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

For the process `H_(2)O(l) (1 "bar", 373 K) rarr H_(2)O(g) (1"bar", 37

1.	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 isA. `DeltaG=0, DeltaS=+ve`B. `DeltaG=0,DeltaS=-ve`C. `DeltaG=+ve,DeltaS=0`D. `DeltaG=-ve,DeltaS=0`
Answer» Correct Answer - a `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`

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

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