Eight grams of water changes from liquid to vapor at a pressure of one atmosphere. In the process, the volume changes from 0.8 cm³ to 674 cm³. The latent heat of vaporization of water is 2.26 * 10⁶ J/kg. Find the work done. Answer in units of J.

work done = 68.2 j

Explanation:

mass (m) = 8 g

initial volume (V1) = 0.8 cm^{3} = 8 x 10^{-7} m^{3}

final volume (V2) = 674 cm^{3} = 6.74 x 10^{-4} m^{3}

latent heat of vaporization = 2.26 x 10^{6} j/kg

pressure = 1 atm = 1.013 x 10^{5} Pa

work done = pressure x change in volume

work done = 1.013 x 10^{5} x ((6.74 x 10^{-4}) - (8 x 10^{-7}))

work done = 68.2 j