\frac{m^3}{kg}$
$\rho=\frac{1}{\upsilon}=\frac{1}{0.003482\frac{m^3}{kg}}=287.19\frac{kg}{m^3}$
