A) \[37.5\,kJ\,mo{{l}^{-1}}\]
B) \[3.75\,kJ\,mo{{l}^{-1}}\]
C) \[42.6\,kJ\,mo{{l}^{-1}}\]
D) \[4.26\,kJ\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
\[\Delta H=work\,done\,=i\times V\times t\] \[=0.50\times 12\times 300=1800\,J=1.8\,kJ\] 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, \[\Delta {{E}_{m}}=\Delta {{H}_{m}}-RT\] \[\Delta {{E}_{m}}=40.6-8.314\times {{10}^{-3}}\times 373.15\] \[=37.5\,kJ\,mo{{l}^{-1}}\]
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