Question

Assuming ideal behavior, calculate the work done when 1.61.6 mole of water evaporates at 373K373K against the atmospheric pressure of 1atm.

Answer 1

Volume of 1.6 mole of water at 373 K in gaseous state= V= nRT/P= 1.6 x 0.082 x 373/ 1Volume of 1 mol = 18 g of liquid water = 18 x 1.6 x 10⁻³L= 0.0288 LW = -P(V2 - V1)= -1(48.93 - 0.0288)
= - 48.90 atm L or = - 4954.8 J.

Answer 2

Explanation:

Volume of 1.6 mole of water at 373 K in gaseous state= V= nRT/P= 1.6 x 0.082 x 373/ 1Volume of 1 mol = 18 g of liquid water = 18 x 1.6 x 10⁻³L= 0.0288