# Potential energy of a pressurized gas canister in space

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• maxolina
I'm not sure if that still applies though.The speed of sound is a function of pressure, temperature, and the gas's composition.f

#### maxolina

Suppose there is a pressurized gas canister in space, at rest. With a mass "m" of gas inside of it at a pressure "P".

Next the valve of the canister is opened. The canister will accelerate in the opposite direction to the valve opening. When all the gas has left the canister, it will be moving with a certain kinetic energy through space.

Is it possible to calculate how much potential mechanical energy the gas has due to its pressure, that will end up transformed into kinetic energy?

Suppose there is a pressurized gas canister in space, at rest. With a mass "m" of gas inside of it at a pressure "P".

Next the valve of the canister is opened. The canister will accelerate in the opposite direction to the valve opening. When all the gas has left the canister, it will be moving with a certain kinetic energy through space.

Is it possible to calculate how much potential mechanical energy the gas has due to its pressure, that will end up transformed into kinetic energy?

I would suspect there is a limit to the kinetic energy the canister will acquire(?), but the actual kinetic energy of the canister will depend on how the gas is released.

• russ_watters
Suppose there is a pressurized gas canister in space, at rest. With a mass "m" of gas inside of it at a pressure "P".

Next the valve of the canister is opened. The canister will accelerate in the opposite direction to the valve opening. When all the gas has left the canister, it will be moving with a certain kinetic energy through space.

Is it possible to calculate how much potential mechanical energy the gas has due to its pressure, that will end up transformed into kinetic energy?
For an ideal gas, the potential energy of a gas does not depend on pressure at all. It is simply equal to the thermal energy of the gas. If you let the canister expel its contents as a well collimated stream with negligible pressure, that stream will be moving at a speed equal to the thermal velocity of the starting contents.

This corresponds to an adiabatic expansion.

There is an interesting formula that results if you consider an isothermal expansion instead. The energy that can be extracted is pressure times volume times the natural logarithm of the ratio between the initial and final pressures:$$E=P_0V_0\ln\frac{P_0}{P_1}=P_1V_1\ln{\frac{P_0}{P_1}}$$

[If ##P_1 > P_0## the formula still works. The negative result means that isothermal compression takes energy]

You can wrap an intuition around this by thinking about an expansion by a factor of 2 as delivering a certain amount of energy. Then an expansion by another factor of 2 delivers the same energy (half the pressure but twice the volume swept out). The number of times you can do this scales with the log of the pressure ratio.

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• russ_watters
This is not a hard problem. The "before" momentum is zero. The "before" energy is the gas pressure times the volume of the tank. The "after" momentum is zero. The "after" kinetic energy is the kinetic energy of the gas plus the kinetic energy of the tank. The gas doesn'tr have a single velocity.

I think the upper bound of the kinetic energy of the canister is found by the gas escaping it at the local speed of sound in the gas. The speed of sound is a function of the pressure in the tank, but I believe it decreases with decreasing pressure. So assuming it is constant ##v_s## ( its value from the initial state of compression).

$$\left( m_c + m_g \right) \frac{dv}{dt} = \rho A v_s^2 \implies M \frac{dv}{dM} = -v_s$$

Where ##M## is the total mass of gas ##m_g## + canister ##m_c##

This would lead to the "burnout velocity":

$$v = v_s \ln \left( \frac{m_c + m_g }{m_c} \right)$$

Where ##m_g## would be given by the ideal gas law:

$$m_g = \frac{1}{RT}V\llap{-} P$$

So substitute all that, and find that the upper bound for the kinetic energy of the canister as a function of pressure is given by:

$$KE(P) = \frac{1}{2} m_c v_s^2 \left( \ln \left( \frac{m_c + \frac{1}{RT}V\llap{-} P }{m_c} \right) \right)^2$$

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The speed of sound is not a function of pressure. However, it is a function of temperature. If temperature is decreasing as the gas escapes (as it should) then the speed of sound will decrease as a result.

https://en.wikipedia.org/wiki/Speed_of_sound#Speed_of_sound_in_ideal_gases_and_air
Oh, I thought the speed of sound in a medium was inversely proportional to the density of the medium. I do recall that it was a function of the temperature for a gas. But the fact that the temperature and Pressure of a gas are not independent is somewhat confusing to me.

Anyway, either "by hook or by crook" that's not changing the outcome...this time!

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For an ideal gas, the potential energy of a gas does not depend on pressure at all. It is simply equal to the thermal energy of the gas. If you let the canister expel its contents as a well collimated stream with negligible pressure, that stream will be moving at a speed equal to the thermal velocity of the starting contents.

This corresponds to an adiabatic expansion.

There is an interesting formula that results if you consider an isothermal expansion instead. The energy that can be extracted is pressure times volume times the natural logarithm of the ratio between the initial and final pressures:$$E=P_0V_0\ln\frac{P_0}{P_1}=P_1V_1\ln{\frac{P_0}{P_1}}$$

[If ##P_1 > P_0## the formula still works. The negative result means that isothermal compression takes energy]

You can wrap an intuition around this by thinking about an expansion by a factor of 2 as delivering a certain amount of energy. Then an expansion by another factor of 2 delivers the same energy (half the pressure but twice the volume swept out). The number of times you can do this scales with the log of the pressure ratio.
What is ##P_1## in the vacuum of space?

What is ##P_1## in the vacuum of space?
Nearly zero.

You may be concerned that this gives rise to infinite energy. However, practical limitations intrude. First, the vacuum of space is not perfect. Second, you eventually run into volumes larger than the observable universe. It is difficult to assemble hermetically sealed devices that large and to find a heat reservoir somewhere else to re-heat the tenuous gas that fills one.

Also, the ideal gas law has questionable applicability in non-stationary space-times.

The fact that we are taking the log of the pressure ratio means that even huge pressure ratios lead to quite modest energy values.

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• erobz