Wednesday, October 3, 2007

Equipartition, Internal Energy, Heat Capacity

The equipartition theorem is an idea we're borrowing from statistical thermodynamics to predict expressions for the internal energy U for ideal gases. Many textbooks embed this concept with gases but Engel-Reid has squirreled it away in Chapter 14. To restate, for each quadratic term in the classical energy expression for a given molecule, there is a contribution of 1/2 kT to the average molecular kinetic energy. Implementing this rule is straightforward for any given structure if we just remember that each translational and rotational degree of freedom contributes 1/2 kT and each vibrational mode contributes a full kT.

As mentioned in class, the equipartition theorem ultimately fails when applied to gases that are nonideal -- of course, applying it to something like liquids or solids is heresy. Moreover it is inadequate when quantum effects are important, for example when the temperature is too low to significantly activate vibrational modes or if particular bonds are too stiff to vibrate at normal temperatures.

For ideal gases, we found that U=U(T); that is, internal energy is function only of temperature for a given sample. In fact it is a linear function of T, meaning that U = (constant)T. Since the heat capacity at constant volume, CV, is the derivative with respect to T, it will always be a constant for an ideal gas. The same can be said for CP, which should be apparent from the relationship CP = CV + nR (or, in molar form, CP,m = CV,m + R). I pulled this relationship out of thin air because we need it now but it will be proven when we are knee-deep in Chapter 3. Again, it is true only for ideal gases and applying it to any other system would be incorrect. Do not fret because more complicated systems (in fact, every system) will be addressed after we are finished laying down the mathematical foundations in Chapters 2 & 3.

A few notes on Problem 02 on hw.2 [P2.6 in Engel-Reid]: The notation for CP,m probably looks weird -- unfortunately it is becoming the standard way to write such equations in textbooks. Here's how I might write the same function:
CP,m = 20.9 + 0.042T ( T in K, CP,m in J K-1 mol-1)
It is important for you to understand why this gas is necessarily nonideal, despite the fact that the question states otherwise. In fact, for reasons stated above, it is not even possible to calculate a value for ∆U, which is also important for you to understand.

Lastly, are you now able to predict CP,m for an ideal diatomic gas absorbed onto the surface of a metal (so that it is constrained to two dimensions)?


Anonymous said...

I'm not sure I'm understanding why the internal energy cannot be calculated in P 2.6. I think it's because T is constant for ideal gases, but the problem says that the temperature is changing. Is that at all on the right track? And so should we not even attempt to solve for the change in U?

rod said...

No, T is not constant for an ideal gas: You can of course specify any temperature for one.

What is true for an ideal gas is this: Both the internal energy U and enthalpy H are functions only of T. Therefore Cv and Cp are constants.

In P2.6, Cpm is clearly not a constant. Therefore this is not an ideal gas; therefore Cpm = Cvm + R is not valid.

Since have no way to get Cvm from Cpm, we can not calculate ∆U by integrating dU = nCvmdT.

You might think that we could get ∆U from the definition of enthalpy: H = U + PV => ∆U = ∆H - ∆(PV). We of course cannot use PV=nRT here and, further, we don't even know an equation of state.

Therefore we cannot calculate ∆U at all.