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.

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.

                     $\frac{(v_{2} - v_{1})}{v_{2}}$

                     = $\frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}$

                     = $\frac{0.116 m^{3}}{1.789 m^{3}}$

                     = 0.065 kg

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