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

Question

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

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)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

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)`