Found this at the gas station yesteday - got me thinking...
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$\begingroup$ Isn't this a chemistry question? Migrate? $\endgroup$– LanceLafontaineCommented Jun 7, 2012 at 19:52
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3$\begingroup$ @LanceLafontaine: Naah, physical chemistry. It just has to do with changing volume by temperature. Remember, it's "gasoline dispensed". It's not talking about any changes to gasoline's chemical properties which may or may not happen in the engine--that's more of a chemistry question. Depending on it's form, if it was asked on chem.SE, we might have kept it. Since it fits here as well, and has an answer, there is really no need to migrate at all. $\endgroup$– ManishearthCommented Jun 8, 2012 at 0:36
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2 Answers
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My belief is that this refers to the density effect alone. Obviously you will get less gas on a hot day since it will be lower density at that temperature.
This is separate from thermodynamic factors. I would venture an educated guess that ICEs will be less efficient on hot days, and I would suspect that this factor would be greater than the density difference. Nonetheless, you could still fill up on a hot day and later drive on a cold day (or vica versa, as you might do after reading this answer). As the temperature changes, the level in your tank will change with it.
As you drive your car, you adjust the throttle to get the needed torque regardless of the density of the gas, so the total mass of fuel is a metric that is much more directly relevant to satisfying your transportation needs.
I can find one reference that gives different densities of crude oils at different temperatures. Here, I find the points $(77^{\circ} F, 1.014 g/cm^3)$ and $(113^{\circ}, 1.002 g/cm^3)$. Crude oil is different from gasoline, but mostly because it contains a wider variety of Carbon chain lengths, and I think the density sensitivity to temperature should be on the same magnitude. I will report a finding that a percentage change in density for a degree Fahrenheit is about:
$$-0.03 \frac{\text{%}}{^{\circ} F}$$
So if the temperature difference between seasons is a fairly typical $30^{\circ} F$ difference, then we're looking at a difference of about $1 \text{%}$ in price. Obviously that comes out to multiple cents per gallon. Anecdotally, I know quite a few Americans who would switch gas stations over smaller amounts. It would also matter a great deal to the gas station's margin.
Again, Wolfram alpha has proven to be useful beyond my expectations. Here is the exact formula to find the density percent difference for octane. Octane is fairly representative (but not exactly) of a fuel with a 100 octane rating. This gives $-0.06395 \text{%}/^{\circ}F$, which is twice what I had for crude oil. So we are looking at closer to a $2 \text{%}$ difference in price for a $30^{\circ}F$ difference in temperature.
Another edit: Basis in equations to show that what I did is legit. I'll use price, some constant, and volume respectively in the first equation.
$$P=\alpha V$$
Next equation has mass, density, and volume respectively.
$$M = \rho V$$
I'll combine these to get the final equation that makes the point.
$$P/M = \frac{\alpha }{\rho} \approx \frac{\alpha}{\rho_0} - \rho \frac{\alpha}{\rho_0^2}$$
$$\Delta (P/M) = - \frac{\Delta \rho \frac{\alpha}{\rho_0^2} }{\frac{\alpha}{\rho_0}} = -\frac{\Delta \rho}{\rho_0} = - \frac{d\rho}{dT} \Delta T$$
The numbers I give above are for $d\rho/dT$. Temperature increases and the effective price you pay for your transportation fuel increases.
answered Jun 7, 2012 at 20:08
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6$\begingroup$ Although in practice the fuel is stored in large underground tanks which don't vary as much as the air temperature $\endgroup$ Commented Jun 7, 2012 at 22:02
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On the other hand most of the car's energy goes to overcome air resistance (at least at highway speeds) and drag is proportional to air density.
And air density is inversely proportional to absolute temperature. So on hot summer days you might not have to burn quite so much - assuming you don't need air conditioning.
answered Jun 7, 2012 at 22:07
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1$\begingroup$ But your engine is much more efficient with cold air - as it is more dense. Wonder where the two curves would balance out :-) $\endgroup$ Commented Jun 7, 2012 at 23:57
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1$\begingroup$ Is it? I can imagine it might have more power since there is more air available - but the fuel system on a modern car will keep the optimal fuel-air mix anyway. So in the summer you have slightly less 'turbo' as it reduces the fuel volume $\endgroup$ Commented Jun 8, 2012 at 3:41