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Standard enthalpy of vaporisationDeltaV(...

Standard enthalpy of vaporisation`DeltaV_(vap).H^(Theta)` for water at `100^(@)C`is`40.66kJmol^(-1)`.The internal energy of Vaporization of water at` 100^(@)C("in kJ mol"^(-1))`is

A

`+37.56`

B

`-43.76`

C

`+ 43.76`

D

`+40.66`

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(1 "mole")(H_(2)O(l)) rarrunderset("1 mole")(H_(2)O(g)), Deltan_(g)=1`
`DeltaU =DeltaH - Deltan_(g)RT`
`= 40.66 xx 1000 J mol^(-1)-1 xx 8.314 JK^(-1) xx 373 K`
`= 40660 - 3101 J mol^(-1)`
`=37559 J mol^(-1)`
`=37. 56 kJ mol^(-1)`
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Knowledge Check

  • Standard enthalpy of vaporisation DeltaV_(vap).H^(Theta) for water at 100^(@)C is 40.66kJmol^(-1) .Teh internal energy of Vaporization of water at 100^(@)C("in kJ mol"^(-1)) is

    A
    `+4376`
    B
    `+40.66`
    C
    `+37.56`
    D
    `-43.76`

  • Standard enthalpy of vapourisation Delta_("vap")H^(@) for water at 100^(@)C si 40.66 kJ mol^(-1) . The internal energy of vaporization of water at 100^(@)C (in kJ mol^(-1) ) is (Assume water vapour to behave like an ideal gas)

    A
    `+37.56`
    B
    `-43.76`
    C
    `+43.76`
    D
    `+40.66`

  • Standard enthalpy of vapourisation DeltaH^(@) forwater is 40.66 KJ mol^(-1) .The internal energy of vapourisation of water for its 2 mol will be :-

    A
    `+43.76`
    B
    `+40.66`
    C
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    D
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