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Work done by the system in isothermal re...

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:

A

`6.75`

B

`9.23`

C

`4.94`

D

`0.2`

Text Solution

Verified by Experts

The correct Answer is:
C

`Delta_(vap)S = (540)/(373)`
`Delta_(f)S = (80)/(273)`
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