[orgo814]

This problem is based off of the first law of thermodynamics

"A sample of 1.00 mol H2O (g) is condensed isothermally and reversibly to liquid water 100 degree Celsius. The standard enthalpy of vaporization of water at 100 degree Celsius is 40.656 kJ/mol. Find w, q, delta U, and delta H for this process."

I had to look at this solutions guide to help me with this problem:

The delta H in this problem is the delta H condensation which is simply the reverse of the delta H vaporization, therefore -40.656 kJ/mol. I understood that.

Since it's reversible process, the work= -NRT ln(Vf/Vi). The book says that since W = -PdeltaV and delta V is negative since the vapor volume (initial volume) is much larger than the liquid volume (final volume). That make sense. But, they derived that W= NRT for that reason for calculating work so I'm a little confused about that. I don't know if this is because with a negative delta V, the work would be positive in the W= -PdeltaV formula so they removed the negative sign.... and there were no volumes given so we couldn't take the natural log. But, I'm not sure if I'm correct in my reasoning. If not, please correct me.

Then, for calculating delta U, I know the formula (and one they use) is U = H + PV which can be looked at as deltaU=  delta H + deltaNRT (since PV= NRT). But, they said that the deltaMole= -1.00 (we were told there is 1.0 mol H2O). I'm not sure why or how they got the negative on there.. we were given no "final moles".

Thanks I look forward to some clarification on this problem.

[orgo814]

Ok I figured out that it was just the change in moles since 2h2 + o2 yields 2H20. So, 2 moles - 3 moles equals -1 mole. But the next problem has the exact same situation but with methanol. I don't know what the reaction would be so I could determine the change of moles..

[Borek]

Initially you had a mole of gaseous water, after the condensation there is no gas at all - so the number of moles of gas changed by -1. It has nothing to do with the water synthesis, reaction is

H₂O(g) H₂O(l)

[Enthalpy]

Much of the question revolves about the volume of the liquid being negligible. At 1 atm and +100°C, liquid water is about 1000 times less bulky than vapour.

No volume given: you often don't need it (and here you could compute it anyway). Many thermo variables depend only on P*V, not P and V separately - and as soon as a gas is nearly perfect, T suffices to define P*V.

A Log ? You may be referring to an adiabatic transformation of a gas... but here it's a liquid-vapour equilibrium and heat is extracted.

The signs... Trying to keep them right all the way long is a too risky method. You better check and correct them only when you really need it: before additions, roots, logs, or at the final result. Here in a condensation, the volume shrinks so water gets work from the rest of the world, for instance from the atmosphere, hence plus.

[orgo814]

Makes sense about the change in moles. Thanks. But with the equation for work, my book says that isothermal reversible expansions use the formula -nrt ln(Vf/Vi) and then in the solutions guide they used -pdeltaV which I thought was for when an external pressure was applied so I'm still confused about that.

[Enthalpy]

O yes, here was the thread where you already mixed up a perfect gas and a gas-liquid transformation, before and after the explanationS. Still mixing up in a second thread.

What could help? In the books, read the text as well, not just look for formulas with mating letters?

[orgo814]

Excuse me, but I'm self teaching myself the material before the fall semester to get ahead. I was confused. I'm not a chemist, I'm an undergraduate and I'm learning it for the first time (by myself with no teacher).