InterviewSolution
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InterviewSolution
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water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from 12 V supply is passed for 300 s through a resistance in thermal contact with it , it found that 0.798 g of water is vaporised . Calculate the molar internal energy change at boiling point (375 .15 K).A. ` 37 .5 kJ mol ^(-1)`B. `3.75 kJ mol ^(-1)`C. `42.6 kJ mol ^(-1)`D. ` 4 .26 kJ mol ^(-1)` |
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Answer» Correct Answer - a `DeltaH`=work done =`ixxVxxt=0.5 A xx12 V xx300 s` `1800 J =+1.8 kJ` molar enthaply of vaporisation, `DeltaH_(m)=(DeltaH)/(moles of H_(2)O)=(DeltaH)/n_(H_(2)^(O))=(1.8kJ)/(0.798/18)=40.6 kJ mol^(-1)` `DeltaH_(m)=DeltaE_(m)+P DeltaV` `DeltaH_(m)=DeltaE_(m)+Deltan_(g)RT` `DeltaH_(m)=DeltaE_(m)+RT` molar internal energy change , `DeltaE_(m)=DeltaH_(m)-RT` `=40.6-8.314xx10^(-3)xx373.15=37.5 kJ mol^(-1)` |
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