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Assuming that water vapour is an ideal gas, the internal energy change (∆U) when 1 mol of water is vapourised at 1 bar pressure and 100 C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol1 and R = 8.3 J mol1 K1 ) will be


A

41.00 kJ mol1

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B

4.100 kJ mol1

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C

4.100 kJ mol1

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D

4.100 kJ mol1

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Solution

The correct option is D

4.100 kJ mol1


H=41kJ mol1 =41000Jmol1

T=100C=273+100=373K

n = 1

U=HnRT=41000(2×8.314×373)

=37898.88J mol1=37.9 kJ mol1


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