For water ∆h°vap = 40.7 kj/mol at 100.°c, its boiling point. calculate ∆s° for the vaporization of 1.00 mol water at 100.°c and 1.00 atm pressure.

Answer : The value of change in entropy for vaporization of water is

Explanation :

Formula used :

where,

= change in entropy  of vaporization = ?

= change in enthalpy of vaporization = 40.7 kJ/mol

= boiling point temperature of water =

Now put all the given values in the above formula, we get:

Therefore, the value of change in entropy for vaporization of water is

mark him as !