# P.Chem1: Internal energy in isothermal, reversible reactions

1. Sep 7, 2011

### Puchinita5

I have these two homework problems, as well as solutions. What I do not understand is why the solution for one is not the solution for the other.

First problem:
A sample consisting of 1.00 mol Ar is expanded isothermally at 0 deg Celc from 22.4 dm3 to 44.8 dm3 reversibly. Calculate q, w, delta U, and delta H. (heat, work, internal energy, and enthalpy)

Second problem:
A sample of 1.00 mol H20(g) is condensed isothermally and reversibly to liquid water at 100 deg Celc. The standard enthalpy of vaporization of water at 100 deg C is 40.646 kJ mol-1. Find w, q delta U, and delta H. (work, heat internal energy, and enthalpy)

Note: both problems refer to perfect gases

For the first problem, the solutions manual says that delta U is 0 since the internal energy of a perfect gas depends only on temperature, and since temperature is constant internal energy must be 0.

The second problem, however this is not the case. Why is the internal energy in the second problem not also zero if it is condensed reversibly and isothermally(which means constant T right?) ??

2. Sep 7, 2011

### Ygggdrasil

(emphasis mine)

The assumption that internal energy depends only on temperature applies only to perfect gasses. In problem two, you have a gas going to a liquid (not a gas). Therefore, because everything does not staying in the gas state, you cannot make the assumption that internal energy depends only on temperature.

A bit more details: The internal energy of a substance is composed of two components: the energy from the interactions between the particles composing the substance and the kinetic energy of the particles composing the substance. For a perfect gas, we assume that there are no interactions between particles, so the internal energy depends solely on the kinetic energy of the particles (which is directly proportional to temperature).

A liquid, however, has fairly strong interactions between the particles, so these interactions contribute greatly to the internal energy of a liquid. Even though the kinetic energy of the particles does not change when you convert a gas to a liquid at a constant temperature, the energy from the interaction between particles does change. Therefore, the internal energy changes.

3. Sep 7, 2011

### Puchinita5

oh wow. That was a great explanation, makes total sense!

If you wouldn't mind answering one more thing for me, for the second problem I'm a little unsure why we can assume (according to the solutions manual) that the external pressure is constant and equals 1.00 atm.

is the external pressure constant because it is reversible? I guess, does reversible mean any "change" in the pressure would be infinitesimally small, so basically constant P? I'm just not sure if my understanding of this is correct.

and are we assuming then that it would equal 1.00 atm because the external pressure is not given, so we just say 1.00 atm cuz that is about earth's pressure at the surface?

or am I just totally making things up lol?

4. Sep 7, 2011

### Ygggdrasil

The pressure has to be 1 atm, because only at that pressure is the condensation of water reversible at 100oC. Changing the pressure will change the boiling point of water (i.e. the temperature at which the gas and liquid phases will be at equilibrium).

Of course, I probably wouldn't expect a student to necessarily remember this and it would probably be much better if the book actually gave you the pressure, but in theory, the question contains enough information to tell you that the pressure has to be 1.00 atm.

5. Sep 7, 2011

### Puchinita5

great, thank you so much! You were very helpful :)