One kilogram of a vapor-liquid water mixture has a quality of 50% at 25 °c approximately what volume does the water (vapor liquid) occupy? a. 44 cubic meters b. 42 cubic meters c. 22 cubic meters d. 1 cubic decimeter

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 = (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 =

                                       =          (as 1 kg = 1000 g, so 0.5 kg = = 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 at a pressure or 3170 Pa.

Hence, on assuming ideal gas behavior of the vapor the equation will be as follows.

                                             =

                                             = 21.72

Whereas we will calculate the volume of liquid water as follows.

                        Volume =

                                      =

                                       = 0.5 L

As 1 L = 0.001 . So, 0.5 L = 0.0005 .

Therefore,     total volume =  Volume of vapor + volume of liquid water

                                           = 21.72 + 0.0005

                                           = 21.7205

                                           = 22 (approx)

Thus, we can conclude that volume occupied by the water (vapor liquid) is 22 .

i think   the answer is c hope i'm right