Explaining internal energy (A level physics)

[VEReade]

I’m having a mental block re. teaching internal energy. Here’s the issue:

For ideal gas, internal energy is entirely kinetic energy. I explain to students this is because there are no interatomic forces of attraction.

Now, a solid...internal energy is sum of kinetic energy and potential energy. The interatomic forces cause potential energy term.

Here’s the issue, work has to be done on particles going from solid to gaseous state. So particles in solid state have less potential energy than when at infinity. Looks like potential energy term is negative. Looks like internal energy could go negative - that sounds bad..!?
Guidance appreciated!?

 

[Charles Link]

The idea of the internal energy being negative is ok. If you look at simple atomic theory, and the Bohr model of the hydrogen atom, all of the computed energy levels are negative. The important parameter here is simply the change in energy that occurs in going from one state to another. What is used as the zero reference energy level is somewhat arbitrary, but in these cases it is the separated particles in a motionless state.

 

[VEReade]

Thanks, that makes sense.

UK A - level physics students used to have to know the potential energy against separation graph. I think it’s been removed from specifications these days - the ones I know anyway. I think this is influencing my thinking as regards teaching internal energy!

As you say, it’s changes in internal energy during , say, phase changes which lead to an understanding of latent heat etc. From a teaching point of view, it is probably best way to approach it...

 

[James Pelezo]

For academic reasons I find separating energy into three distinctive issues; that is, Kinetic Energy, Potential Energy (or, Positional Energy) and Internal Energy. I must confess that I'm a chemistry professor, not a physics prof, so my distinctions between these energy 'types' is chemistry oriented. Here's my definitions and examples ...
Kinetic Energy is the mechanical energy of motion, Potential energy is mechanical stored energy and Internal energy is the energy content of a system due to its state of existence and chemical structure. That is, state of existence meaning solid, liquid or gas and chemical structure meaning the substances molecular geometry. This is generally a macro molecular interpretation as opposed to a kinetic micro molecular particle level interpretations, but does give a more physical illustration of changes in internal energy.

Consider Water (good old H₂O)... It can exist as solid (ice), liquid (water) or gas (steam) but its chemical structure is always bent angular with two covalent bond and two non-bonded pair of electrons around the central element oxygen. To change the structure would be to change the molecule and the internal energy of water that supports this structural configuration.

https://www.google.com/search?rlz=1...TiAhXlmq0KHUmSCFAQ7Al6BAgJEA0&biw=1536&bih=722

Typically, for chemical reactions, changes in internal energy is calculated from the 1st Law equation ΔU_Rxn = Heat of Reaction (q) + Work of Rxn (w) = ΔH_Rxn + Pressure(P)⋅ΔMolar Volume of gasses (V), or simply ΔU = ΔH + P⋅ΔV_m.

Heat of Reaction (or, Enthalpy of Rxn) can be calculated using the Hess's Law Equation ΔH_Rxn = Σn⋅ΔH_f^o(Products) - Σn⋅ΔH_f^o(Reactants) which shows a clear differential numerically in energy content.

Work of Reaction depends upon changes in molar volumes of gasses (ΣV_m(products) - ΣV_m(reactants). If ΔV_m > 0 => work is exothermic expansion and is subtracted from the Heat of Reaction; if ΔV_m < 0 => work is endothermic compression and is added to the Heat of Reaction and if If ΔV_m = 0 => work is isothermic (w = 0) and ΔU_Rxn = ΔH_Rxn.

Example (Open System, ΔV_m ≠ 0)
3H₂(g) + N₂(g) => 2NH₃(g); ΔH_Rxn = -91.8 Kj
The reaction shows 4V_m(reactants) → 2V_m(products) => decrease in 2V_m net change in molar volume and is a compression reaction. Therefore, work is endothermic (w > 0 ) ~ +5.0 Kj. Therefore, the change in Internal Energy (ΔU) = ΔH + P⋅ΔV_m = -91.8Kj + (+5.0Kj) = -86.8Kj (net).
Considering structural differences between reactants and products, one can see a clear difference in structure between H - H and O=O giving a pyramidal structure for gaseous ammonia. That is, the internal energy of ammonia is 91.8Kj less than the internal energy of H₂(g) and O₂(g).

https://www.google.com/search?rlz=1....0.60.534.10...1...1..gws-wiz-img.xzAEwcWMe5g
Clearly different structure than H - H and O = O and thus different internal energy holding the structure in a pyramidal geometry.