For the process `H_(2)O(l) rarr H_(2)O(g)` at `T=100 ^(@)C` and 1 atmosphere pressure, the correct choice isA. `DeltaS_("system") gt 0` and `DeltaS _("surroundings") gt 0`B. `DeltaS_("system") gt 0` and `DeltaS _("surroundings") lt 0`C. `DeltaS_("system") lt 0` and `DeltaS _("surroundings") gt 0`D. `DeltaS_("system") lt 0` and `DeltaS _("surroundings") lt 0`

For the process `H_(2)O(l) rarr H_(2)O(g)` at `T=100 ^(@)C` and 1 atmo

1.	For the process `H_(2)O(l) rarr H_(2)O(g)` at `T=100 ^(@)C` and 1 atmosphere pressure, the correct choice isA. `DeltaS_("system") gt 0` and `DeltaS _("surroundings") gt 0`B. `DeltaS_("system") gt 0` and `DeltaS _("surroundings") lt 0`C. `DeltaS_("system") lt 0` and `DeltaS _("surroundings") gt 0`D. `DeltaS_("system") lt 0` and `DeltaS _("surroundings") lt 0`
Answer» Correct Answer - B At `T= 100^(@)C` and 1 atmosphere presssure (i.e.,boilingpoint ) ,thereexists the equilibrim `H_(2)O(l) hArr H_( 2)O (g)`, For this conditions,`DeltaS_("total") = 0` Hence,`DeltaS_("system") + Delta_("surroundings") = 0` or `DeltaS_("system") = - DeltaS_("surroundings")` But as `H_(2)O(l)` changes into `H_(2)O(g)`, `DeltaS_("system") = + ve , i.e., gt 0` `:. DeltaS_("surroundings")` willbe`lt 0`.

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