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The value for `DeltaU` for the reversible isothermal evaporation of 90 g water at `100^(@)C` will be `(DeltaH_("evap")" of water "=40.8" kJ mol"^(-1), R=8.314"J K"^(-1)"mol"^(-1))`A. 4800 kJB. `188.494 kJ`C. `40.8 kJ`D. `125.03 kJ` |
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Answer» Correct Answer - B `DeltaH" for 18 g water = 40.8 kJ"` `"For 90 g water "=(40.8)/(18)xx90=204kJ` `"n for 90 g water "=90//18=5` `Deltan_(g)=5-0=5` `DeltaH=DeltaU+Deltan_(g)RT` `DeltaU204000-(5xx8.314xx373)=188494 or 188.494 kJ` |
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