Understanding Vapor Pressure in Open Systems

[James Pelezo]

You have not answered my questions:
Gas was mentioned only to show that your earlier notion (about different temperatures at different points) is wrong. Fact that kinetic energies of molecules in a phase are different doesn't imply that the temperatures in different places are different. It clearly doesn't work this way for gases, why should it be true for liquids?

OK, two points seem to be in this issue of discussion, energy distribution in gas phase particles and energy distribution in liquid phase particles. Gases 1st: While it's true the Boltzmann-Maxwell distribution of particle energies vs numbers of particles is an illustration of particle energy distributions, it is a common understanding that the Kinetic Energy of a mole of any gas equals (3/2RT). This suggests that Kinetic Energy (homogeneous gas-phase system) is directly proportional to Temperature and changes in Kinetic Energy would, in fact, demonstrate different temperature values. The average kinetic energy of a molecule, which by definition is (1/2mv²), is obtained by dividing (3/2RT) by Avogadro's number, N_A. This gives gas phase kinetic energy as (1/2mv² = (3/2RT/2N_A) that is, according to this equation, temperature dependent.

2nd, liquid phase: In a sample of homogeneous a single component system in liquid phase, the energy content is described by (mc∆T) and has a temperature change factor included. However, when a system is in phase transition;i.e., liquid phase => vapor phase, energy input goes exclusively to providing the kinetic energy to effect vaporization and temperature change is not observed under such circumstance. The results of a common experimental investigation of this issue is verified in the 'Heating Curve' for a given mass of water. Tracing, say energy change as a function of temperature change of a finite mass of water from below freezing to a few degrees above boiling (liquid-gas transition) reveals that single phase components demonstrate a temperature change with the input or extraction of heat. However, when two phases are in contact, no temperature change is observed and heat content (Kinetic Energy) for the entire mass quantity of interest is given by (m∆H_t), where ∆H_t is the phase change transition constant. This energy is uniform and distributed homogeneously throughout the liquid-phase system so long as energy input continues and no changes in atmospheric pressure are imposed. Here is a link that may better clarify energy changes observed experimentally. http://chemistry.bd.psu.edu/jircitano/heatcurv.html I specifically direct your attention to 'Step 2' statement, "The heat does not increase the kinetic energy of the molecules, so the temperature remains constant. As long as there are H-bonds to break (as long as there is solid present), the temperature cannot increase". Also, here's another link that applies the Boltzmann-Maxwellean Distribution to factors required for reaction initiation including phase transition issures. http://www.chemunlimited.com/The Arrhenius Equation.html
Hope this helps, jp

 

[Chestermiller]

I don't think this helps because we all agree with this. What we don't agree with is that the kinetic energies of all the molecules in a thermally homogeneous liquid are identical to one another, particularly at the boiling point. This was our (Chet, Borek, Russ) understanding of what you were saying. If it wasn't what your were saying, please excuse us for our misinterpretation. Please clarify.

Chet