This is a very confused discussion. Gas being forced through a nozzle, after which it has a lower pressure, is an *irreversible* process in which the entropy increases. This has nothing to do with adiabatic expansion. It has everything to do with the **Joule-Thomson effect**.

The change in temperature following the drop in pressure behind the nozzle is proportional to the Joule-Thomson coefficient, which can be related to the (isobaric) heat capacity of the gas, its thermal expansion coefficient, and its temperature. This is a famous standard example in thermodynamics for deriving a nontrivial thermodynamic relation by using Maxwell relations, Jacobians, and whatnot. Interestingly, it is not certain that the temperature drops. For an ideal gas – which seems to be the only example discussed so far in this thread – it wouldn't, because the Joule-Thomson coefficient exactly vanishes. This is because the cooling results from the work which the gas does against its *internal* van der Waals cohesive forces, and there are no such forces in an ideal gas.

For a real gas cooling can happen, but only below the inversion temperature. For instance, the inversion temperature of oxygen is about $1040$ $K$, much higher than room temperature, so the JT expansion of oxygen will cool it. $\text{CO}_2$ has an even higher inversion temperature (about $2050$ $K$), so $\text{CO}_2$ fire extinguishers, which really just spray $\text{CO}_2$, end up spraying something that is very cold. Hydrogen, on the other hand, has an inversion temperature of about $220$ $K$, much smaller than room temperature, so the JT expansion of hydrogen actually *increase* its temperature.

In principle, **YES.**

Remember, temperature is related to the velocity distribution of the particles inside the gas. The key word here is $collision$.

Accelerating the train and suddenly stopping it is akin to **shaking a container that contains fluid once**. The sudden acceleration and deceleration of the contained will impart momentum from the wall of the container to the particles inside. I see two cases.

(a) If the container contains liquid and is not completely full, you could increase its temperature because you are imparting momentum of the particles inside the liquid by virtue of them sloshing inside the container.

(b) If the container contains liquid and is completely full, you could increase its temperature in the same principle but not significantly. Similarly, for a gas, while in principle you could have the same effect as in a container filled with gas, the density of the gas is so much lower than the liquid, the effect would have to be negligible.

## Best Answer

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

effectiveprice you pay for your transportation fuel increases.