Respuesta :
Answer: Option (C) is the correct answer.
Explanation:
It is given that total mass is 1 kg and quality is 50%. Hence, the weight of liquid and vapor is 0.5 kg each.
Since, temperature is given as [tex]25^{o}C[/tex] = (25 + 273.15) K = 298.15 K
Molecular mass of water = 18 g/mol
Therefore, calculate the number of moles of water as follows.
        No. of moles of water vapor = [tex]\frac{mass}{\text{molar mass}}[/tex]
                    = [tex]\frac{500 g}{18 g/mol}[/tex]      (as 1 kg = 1000 g, so 0.5 kg = [tex]0.5 \times 1000 g[/tex] = 500 g)
                    = 27.78 mol
This will also be equal to the number of moles of liquid water present.
According to the steam tables, water exists in its saturated state at [tex]25^{o}C[/tex] at a pressure [tex]P_{sat} = 3.17 kPa[/tex] or 3170 Pa.
Hence, on assuming ideal gas behavior of the vapor the equation will be as follows.
                [tex]V_{vapor} = \frac{nRT}{P_{sat}}[/tex]
                       = [tex]\frac{27.78 mol \times 8.314 J/K mol \times 298.15 K}{3170 Pa}[/tex]
                       = 21.72 [tex]m^{3}[/tex]
Whereas we will calculate the volume of liquid water as follows.
            Volume = [tex]\frac{mass}{Density}[/tex]
                   = [tex]\frac{0.5 kg}{1 kg/l}[/tex]
                    = 0.5 L
As 1 L = 0.001 [tex]m^{3}[/tex]. So, 0.5 L = 0.0005 [tex]m^{3}[/tex].
Therefore, Â Â total volume = Â Volume of vapor + volume of liquid water
                      = 21.72 [tex]m^{3}[/tex] + 0.0005 [tex]m^{3}[/tex]
                      = 21.7205 [tex]m^{3}[/tex]
                      = 22 [tex]m^{3}[/tex] (approx)
Thus, we can conclude that volume occupied by the water (vapor liquid) is 22 [tex]m^{3}[/tex].
