# Ideal gas free expansion but cools down

Disclaimer: I don't think this is a duplicate of the existing questions on why air cools down when expanded.

I have recently been deflating my bike tyres from 2 bars to 1 atm. As the air exitted the nozzle, the metallic part substantially cooled down. I have also seen that this happens when a plastic bottle is filled with pressurized air ~2 bars, and condensation appears when presure is released.

I am familiar with the Joule-Thomson expansion which predicts that for most non-ideal gases, expansion implies they cool down. However, ideal gases should not cool down. However, I have been told that at ~2 bars and 20ºC the atmospheric air pretty much behaves like an ideal gas.

Consequently, I am finding it hard to figure out why the nozzle of the bike tyre cooled as much (about 2 degrees maybe?) when I released the air. Does it have to do with the water disolved on it?

PS: Wikipeda mentions that 3ºC are predicted for a gas at ~22 bars if I am not mistaken.

• The metal on the nozzle is a good thermal conductor. Prior to releasing the air the inside of the tire is in equilibrium with the outside (heat bath). Upon releasing the air, the pressure will drop and so will the temperature inside the container ($T = PV/Nk$). In reality it's more complicated then this since the N is changing as well. This is moving to another (N, V, T) ensemble as this situation would be well modeled by the canonical ensemble. But yeah the temperature difference outside the metal and inside the tube will cause it to "feel" cold, as heat moves from outside to inside. Jun 2, 2021 at 14:37
• Good point! However, I was not expecting the temperature inside the tyre to change, since it is an ideal gas freely expanding. Jun 2, 2021 at 14:40
• Volume in the tire is fixed so it isn't freely expanding. If the tire were allowed to expand with a pressure decrease then for an ideal gas there would be no temperature change. Jun 2, 2021 at 14:43
• Interesting, so what could be the mechanism that is driving this temperature drop? (temperature = kinetic energy of the particles = constant?) Jun 2, 2021 at 14:50
• Why would the kinetic energy of the particles be constant? Average kinetic energy for an ideal gas is $E=(3/2)NkT$. If the number of particles is reduced and all else is constant the average energy will drop. When a new equilibrium is established the lower average energy will equate to a lower temperature. The process of removing air is moving from one equilibrium to another, so strictly speaking these equations only hold in the start and end states. Jun 2, 2021 at 14:59

Take air to be an ideal gas. Decouple the process into adiabatic expansion within the tire (for any conceptual region) and Joule–Thomson expansion at the valve. The former cools the gas; the latter doesn't. The cooler gas exiting past the metallic valve cools it by convection.

• Very interesting combination of both processes, I think that this answer also points in the right direction! Jun 2, 2021 at 20:23

The expansion process that the air in the tire experiences is very different from the expansion process encountered by the gas in free expansion. In free expansion, there is no resistance to the gas expanding, and it does no work. In the case of the tire, there is significant flow resistance to the gas passing through the valve, and a significant pressure drop occurs across the valve. In order to overcome this flow resistance and to force a parcel of gas through the valve, the gas behind it must do a significant amount of work. This work that has been done by the gas remaining inside the tire at any point in time results in a decrease in its internal energy per unit mass. This is reflected in a decrease in its temperature.

If we apply the open system (control volume) version of the first law of thermodynamics to this system and neglect heat transfer, we obtain:$$d(nu)=hdn$$where n is the number of moles of gas remaining in the tire at any time, u is the internal energy per unit mass of the gas in the tire, and h is the enthalpy per unit mass of the gas entering the valve (and equal to the enthalpy per unit mass of the gas within the tire). For an ideal gas, with no loss of generality, we can write $$u=C_vT$$ and $$h=C_pT$$. Substituting into the above equation then yields: $$nC_vdT=(C_p-C_v)Tdn$$or, equivalently, $$d\ln{T}=(\gamma-1)d\ln{n}$$where $$\gamma=C_p/C_v$$. The number of moles of gas in the tire n is related to the tire volume V and the molar volume of the gas v by: $$n=V/v$$ Therefore we have $$\frac{d\ln{T}}{d\ln{v}}=-(\gamma-1)$$This is just a well-known relationship for the adiabatic reversible expansion of an ideal gas.

• I liked the first paragraph a lot, it gave me much insight! And an idea of how to pictorically represent this with particles... Jun 2, 2021 at 20:19
• For more insight, see the diagram in Example 6.10, Considering Air Leaking From a Tank, Fundamentals of Engineering Thermodynamics, Moran et al. This book is available online. Jun 2, 2021 at 21:32
• Never heard of this book but I like its pedagogical approach so much! Yes, I think I had something like that diagram in mind: 1.- the bulk of air pushing against the air that leaves the valve 2.- the energy of the air that exits (and all that has been transfered to it while being pushed) gets lost in the atmosphere... Jun 2, 2021 at 21:45
• This book is really excellent, particularly Chapter 6 that deals with entropy and entropy balances. Jun 2, 2021 at 21:52

Based on the two previous answers, this is my attempt from a microscopical point of view:

• The portion of air that happens to be near the valve acts as a "wall". The rest of the air inside the tyre "pushes" this wall out. Thus, particles inside the tyre cool down because they are transferring Kinetic Energy to the particles in the wall. Adiabatic expansion

• The particles near the valve eventually escape the tyre, so the energy that has been transfered to them also exists the tyre. Joule effect

Thus the Kinetic Energy of the particles inside the tyre gets lower and lower $$\Rightarrow$$ lower temperature.

(Example 6.10 from Fundamentals of Engineering Theormodynamics by Moran)

Edit: given that the particles of the gas interact (they "push" each other...) this implicitly asumes a Real Gas (not Ideal)