Standard enthalpy of vapourisation \Delta _{vap}H^o for water at 100 oC is 40.66 kJ mol-1. The internal energy of vapourisation of water at 100 oC (in kJ mol-1) is:

  • Option 1)

    +37.56

  • Option 2)

    -43.76

  • Option 3)

    +43.76

  • Option 4)

    +40.66

 

Answers (1)

 

Enthalpy of Vapourisation -

Amount of enthalpy change to vapourise 1 mole liquid completely at its boiling point.

- wherein

At\: 100\: c^{\circ}

H_{2}O_{(l)}\rightarrow H_{2}O_{(g)}

\Delta H_{vapourisation}= 40.6\, KJ/Mole

 

 H_{2}O (l)\overset{100^{\circ}C}{\rightarrow}H_{2}O(g)

\Delta _{vap}H^{\circ}=\Delta _{vap}E^{\circ}+\Delta n_{g}RT

For the above reaction

\Delta n_{g}=n_{p}-n_{r}=1-0=1

\therefore40.66 KJ mol-1 = \Delta _{vap}E^{\circ}+1\times 8.314\times 10^{-3}\times 373

or \Delta _{vap}E^{\circ}= 10.66 KJ mol-1 - 3.1 KJ mol-1

= +37.56 KJ mol-1 


Option 1)

+37.56

this is the correct option

Option 2)

-43.76

this is the incorrect option

Option 3)

+43.76

this is the incorrect option

Option 4)

+40.66

this is the incorrect option

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