(A) Since the water vapor and water are at equilibrium, the water vapor pressure is a function of the temperature (and vice versa). 5.00*10^5 N/m^2 is 4.94 atm. The temperature at that water vapor pressure is 151.8 C. You can obtain that from tables of H2O saturation pressure and temperature.
(C) The specific volume of the vapor is 374.8 cm^3/g at that temperature and pressure. (That also comes from saturated vapor tables, but can also be estimated from the perfect gas law.) The 1.0 m^3 volume is at least 99.7% vapor because there are only 3 kg of water in a container that can hold 1000 kg of liquid. Therefore the mass of water as vapor is 10^6 cm^3/374.8 cm^3/g = 2.67 kg.
(B) Only 0.23 of the available 3 kg of water is in the liquid state.
A cylinder contain 3 kg of water and water vapor mixture in equilibrium at a pressure of 500 KN/m^2. If the volume of the cylinder is 1.0 m^3, calculate the following.
A)temperature of the mixture
B) the volume and mass of the liquid
C)The volume and mass of the vapor
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