Calculate w and `DeltaU` for the conversion of 1 mole of water at `100^(@)C` to steam at 1 atm pressure. Heat of vaporisation of water at `100^(@)C` is `40670 Jmol^(-1).`

Explanation

Volume of 1 mole of water at 1 atm presure, `V _(1) = 18 mL = 18 xx 10 ^(6) m ^(3).`
Volume of 1 mole of steam at `100^(@)C, V _(2) = (22.4 L xx 373 K)/(273 K) = 30.6 L = 0.0306 m ^(3).`
Work done , `w = P (V _(2) - V _(1)) = 101325 Nm ^(-2) (0.036 m ^(3) - 18 xx 10 ^(6) m ^(3)) = 101325 Nm ^(-2) xx 0.036 m ^(3) = 3100J`
Since conversion of water into steam is accompanied by increase in volumme, work is done by the system on the surrounding Hence w is negative by convertion. Thus, `w =-3100 J`
According to First Law,` Delta U = q + w = 40670 J + (-3100J) = 375 70 J.`