# Based on their respective number of degrees of freedom, what's the most efficient fuel; hydrogen or gasoline?

The reaction of octane's combustion is:

$$\ce{2 C8H18 + 25 O2 -> 18 H2O + 16 CO2}$$

I am asked for computing the number of degrees of freedom of the octane given that its specific heat is $$\pu{256 JK^{-1}mol^{-1}}$$.

Then I have to compare the number of degrees of freedom obtained for octane with those for $$\ce{H2}$$. Based on this I have to argue what's the most efficient fuel.

What I've done

I assumed octane stores thermal energy only in quadratic degrees of freedom and these do not depend on $$T$$. Using the equipartition theorem one gets:

$$C_V = \frac{\partial U}{\partial T} = \frac{Nfk_\mathrm B}{2} = \frac{nfR}{2}$$

Where: $$k_\mathrm B = \pu{1.381 \times 10^{-23}JK^{-1}}$$ and $$R = \pu{8.315 JK^{-1}mol^{-1}}$$

Then for octane:

$$C_V = fR$$

$$f = 31$$

Aren't these too many degrees of freedom?

For $$\ce{H2}$$ there are $$3$$ translational + $$2$$ rotational degrees of freedom; $$f = 5$$

I have worked with Otto's engine model (real heat engines; internal combustion):

$$\eta = 1 - \left(\frac{V_2}{V_1} \right)^{\gamma -1}$$

Usually the compression ratio is of the order:

$$\frac{V_1}{V_2} = 8$$

Where $$\gamma$$ is the adiabatic exponent:

$$\gamma = \frac{f + 2}{f}$$

Thus, the efficiency of a gasoline engine:

$$\eta = 0.12$$

While the efficiency of an hydrogen engine:

$$\eta = 0.56$$

My point is that the efficiency I get for gasoline is not what I expected (at least $$0.20$$). So I must be missing something on calculating the degrees of freedom for octane.

Thus hydrogen is a more efficient fuel.

• Once again, it is remarkable that this question (as many textbooks and lectures) selects the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion, although gasoline actually does not contain much octane (typically less than 1 %).
– user7951
Apr 29, 2019 at 11:40
• I'm also not sure that a purely theoretical definition of "efficiency" is remotely useful. In the real world we have to think of ideal like "how much energy can we extract per unit mass or unit volume" and we have to worry about the mass and volume of the tank required to store the fuel. Hydrogen is bad for both whatever the theory says about the combustion reaction. Apr 29, 2019 at 15:40
• @NightWriter 2 cancels out with $n= 2$ Let me know if you don't see it. May 4, 2019 at 17:04
• I don't think that the degrees of freedom are the problem. You calculated them using an experimental value, so you get an effective number of active degrees of freedom, at that given temperature. Using $\gamma = 1.41$ for $\ce{H2}$ and $\gamma = 1.05$ for octane, I got $\eta = 0.57$ and $0.10$, respectively. I guess that the low value for octane efficiency comes from the fact that you have a mixture of gases in the engine, so, realistically, $\gamma$ will be closer to $1.40$ (value for air) and the efficiency will be larger than $0.10$. May 4, 2019 at 18:40
• @JD_PM Just on the right side. I've never used this model, so I don't know why you expect butane to be above 0.20 efficiency, but this correction will produce your expected value above 0.20 efficiency. You can read this here: en.wikipedia.org/wiki/… . There is no $n$ on the right-side, so your $f$ comes out between 15-16, and your units in the last step will match. May 7, 2019 at 11:56

It sounds like most of your confusion is coming from the fact that you are assuming that the vibrational degrees of freedom are completely frozen out at moderate temp. This is a common misconception. To see that this assumption is not true, we can look at the simple example of carbon dioxide. It is a linear molecule, so it has three degrees of translational freedom and two of rotation. If vibrational modes do not contribute at all, we expect that $$\overline{C}_v=\frac52 R$$ and $$\overline{C}_p=\frac72 R$$, so $$\gamma=\dfrac{C_p}{C_v}=1.4$$. If we include the 4 vibrational degrees of freedom, each of which contributes 1R to $$\overline{C}_v$$, we would have $$\dfrac{C_p}{C_v}=1.15$$. The experimental value at $$20$$ $$^\circ C$$ is about 1.3, indicating that the vibrational modes are already contributing to the heat capacity at this temperature. A plot of the heat capacity as a function of temp shows a gradual increase across a very wide range of temperature.

More rigorously, the contribution of a vibrational degree of freedom to the molar heat capacity is

$$R\cdot\left(\frac{\Theta}{T}\right)^2\cdot\frac{e^{\Theta/T}}{(e^{\Theta/T}-1)^2},$$

where the "vibrational temperature" $$\Theta=\dfrac{hcv}{k_B}$$. For the lowest energy vibrational normal modes of $$\ce{CO2}$$, $$v=667 cm^{-1}$$, so $$\Theta = 961 K$$. Although this is quite a bit higher than typical temperatures of interest, the above expression tells us that these vibrations contribute $$\sim 0.45 R$$ to $$\overline{C}_v$$ at 300 K. Since there are two modes at this frequency in $$\ce{CO2}$$, we expect that $$\overline{C}_v$$ and $$\overline{C}_p$$ are both increased by 0.9R at 300 K relative to the values excluding vibrations. This correction gives a value of $$\dfrac{C_p}{C_v}$$ very close to observed experimental value.

Going back to your case of octane, there are a number of normal modes with frequencies around 720 cm$$^{-1}$$, certainly low enough to contribute at 300 K. We have 18 + 8 = 26 total atoms, so we expect a maximum of 3*26=78 degrees of freedom. Since octane is not linear (in the strict sense of all atoms in a single line), there are 6 non-vibrational DOF. Thus, we expect that the number of degrees of freedom calculated using your method based on the experimental specific heat should be somewhere between these numbers.

Your calculation, however, needs to be corrected to account for the fact that vibrational degrees of freedom contribute 1R rather than R/2, so the equation becomes

$$\overline{C}_v = 3 R + (f-6)R.$$

From that, I get $$f=33.8$$, well within our expected range. You can repeat the rigorous calculation using the exact frequency of each normal mode of octane and see how close the result is to the experimental value.

UPDATE: To address the efficiency calculation, note that including the vibrational modes changes your equation for $$\gamma$$. Now we have that $$\overline{C}_v=R(3+f-6)$$ for (nonlinear) octane and $$\overline{C}_v=R(\frac52)$$ for hydrogen (ignoring its one vibrational mode). If we use the apparent degrees of freedom of octane of 33.8, $$\gamma_{octane}=1.0325$$ and $$\gamma_{H2}=1.4$$. [Note also that there is really no reason to calculate those apparent degrees of freedom except for curiosity, since we can just add R to the given $$\overline{C}_v$$ to get $$\overline{C}_p$$ and divide to get $$\gamma_{octane}$$.] From that I get an efficiency of 6.5% for the octane and 56% for the hydrogen.