# Internal energy

1. Apr 3, 2014

### edanzig

"What is the internal energy of 10 moles of helium at 100 K?"
The same question is asked about 10 moles of oxygen. I'm going to guess PV=nRT which would give 8,314 J as the PV value. Because (at least in chemistry this is true) delta H = delat E +/- work and work = PV, so PV would somehow resemble "internal energy." The answer given is 12,473 J which is 3/2 of the calculation. I know I am on the right track but it would be helpful if someone could just throw it together for me. Thanks

2. Apr 3, 2014

### collinsmark

This isn't my best area, but I'll try to get you started since nobody has responded yet.

The first question to ask yourself is, "is helium a monoatomic or diatomic gas?" (That actually matters.)

Using that you can find CV , the heat capacity at constant volume. And from there, you can determine the internal energy.

This article can get you going, or I'm sure you can find the equivalent material in your textbook or coursework:
http://en.wikipedia.org/wiki/Heat_capacity_ratio

The reason (or at least a reason) that the internal energy is not simply PV is that some additional energy is stored in the form of microscopic rotational energy of the atoms/molecules. The internal energy is PV for an ideal gas [Edit: here, by "ideal gas" I mean a gas that is neither monoatomic nor diatomic*], but that simple relationship doesn't hold when actual matter consisting of atoms and molecules are involved.

*[It is possible to model "ideal" monoatomic and "ideal" diatomic gases, however, as is done in the link I provided. That's probably the approach you should use for this problem.]

Last edited: Apr 3, 2014
3. Apr 3, 2014

### Matterwave

One can use the equipartition theorem to get that the internal energy of a perfect gas is equal to $$U=\frac{f}{2}Nk_B T$$

Where f is the number of degrees of freedom of the gas. It's 3 for a monoatomic molecule (3 translational degrees of freedom), and 4 (5? I can't remember) or higher for diatomic molecules as you start to get the vibration and rotational modes.

N in the above is the total number of molecules. Given the moles n, one has the equivalent formula:

$$U=\frac{f}{2}nRT$$