Question

Calculate the work done when $1.0$ mole water at $373K$ vaporizes against an atmospheric pressure of $1.0$ atmosphere. Assume ideal gas behaviour.

Answer 1

Volume of water before vapourization,

1 mole of water = 18 g of water
Volume of water $=\frac{18g}{1g/ml}$ [$\because V=\frac{m}{d}$ ]

$V_1=18ml$

Volume of water after vapourization,

$V_2=\frac{nRT}{P}=\frac{1.0\times 0.0821\times 373}{1.0}=30.6litre$

$V_1$ is negligible w.r.t. $V_2$

$w=-P_{ext}\times \Delta V=-\left(1.0\right)\times \left(30.6\right)litre-atm$

$=(-30.6)\times 101.3J=-3098.3J$