Does ΔΕ(total) = ΔE (internal) for Ideal Gases?

[jaybird12]

The only way we can have a transfer of energy (any type of energy) is through heat or work. ΔE = q + w. For an ideal gas, TPR says (TPR Gen. Chem. pg. 215) that Eint is proportional to temperature (which makes sense, because as EK Chem pg. 71 & 85 show, the only part of internal energy that can change for the monatomic particles of an ideal gas is the translational energy. So any change in internal energy means a change in temperature). But then on TPR pg. 219, it says that if ΔT = 0, then ΔΕ= 0 (and so q = -w). How can we say that all of ΔE = 0? Shouldn't we say that only ΔEint = 0? With ideal gases, is ΔE = ΔΕint?

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[Cawolf]

Generally speaking ΔΕ will be referring to the energy of the system. So I think here that they are referring to the same thing.

 

[jaybird12]

So gases aren't considered to have any PE, KE, or any other energy besides internal energy?

 

[Cawolf]

Internal energy is the total energy of a system - including kinetic and potential.

 

[jaybird12]

Hmm, well, according to my EK book and Wikipedia, Internal Energy is the energy of molecules, a microscopic energy, and it excludes the macroscopic Kinetic Energy and Potential Energy.

 

[Cawolf]

I could most likely be wrong, I am no chemistry expert by any sense.

Let me explain how I see it though.

It is fact that the temperature of a gas is proportional to it's internal energy. Using the ideal gas law, you can state that T = PV/nR. Therefore internal energy (E) is proportional to velocity, since since kinetic energy (KE) is proportional to the square of velocity - these are related.

This can also be seen in formula that KE(avg) = 3/2RT per mole of ideal gas.

We also know that the potential energy (PE) of an ideal gas is approximately zero. There are no nuclear forces or chemical forces between the molecules of the gas, the gravitational PE is essentially zero, and we assume no electrostatic interactions by nature of an ideal gas.

These arguments support that fact that in an ideal gas, E = KE and PE ~0.

 

[jaybird12]

New insight: in EK Chem pg. 71 about internal energy, it says at the end that in "a closed system at rest [ie, no KE] with no electric or magnetic fields [ie, no PE], the only energy change will be in internal energy". This describes an ideal gas, so then yeah, for an ideal gas, ΔE = ΔΕint, and that's why any ΔΕ (whether it be PV work or q) goes straight into ΔEint and thus the temperature.
Thanks for helping out!