One mole of solid iron was vaporized in an oven at `3500` K . If iron boils at `3133`K abd enthalpy of vaporization is `349 KJ mol^(-1)` , determine `DeltaS_("system"), DeltaS_("surrounding")` and `DeltaS_("universe")`. (Oven is considered as surroundings).

One mole of solid iron was vaporized in an oven at `3500` K . If iron

1.	One mole of solid iron was vaporized in an oven at `3500` K . If iron boils at `3133`K abd enthalpy of vaporization is `349 KJ mol^(-1)` , determine `DeltaS_("system"), DeltaS_("surrounding")` and `DeltaS_("universe")`. (Oven is considered as surroundings).
Answer» Correct Answer - `DeltaS_("system")=111.4 JK^(-1), " " DeltaS_("surr")=-99.71 JK^(-1), `DeltaS_("surr")=+11.69 JK^(-1)` At boiling point , `" "DeltaS_("system")=(DeltaHvap)/(t_(b)) = (349xx10^(33))/(3133) = 111.4 JK^(-1)` Heat change in surrounding `=-349 KJ` `DeltaS_("surr") =-(349xx1000)/(3500) =-99.71 JK^(-1)` `implies " "DeltaS_("univ") = DeltaS_("sys") + DeltaS_("surr") = 111.4=99.71 =+ 11.69 JK^(-1)`

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One mole of solid iron was vaporized in an oven at `3500` K . If iron boils at `3133`K abd enthalpy of vaporization is `349 KJ mol^(-1)` , determine `DeltaS_("system"), DeltaS_("surrounding")` and `DeltaS_("universe")`. (Oven is considered as surroundings).

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