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latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10 .0 kcal/ mol . What will be the change in internal energy (`DeltaE`) of 3 moles of liquid at same temperature ?A. 30 kcalB. `-54 kcal`C. 27.0 kcalD. 50 kcal |
Answer» Correct Answer - c `DeltaH=DeltaE+Deltan_(g) RT, 30 = 30 Delta + 3xx2xx500xx10^(-3)` `DeltaE=27 kcal` |
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