In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).
Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:
[tex]x=\frac{m_{steam}}{m_{total}}[/tex]
Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:
[tex]x=\frac{2kg}{10kg} =0.20[/tex]
Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:
[tex]v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg[/tex]
This means that the volume of the container will be:
[tex]V=10kg*0.4727m^3/kg\\\\V=4.73m^3[/tex]
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