CHM 5414 Lecture Notes
29 October 1996

Flames are even more fascinating than soap bubbles. The ancients were so enamored of Fire that they elevated it to one of the four elements (earth, air, fire, water). In fact, that philosophy persisted through the Phlogiston theory of conflagration up until only a couple of hundred years ago!

But even knowing the dance of fuel and oxidizer, exothermicity and buoyancy, doesn't dash the wonder of flame. That couldn't be 'cause all us chemists are closet pyromaniacs, could it?

In flames, heat and work play off against one another in a most satisfying way which leads us below to general expressions for adiabatic processes!

In the last lecture, we found that the energy of an ideal gas depended not on its pressure but on its temperature. Compressed or exanded in thermal contact with surroundings, ideal gases have fixed energy. They are worked upon or do work, but that work is exactly compensated for by heat transfer:

dE = q + w = 0 or q=-w
It's easy to imagine the work involved as -P_extdV, but the "heat transferred" doesn't change the gas's appearance; instead it bumps molecular microstates up or down their quantum energy level ladders invisibly save for instruments which measure temperature.

Let's imagine a non-isothermal process; temperature is now permitted to change, and there must be some sensitivity of thermodynamic values to that like:

dE = (E/T)_V dT = C_V dT
We've chatted about C_V before from a statistical mechanical point of view. It's magnitude reflects the number of ways a system can hide energy (available degrees of freedom) as the temperature increases. If few modes are available in the partition function sense, the temperature can shoot up with trivial energy additions. (Hey, it's reciprocal; if dE is insensitive to dT due to a small C_V, then dT is super-sensitive to dE, right?)

On the other hand, we know that z increases with temperature, making additional states (and indeed entire modes) turn active. At solid state temperatures, only the phonon (solid vibration) modes are active. At liquid state temperatures (T>_trans), those phonons have become cramped translations, and rotations have started to become active. By gas temperatures, both translations and rotations are fully active (T>_rot), and vibrations are slowly beginning to turn on. "Fully active" means that the energy is as expected from the Equipartition Theorem, ½RT for each available mode (a full RT for each vibration since it incorporates both kinetic and potential energies). So for a rigid 3-d polyatomic gas, 3 translations + 3 rotations yields <E>=3RT which tells us C_V in that T range will be about 3R. (All of this for a mole, of course; C_V is an extensive variable just like E.)

Just for jollies, a monatomic gas has no rotations, so its C_V is expected to be about (3/2)RT around room temperature.

We could, of course, calculate C_V from z as we have E by simply differentiating the statistical mechanical expression for molar E with respect to T (at constant V).

C_V = 2 RT (ln z/T)_V + RT²(²ln z/T²)_V,P
but that's not terribly practical. Instead, we usually take advantage of the fact that at room temperature, we're still waiting for the vibrations to get interested (T<_vib), so C_V is changing only rather slowly with T. Indeed, the variation is fit to a T polymial usually of the form:

C_V = a + bT + c/T²
Those parameters aren't fundamental in any way; this is just curve-fitting here. But it is a particularly simple curve, easily integrated:

dE = _T1^T₂ a + bT + c/T² dT = aT + ½bT² - c(1/T)

As chemists, we're more enthusiastic for dH which also has a temperature dependence. But while dE=q-PdV measures q_V, dH=q+VdP measures q_P, and we're not surprised that it's temperature dependence,

dH = (H/T)_P dT = C_P dT

bears the _P subscript. It too is very weakly T dependent at room temperature (for the same reason), and gets expanded into the very same polynomial, with different coefficients; so

dH = a'T + ½b'T² - c'(1/T)

and it's fair to ask how C_V and C_P relate. The answer's easy for an ideal gas. dH=dE+d(PV)=dE+nRdT, so C_P=C_V+R. Your mileage may vary if you're not ideal.

Why do we want to know this???

Because the interesting chemical reactions are either exothermic, providing lots of q, or endothermic, requiring some q, and almost never thermoneutral. Therefore, either the reactants or the products are obliged to absorb the reaction q, and the resultant dT may influence the equilibria. This is important to know, if we're making our livelihood off chemical produce. :-)

Just as we can find the H ° of our favorite reaction:

a A + b B c C + d D
as
H° = c H_f°(C) + d H_f°(D) - a H_f°(A) - b H_f°(B)
so to can we define a
C_P = c C_P(C) + d C_P(D) - a C_P(A) - b C_P(B)
which permits us to evaluate H at something other than standard temperature. If the C_P heat capacities are really constant over the temperature difference from 25°C to T, then each H varies by its own C_PT. And the nonstandard temperature H we seek is:

H(T) = H ° + C_PT

and similarly with E if we use C_V.

But what of S? It too must be temperature dependent via, say, dH=TdS+VdP, which for isobaric (fixed pressure) processes (for which H is famous), means dS=(dH)/T=C_P(1/T)dT. And for reasonably constant C_P, dS=C_Pd lnT or S=C_Pln(T₂/T₁). And if we were masochistic, we could now find the temperature dependence of A (using C_V, of course) and that for G as well (using C_P).

