Work done by the system in isothermal reversible process is `w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1))`. Also in case of adiabatic reversible process work done by the system is given by: `w_(rev.) = (nR)/(gamma -1) [T_2 - T_1]`. During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy `DeltaS = q_(rev.)/T`. The entropy changes also occurs during transformation of one state to other end expressed as `DeltaS = DeltaH/T`. Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and `DeltaG=DeltaH - T DeltaS`.
The heat of vaporisation and heat of fusion of `H_2O` are `540 cal//g` and `80 cal//g`. This ratio of `(DeltaS_(vap.))/(DeltaS_("fusion"))` for water is:

Options

A. `6.75`
B. `9.23`
C. `4.94`
D. `0.2`

Answer

c

Explanation

`DeltaS_(v)=(DeltaH_(v))/373=540/373`
`DeltaS_(f)=(DeltaH_(f))/273=80/273`
`:. (DeltaS_(v))/(DeltaS_(f))=540/373xx237/80=4.94`