Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
Initial volume [tex](v_{1})[/tex] = 1.673 [tex]m^{3}[/tex]
Final volume [tex](v_{2})[/tex] = 1.789 [tex]m^{3}[/tex]
As, the amount of water vapor condensed will be as follows.
[tex]\frac{(v_{2} - v_{1})}{v_{2}}[/tex]
= [tex]\frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}[/tex]
= [tex]\frac{0.116 m^{3}}{1.789 m^{3}}[/tex]
= 0.065 kg
Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.
