For the process `H_(2)O(l)toH_(2)O(g)` at `t=100^(@)C` and 1 atmosphere pressure, the correct choice is:A. `DeltaS_("system") gt 0` and `DeltaS_("surrounding") gt 0`B. `DeltaS_("system") gt 0` and `DeltaS_("surrounding") lt 0`C. `DeltaS_("system") lt 0` and `DeltaS_("surrounding") gt 0`D. `DeltaS_("system") lt 0` and `DeltaS_("surrounding") lt 0`

For the process `H_(2)O(l)toH_(2)O(g)` at `t=100^(@)C` and 1 atmospher

1.	For the process `H_(2)O(l)toH_(2)O(g)` at `t=100^(@)C` and 1 atmosphere pressure, the correct choice is:A. `DeltaS_("system") gt 0` and `DeltaS_("surrounding") gt 0`B. `DeltaS_("system") gt 0` and `DeltaS_("surrounding") lt 0`C. `DeltaS_("system") lt 0` and `DeltaS_("surrounding") gt 0`D. `DeltaS_("system") lt 0` and `DeltaS_("surrounding") lt 0`
Answer» Correct Answer - B `H_(2)O(l) rarr H_(2)O(g) "at" T=100 ^(@) C, 1atm` equilibrium exists . `therefore DeltaG=0 , DeltaH-TDeltaS=0` `DeltaH=TDeltaS gt 0` for system m since evaportation is endothermic `(DeltaS)_("system") gt 0 , also (DeltaS)_("surrounding")=(q_("surr"))/(T_("surr"))` Heat gained by system = heat lost by surroundings `therefore q_("surr") lt 0 therefore (DeltaS)_("surr") lt 0`

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