Question

Water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporised. Calculate the molar internal energy change at boiling point (373.15 K).

• A. $ 37.5\,kJ\,mo{{l}^{-1}} $
• B. $ 3.75\,kJ\,mo{{l}^{-1}} $
• C. $ 42.6\,kJ\,mo{{l}^{-1}} $
• D. $ 4.26\,kJ\,mo{{l}^{-1}} $

Answer 1

Answer: (A)

$ \Delta H=work\,done\,=i\times V\times t $
$ =0.50\times 12\times 300=1800\,J=1.8\,kJ $
Molar enthalpy of vaporisation,
$ \Delta {{H}_{m}}=\frac{\Delta H}{moles\,of\,{{H}_{2}}O}=\frac{\Delta H}{{{n}_{{{H}_{2}}O}}} $
$ =\frac{1.8\,kJ}{\frac{0.798}{18}}=40.6\,kJ\,mo{{l}^{-1}} $
$ \Delta {{H}_{m}}=\Delta {{E}_{m}}+p\Delta V $
$ \Delta {{H}_{m}}=\Delta {{E}_{m}}+\Delta {{n}_{g}}RT $
$ \Delta {{H}_{m}}=\Delta {{E}_{m}}+RT $
$ [\therefore \,\ \Delta {{n}_{g}}=1for\,{{H}_{2}}O(l){{H}_{2}}O(g)] $
$ \therefore $ Molar internal energy change,
$ \Delta {{E}_{m}}=\Delta {{H}_{m}}-RT $
$ \Delta {{E}_{m}}=40.6-8.314\times {{10}^{-3}}\times 373.15 $
$ =37.5\,kJ\,mo{{l}^{-1}} $