# Speed of pressurized gas escaping into vacuum

Does speed of pressurized gas escaping through a narrow nozzle into vacuum depend on the pressure?

I've asked a question on Space.SE regarding utilizing gas at extreme pressures for propulsion. My idea was that the higher the pressure of the gas, the higher specific impulse would be achieved, because it would increase speed of the escaping particles.

An answer cites the Choked Flow article, claiming The exhaust speed of a rocket is limited by the speed of sound. The answer was criticized as the linked article only gives the case of atmospheric pressure on one side and vacuum on the other as where this would apply, and my question was about way higher pressures.

I'm finding this answer difficult to accept too, because speed of sound should be relative to the medium; a co-pilot of a supersonic plane can still communicate with the pilot, because the air enclosed in the cockpit moves with them. So, in a long pipe, speed of sound in decompressing gas should be relative to the local speed of the gas. Still, I don't have any solid background or source to back it up - thus the question. Is the speed still limited to speed of sound in case of very high pressures? If so, what effect limits it?

• I agree that "the speed of sound" answers is not really satisfying. It is rather tricky to continue acceleration of a gas when reaching the sound barrier, but not impossible. The gas stream coming out of a (stationary) jet engine can be supersonic. Commented Apr 20, 2015 at 10:38

Simple estimate: Bernoulli's equation states that along a streamline $$\frac{1}{2}v^2+\frac{w}{\rho} = {\it const},$$ where $$\rho$$ is the mass density and $$w={\cal E}+P$$ is the enthalpy. For an ideal gas, $${\cal E}=(3/2)P$$ and $$w=(5/2)P$$. Clearly, if you start inside the container with $$v=0$$ and $$P=P_0$$ the final speed will be proportional to $$\sqrt{P_0}$$. Indeed, for an ideal gas $$v\sim \sqrt{(5T)/(2m)}$$, which is bigger (but not parametrically larger) than the speed of sound $$c_s\sim \sqrt{(5T)/(3m)}$$.
I would say that the speed of the gas escaping to vacuum corresponds to the gas temperature. What happens is that you only remove one wall (of a box) and the gas molecules that would bounce off the (removed) wall they just start to freely escape. With the velocity given by $kT$. The opposite wall however still feels the gas pressure.