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Calculate the entropy change involved in convertion of 1 mole of water at 373 K to vapours at the same temperature. Latent heat of vaporisation of water = 2.257 kJ g^(-1). |
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Answer» Solution :`k = 1.8 xx 10^(-7)` `DELTA H = 2.257 xx 18 KJ = 40.626 kJ mol^(-1)` `Delta S_("vap") = (Delta H_(vap))/("B.pt. in kelvin") = (40.626 kJ mol^(-1))/(373 K) = 0.1089 kJ mol^(-1) = 108.9 kJ^(-1) mol^(-1)`. |
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