# Balloon of ideal gas pops in vacuo - K.E. per atom unchanged, velocities become locally correlated?

My question is: is the following thinking pretty much correct?

I'm interested hearing about anything here that is fundamentally wrong. Of course there is a bit of hand waving and approximation.

If a balloon of ideal gas in equilibrium at a finite temperature T suddenly bursts while in space, the gas cloud will immediately begin to expand.

If the initial average kinetic energy per particle was $\frac{3}{2}{k_b}T$, the initial total kinetic energy would be just $\frac{3}{2}N{k_b}T$. There shouldn't be much loss of energy anywhere, so I'm assuming the final total kinetic energy would be about the same.

The much-larger gas cloud is now cold "due to expansion", but what's really happened is that the current position of each atom is determined by it's velocity at the last scatter event, when the mean free path became larger than the diameter of the cloud. Now, if a group of atoms are still near each other, it's because their velocities have been similar for a while.

That means the relative velocity of a group of nearby atoms with respect to each other is much lower than the average velocity they are moving, and therefore the local temperature is low.

my question: The average kinetic energy per atom hasn't changed. However their velocities are now highly correlated locally, and so we say the temperature has decreased - it is "cold".

• "There shouldn't be much loss of energy anywhere" - right never a loss but energy does tend to move about when equilibria are suddenly upset. Where did the energy go that was stored in the stretched material of the balloon? – docscience Apr 14 '16 at 16:02
• understood. My point is the balloon was also storing energy. Did some of that energy after the pop transfer to the gas? – docscience Apr 14 '16 at 16:11
• @docscience, Re, energy transfer from balloon fabric to the gas: Look on YouTube for high-speed video of popping latex balloons. It's very fast, and what it looks like is about what you would expect if, instead of being filled with gas, the balloon was stretched around a rigid, invisible, frictionless ball. A tear starts, and quickly grows, and the balloon "unwraps itself" from the ball. It's like the balloon shrinks down to its un-stretched state before the air even knows to start "escaping". I'd be suprised if much energy is transferred to the air. – Solomon Slow Apr 14 '16 at 18:02

## 2 Answers

For an ideal balloon-pop (no interactions with the walls of the balloon) $dU = 0$. In the case of an ideal gas, $U = U(T) \propto NkT$. So the temperature, strictly speaking, does not change when the balloon expands. You end up with a gas at the same temperature but with a lower density.

• So the balloon does not expand in my question, it simply pops - ceases to be a boundary. The gas will certainly expand and cool. – uhoh Jun 17 '16 at 5:08
• @uhoh And why will it cool? – oscarafone Jul 9 '16 at 5:52

After writing this answer I recalled asking this question two years ago. I'll answer my own question just to wrap this up.

My question is: is the following thinking pretty much correct?

Yes. For a given small region of the expanding gas, the temperature would be defined by the distribution of velocities around the collective center of mass velocity of that region.