Skip to main content
addition thoughts on the question.
Source Link
jimchmst
  • 2.4k
  • 4
  • 9

IF the system is isolated and the source of the gas part of the room at the same T and V of the room constant the expanding gas can do no work and there will be no change in T from the simple mixing. The massive Kinetic energy from the directed air stream will eventually turn into heat. This can be eliminated by constraining the gas to diffuse randomly. There will be a Joule-Thompson effect either releasing or absorbing heat depending on the temperature. Finally, the entropy of the expanding, released, gas will increase, this means the entropy of the Universe will increase.

I imagine a pressurized helium balloon with a non-plastic wall but porous to helium, so no polymer crystal effects, slowly releasing the helium until the partial pressures equalize increasing the overall pressure at constant volume.

A correction: The compressed gas has potential energy equal to the work done in filling the tank. Upon release this is converted to KE, the wind generated. Eventually, as the wind is stilled, this becomes heat. This cannot be eliminated by a diffusion process if there is a pressure difference.

IF the system is isolated and the source of the gas part of the room at the same T and V of the room constant the expanding gas can do no work and there will be no change in T from the simple mixing. The massive Kinetic energy from the directed air stream will eventually turn into heat. This can be eliminated by constraining the gas to diffuse randomly. There will be a Joule-Thompson effect either releasing or absorbing heat depending on the temperature. Finally, the entropy of the expanding, released, gas will increase, this means the entropy of the Universe will increase.

I imagine a pressurized helium balloon with a non-plastic wall but porous to helium, so no polymer crystal effects, slowly releasing the helium until the partial pressures equalize increasing the overall pressure at constant volume.

IF the system is isolated and the source of the gas part of the room at the same T and V of the room constant the expanding gas can do no work and there will be no change in T from the simple mixing. The massive Kinetic energy from the directed air stream will eventually turn into heat. This can be eliminated by constraining the gas to diffuse randomly. There will be a Joule-Thompson effect either releasing or absorbing heat depending on the temperature. Finally, the entropy of the expanding, released, gas will increase, this means the entropy of the Universe will increase.

I imagine a pressurized helium balloon with a non-plastic wall but porous to helium, so no polymer crystal effects, slowly releasing the helium until the partial pressures equalize increasing the overall pressure at constant volume.

A correction: The compressed gas has potential energy equal to the work done in filling the tank. Upon release this is converted to KE, the wind generated. Eventually, as the wind is stilled, this becomes heat. This cannot be eliminated by a diffusion process if there is a pressure difference.

Source Link
jimchmst
  • 2.4k
  • 4
  • 9

IF the system is isolated and the source of the gas part of the room at the same T and V of the room constant the expanding gas can do no work and there will be no change in T from the simple mixing. The massive Kinetic energy from the directed air stream will eventually turn into heat. This can be eliminated by constraining the gas to diffuse randomly. There will be a Joule-Thompson effect either releasing or absorbing heat depending on the temperature. Finally, the entropy of the expanding, released, gas will increase, this means the entropy of the Universe will increase.

I imagine a pressurized helium balloon with a non-plastic wall but porous to helium, so no polymer crystal effects, slowly releasing the helium until the partial pressures equalize increasing the overall pressure at constant volume.