But while C_whatever is critical for evaluating thermo energy/entropy changes at different temperatures via Hess's Law, a variant of it is interesting characterizing the second great class of thermodynamic processes:

Adiabatic Processes

Instead of perfect thermal coupling (isothermal) of system to surroundings, adiabatic processes require perfect insulation! This is greeted with greater suspicion; after all, doesn't everything have some finite, even if tiny, heat conductivity? Thermos bottles don't keep coffee hot forever.

True. There's even a physical law due to Isaac Newton on the subject called Newton's Law of Cooling (equally applicable to heating). It states that the rate of cooling (heating) is directly proportional to the current T between system and surroundings, regardless of the thermal conductivity coefficients! While it's meaningless to approach adiabaticity by minimizing the very T we seek to establish, Newton's Law suggests that, if a rate is involved, simply look to extremely short processes as necessarily adiabatic since time is required to bleed away heat.

The astute among you will realize that Newton's Law refers only to thermal conduction (through matter) not convection (through matter transport) or via radiation. Photons are a good deal more prompt than any other heat transfer mechanism, but they're remarkably inefficient. Radiative cooling is truly lethargic.

What has this to do with approaching adiabatic processes? We might look to rapid ones. Explosions come to mind, if one is of the mad scientist brand of chemist. Mea culpa. Mea culpa. Mea maxima culpa.

Not only are explosions rapid, effectively trapping their exothermicity among the products, but they're also over before the (typically gaseous products have begun to expand. So they've not yet done Pressure-Volume work, and E is the interesting quantity since it equates to heat at V=0.

If we imagine instantaneous completion of an explosion, the products must distribute the enormous exothermicity among themselves. And Hess encourages us to imagine that the reaction proceeds at 25°C releasing E° after which that E° is reabsorbed by all the products (and anything else in the vicinity...unreacted reagents, non-reactive diluents, etc.) until the temperature has risen to accommodate it all. Since for exothermicity, E<0, the appropriate relation is:

E° = - [ c C_V(C) + d C_V(D) ] T
or

T_explosion = 298°K - E°/[sum of product C_V]
Similarly, but with less justification, a flame temperature might be approximated as:

T_flame = 298°K - H ° / [sum of product C_P]
since the flame products do expand against atmospheric pressure while reacting.

A few trials with these expression convince one that adiabaticity is wildly overestimated. The oxyacetylene torch, for example, is calculated to be hotter than the photosphere of the sun! Sure. Still comparisons of even these hopelessly approximate temperatures are useful to rank real flame temperatures.

Adiabatic Expansion

Isothermal expansions have been covered in the previous notes. If they could be imagined carried out reversibly, the P_ext would have to be P_internal - at all points in the process. That makes P_ext=nRT/V for an ideal gas undergoing isothermal expansion.

P = constant×V ^-1 along an isotherm giving rise to the hyperbola in the chart at right. But an adiabatic expansion cannot take those paths; the energy the gas loses to Pressure-Volume work is not made up by heat transfer since that is presumed to be impossible. Thus the internal energy and the temperature of the ideal gas must decline. While the adiabatic expansion might begin at point A, it must end at point B on a different isotherm.

The black curve in the figure at the left is an adiabat for a monatomic gas. Clearly it falls more steeply than V ^-1. The adiabatic expansion starts at V=2 L where the adiabat crosses the 373 K isotherm in this example, and it ends at V=6.3 L where it crosses the 173 K isotherm. Life would be simpler if it dropped from 373 K to 173 K at constant V_initial then slid down the 173 K isotherm. The reason that's easier is that E=0 along the isotherm but changes by C_VT along the fixed V path.

But both paths reach the same (final) state! And the adiabat gets there by doing work on the atmosphere in an amount we can imagine to be reversible. We can equate this reversible work with the C_VT and learn how V relates to T under adiabatic expansion.

We imagine the process proceeding in baby steps along each isotherm over which the adiabat passes; that is, we drop a little (C_VdT), then slide a little (-PdV), repeated in small increments, shadowing the adiabat with infinitesimal drops and slides. Along the adiabat steps, dE=-PdV; along the drop/slide alternate, dE=C_VdT+0. Since they get the same place, they must be equal:

nC_V dT = - nRT dV/V
C_V dT/T = - R dV/V
C_V ln(T₂/T₁) = + R ln(V₁/V₂)
ln(T₂/T₁) = (R/C_V) ln(V₁/V₂)
ln(T₂/T₁) = ln(V₁/V₂)^R/C_V
(T₂/T₁) = (V₁/V₂)^R/C_V or
T₂V₂^R/C_V = T₁V₁^R/C_V = const = T V^R/C_V
Sustitution of the ideal gas equation into this formula and recognition that C_P = C_V+R for an ideal gas gives two further adiabatic relations:

P / T^C_P/R = constant and the one we need:
P V^C_P/C_V = constant or, given the values expected for heat capacities of an ideal monatomic gas, C_P/C_V=(5/2)÷(3/2)=5/3, the black adiabat above is the equation P=const/V^5/3 which does indeed fall faster than 1/V.

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

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