[Pranav]

Q. When 1 mole of ice melts at 0°C and at constant pressure of 1 atm. 1440 calories of heat are absorbed by the system. The molar volumes of ice and water are 0.0196 and 0.0180 litre respectively. Calculate ΔE and ΔH for the reaction.

Attempt:

According to the definition, ΔH is the heat supplied at constant pressure, hence in this case ΔH is 1440 calories.

As ΔE=ΔH-PΔV, ΔE is less than 1440 calories but the answer key states that ΔE and ΔH, both are equal to 1440 calories.

Any help is appreciated. Thanks!

[Pranav]

Anyone?

[curiouscat]

Patience.

[Pranav]

Still no one? -_-

[Corribus]

You are technically not wrong in your reasoning, but try putting in your real numbers and seeing what the result is.  Then think about why the answer is what the answer is. 

[Pranav]

You are technically not wrong in your reasoning, but try putting in your real numbers and seeing what the result is.  Then think about why the answer is what the answer is.

ΔE=ΔH-PΔV=1440-1*0.0016*24.217≈1439.96 calories. Ah yes, there is negligible work done here. Thank you Corribus.

[Corribus]

Yes, in the case of ice melting, the heat of the phase change so far exceeds the pressure-volume work that the latter is pretty much negligible.  Water is a strange beast all around, so it's easy to become confused.  The way the homework question was framed didn't help.  That is the devil of making approximations!