# Thread: another basic enthalpy question

1. HI

I have a basic enthalpy question to solve. Here it is

The enthalpy change per mole when vaporising water at 100 degrees C and 1 bar is 40.7 kj per mol.

The molar volume of the liquid is 18cm^3 and the molar volume of the gas is 24 dm^3 so you can assume the initial volume of the liquid is negligible.

What is the work done on the surroundings?

> I think this is 2.4

How does this answer relate to the formula (where d = delta) DH = q +PDV

>I believe when reactions involve gases they are carried out, you can't just measure the heat change, you have to take into accoun the PDV term as the change in volume is note worthy. So then i believe you get

40.7 = q + 2.4
q = 38.3

So this means the reaction would have consumed 38.3 units of heat.

How does thsi answer relate to the formula DU = q + w?

The system absorbed 38.3 units of heat, of which 2.4 were used for expansion to the internal energy change is

35.9 = 38.3 - 2.4

I don't think i have got this right. I think i have got my terms all muddled up

thanks

2.

3. I'm not trying to get someone to do my homework by the way. I'm trying to teach myself thermodynamics from a book.

4. I think when doing thermodynamics it's very important to keep in mind how your equations relate to a 'real' system. Otherwise it becomes too abstract.

In your post, for instance, I don't understand what transformation you're examining (what is the work done on the surrounding - OK, but is that when 1 mol of water is vaporised? or what else? and under what conditions? because as you know enthalpy change relates to heat in a certain way only at constant pressure), and I find very wrong not to see any units (2.4 what?).

I think you should formulate your problem more clearly - you might find out that you already know the answer. Very often failure in problem solving is due to bad formulation.

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