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If water vapour is assumed to be a perfe...

If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mol of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)`.Calculate the internal energy change, when
(a) 1 mol of water is vaporised at 1 bar pressure and `100^(@)`C.
(b) 1 mol of water is converted into ice.

Text Solution

Verified by Experts

(i) For vaporisation of water, the change is `: H_(2)O (l ) rarr H_(2)O(g)`
`Delta n_(g) = 1 -0=1`
` Delta H = Delta U + Delta n_(g) RT`
or `Delta U = Delta H - Delta n_(g) RT = 41.00 kJ mol^(-1) - ( 1 mol) xx ( 8.314 xx 10^(-3)kJ K^(-1) mol^(-1)) ( 373 K) `
`= 41.00 - 3.10 k J mol^(-1) = 37.90 kJ mol^(-1)`
(ii) For conversion of water into ice, the change is `H_(2)O(l) rarr H_(2)O(s)`
In this case, the volume change in negligible. Hence, `Delta H = Delta U = 41.00 kJ mol^(-1)`
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Knowledge Check

  • Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and 100^(@)C is 41 kJ mol^(-1) (If water vapour is assumed to be perfect gas). Find out of the internal energy change. If 1 mole of water is vaporised at 1 bar pressure and 100^(@)C .

    A
    `+37.904kJmol^(-1)`
    B
    `-37.904kJmol^(-1)`
    C
    `44.096kJmol^(-1)`
    D
    `-44.096kJmol^(-1)`

  • The enthalpy of vaporisation of water at 100^(@)C is 40.63 KJ mol^(-1) . The value Delta E for the process would be :-

    A
    `37.53 KJ mol^(-1)`
    B
    `39.08 KJ mol^(-1)`
    C
    `42.19 KJ mol^(-1)`
    D
    `43.73 KJ mol^(-1)`