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( _ vap H ^ ) for water is (+40.79 ~kJ ~mol ⁻¹ ) at 1 bar and (100 ^ C ). Change in internal energy for this vapourisation under same condition is ( ) ( kJ mol ⁻¹ ). (Integer answer) (Given ( R =8.3 JK ⁻¹ ~mol ⁻¹ ))

Correct answer

38

Step-by-step solution

aligned & H ₂ O ( ) H ₂ O ( g ) H _ vap ^0=40.79 ~kJ / mole & H _ vap ^0= U _ vap ^0+ n _ g RT & 40.79= U _ vap ^0+ 1 8.3 373.15 1000 & U _ vap ^0=40.79-3.0971 & =37.6929 & U _ vap ^0 38 aligned

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