Question

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

Answer 1

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 $25^{o}C$ = (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 = $\frac{mass}{\text{molar mass}}$

                                       = $\frac{500 g}{18 g/mol}$          (as 1 kg = 1000 g, so 0.5 kg = $0.5 \times 1000 g$ = 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 $25^{o}C$ at a pressure $P_{sat} = 3.17 kPa$ or 3170 Pa.

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

                               $V_{vapor} = \frac{nRT}{P_{sat}}$

                                             = $\frac{27.78 mol \times 8.314 J/K mol \times 298.15 K}{3170 Pa}$

                                             = 21.72 $m^{3}$

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

                        Volume = $\frac{mass}{Density}$

                                      = $\frac{0.5 kg}{1 kg/l}$

                                       = 0.5 L

As 1 L = 0.001 $m^{3}$. So, 0.5 L = 0.0005 $m^{3}$.

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

                                           = 21.72 $m^{3}$ + 0.0005 $m^{3}$

                                           = 21.7205 $m^{3}$

                                           = 22 $m^{3}$ (approx)

Thus, we can conclude that volume occupied by the water (vapor liquid) is 22 $m^{3}$.