For the process `H_(2)O(l)toH_(2)O(g)` at `t=100^(@)C` and 1 atmosphere pressure, the correct choice is:

Options

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

B

Explanation

`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`