Calculate w and `Delta U` 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 50670 `J mol ^(-1).`

Explanation

Volume of 1 mole of water 1 atm pressure, `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 N m ^(-2) (0.0306 m ^(3) - 18 xx 10 ^(-6) m ^(3)) = 101325 Nm ^(-2) xx 0.0306 m ^(3) = 3100 J`
Since converson of water into steam is accompmed by increase in volume, work is done by the system on the surroundings. Hence, w, is negative by convertion. Thus, `w =- 3100J`
According to First Law, `Dlta U = + w = 40670 J + (-3100 J) = 37570 J.`