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18.0 g of water completely vaporises at ...

18.0 g of water completely vaporises at `100^(@)C` and 1 bar pressure and the enthalpy change in the process is `40.79 kJ mol^(-1)`. What will be the enthalpy change for vaporising two moles of water under the same conditions ? What is the standard enthalpy of vaporisation for water ?

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`18.0` g of water `=` 1 mole
`:. ` Standard enthalpy of vaporisation at`100^(@)C` and 1 bar pressure `=40.79 kJ mol^(-1)`
Enthalpy change in vaporisationof 2 moles of water `= 2 xx 40.79 kJ = 81. 58 kJ`
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Knowledge Check

  • The internal energy change for the vapourisation of one mole water at 1 atm and 100°C will be (Enthalpy of vapourization of water is 40.66 kJ

    A
    Zero
    B
    37.56 kJ
    C
    -37.56 kJ
    D
    40.6 kJ
  • If the enthalpy change for the transition of liquid water to steam is 300kJ mol^(-1)" at "27^(@)C , the entropy change for the proces would be

    A
    `1000 JK^(-1)mol^(-1)`
    B
    `10 JK^(-1)mol^(-1)`
    C
    `1.0 JK^(-1)mol^(-1)`
    D
    `0.1JK^(-1)mol^(-1)`
  • If the enthalpy change for the transition of liquid water to steam is 30 kJ mol^(-1) at 27^(@)C , the entropy change for the process would be

    A
    `100 J mol^(-1)K^(-1)`
    B
    `10J mol^(-1) K^(-1)`
    C
    `1.0 J mol^(-1) K^(-1)`
    D
    `0.1 J mol^(-1) K^(-1)`
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