For the process `H_(2)O(l) (1 "bar", 373 K) rarr H_(2)O(g) (1"bar", 373 K)` the correct set of thermodynamic parameters is

Options

A. `DeltaG=0, DeltaS=+ve`
B. `DeltaG=0,DeltaS=-ve`
C. `DeltaG=+ve,DeltaS=0`
D. `DeltaG=-ve,DeltaS=0`

Answer

a

Explanation

`underset((1 "bar", 373K))(H_(2)O_((l)))hArrunderset((1 "bar", 373 K))(H_(2)O_((g)))`
At `100^(@)C H_(2)O_((l))` is in equilibrium with `H_(2)O_((g))`,therefore
`DeltaG=0`
Because liquid molecules are converting into gases molecules therefore `DeltaS=+ve`