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Assuming that water vapour is an ideal gas, the internal energy change (\Delta U) when 1 mol of water is vapourised at 1 bar pressure and 1000C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol-1 and R = 8.3 J mol-1 K-1) will be

  • Option 1)

    41.00 kJ mol-1

  • Option 2)

    4.100 kJ mol-1

  • Option 3)

    3.7904 kJ mol-1

  • Option 4)

    37.904 kJ mol-1

 

Answers (1)

best_answer

As we learnt in

Molar heat capacity for isobaric process C(p) -

dH=n\: C_{P}\: dT
 

- wherein

dH=dE+d(PV)

or

dH=dE+n\, R\, dT

 

 Given \Delta H=41KJmol^{-1}=41000KJmol^{-1}

T=100^{\circ}C=273+100=373K

for n = 1

\Delta U=\Delta H-\Delta nRT=41000-(2\times8.314\times373)

=\:37898.88\:Jmol^{-1}\:=37.9\:KJmol^{-1}


Option 1)

41.00 kJ mol-1

This option is incorrect.

Option 2)

4.100 kJ mol-1

This option is incorrect.

Option 3)

3.7904 kJ mol-1

This option is incorrect.

Option 4)

37.904 kJ mol-1

This option is correct.

Posted by

prateek

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