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.

Step-by-step 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}}`

=
`(500 g)/(18 g/mol)` (as 1 kg = 1000 g, so 0.5 kg =
`0.5 * 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) = (nRT)/(P_(sat))`

=
`(27.78 mol * 8.314 J/K mol * 298.15 K)/(3170 Pa)`

= 21.72
`m^(3)`

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

Volume =
`(mass)/(Density)`

=
`(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)`.