Question

One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It is cooled to a temperature at which the specific volume of water vapour is 1.789 m. The amount of water vapour condensed in kilograms is (a) 0.0 (b) 0.065 (c) 0.1 (d) 1.0

Answer 1

Answer: Option (b) is the correct answer.

Step-by-step explanation:

The given data is as follows.

Initial volume
`(v_(1))` = 1.673
`m^(3)`

Final volume
`(v_(2))` = 1.789
`m^(3)`

As, the amount of water vapor condensed will be as follows.

`((v_(2) - v_(1)))/(v_(2))`

=
`((1.789 m^(3) - 1.673 m^(3)))/(1.789 m^(3))`

=
`(0.116 m^(3))/(1.789 m^(3))`

= 0.065 kg

Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.