Answer:
1) LIQUID, 2) LIQUID, 3) GASEOU, 4) GASEOUS, v = 0.76 m³ / kg
Step-by-step explanation:
In this exercise it is asked to determine the state of the water, that is, if it is solid, liquid or gas. For this we must use a phase diagram of water which is a graph of Pressure versus Temperature. Let's describe the water diagram
* Below 0ºC and normal pressure the water is in a solid state
* Below this temperature and at low pressure it becomes a gaseous state
* Above 0.01ºC and normal pressure is in liquid state
* Above 0.01ºC and low pressure is in a gaseous state
there are two important points
* The triple point at t + 0.01ºC and P = 0.006 atm where the three states coexist
* The critical point T = 374ºC and P = 218 atm where water decomposes into hydrogen and oxygen
Specific volume is related to density
v = 1 / ρ
ρ = 1 / v
the density of water is approximately 1000 kg / m³ in the liquid state at t = 4ºC and decreases with increasing pressure 960 kg / m³ at T = 100ºC (but without changing to the gaseous state
With the above considerations we can answer the questions
1) P = 300 kPa = 3 105 Pa
v = 0.5 m³ / kg
atmospheric pressure is Patm = 1.01 105 Pa
P = 3 105 Pa (1 atm / 1.01 105 Pa) = 3 atm
From the phase diagram described, the water can be in two liquid or gaseous states, depending on the temperature, as indicated by the speed of the same state, the water must be LIQUID
the parameter v that you indicate is the
with the other parameter we can calculate the density
rho = 1 / 0.5
rho = 2.0 kg / m³
in a thermodynamic system the three basic properties are: pressure, volume and temperature,
we can calculate the body temperature
The body must be at a temperature between 0 <T <100ºC
2) P = 28 Mpa = 28 106 Pa
P = 28 106 Mpa (1 atm / 1.01 105 Pa) = 280 atm
T = 200ºC
When examining the diagram it can be seen that the water is in the range of the LIQUID state
3) P = 1MPa
P = 1 106 Pa (1atm / 1.01 105Pa) = 10 atm
T = 405ºC
In this case, the only accessible state is the GASEOUS
4) T = 100ºC
x = 60%
Examining the phase diagram at this temperature depending on the pressure the possible states are Vapor and liquid, for pressures below 1 atm the state is GASEOUS
for the gaseous state we can use the ideal gas equation
PV = nR T
let us perform the calculation for a mole of gas n = 1, the ideal gas constant is R = 8,206 10-2 atm / mol K
V = nRT / p
V = 1 8,206 10-2 100/1
V = 8,206 m³
v = V / m
The mass of the water is 18 Kg / mol, which indicates that 60% is in the gaseous state, so the mass in this state is
m = 0.60 18
m = 10.8 kg
v = 8.206 / 10.8
v = 0.76 m³ / kg