What amount of work is performed when 1 kg of water turns into steam at `100^(@)C` ? How much energy is spent to break the bonds between the molecules?

Explanation

Since evaporation takes place at a constant pressure, `W= p(v_(0)^("vap")-v_(0)^("liq"))` where `v_(0)=1//rho` is the specific volume, i.e. the volume of 1 kg of the substance. Hence
`W=p((1)/(rho_("vap"))(1)/(rho_(11q)))`
The density of steam at `100^(@)C` may be found from Table 35.1 (see 35.3). Since the specific volume of va pour is almost a thousand times gronter than the specifio volume of liquid, it follows with better than 1% accuracy that `W=p//rho_(vap)`
The energy spent on breaking the bonds between the molecules can be found from the first law of thermodynamics: `Delta = L - W` where L is the specific heat of evaporation.