What happens to temperature when compressing air adiabatically to 200 atm?

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Discussion Overview

The discussion revolves around the effects of adiabatic compression of air, specifically when compressing air from 1 atm to 200 atm starting at a temperature of 273K. Participants explore the resulting temperature changes, the application of relevant thermodynamic formulas, and the implications of heating air under pressure. Additionally, there are related inquiries into gas expansion and its thermodynamic effects.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant calculates the temperature after adiabatic compression using the formula T2/T1 = (P2/P1)^([g-1]/g) and arrives at approximately 1240 K, questioning the correctness of this result.
  • Another participant confirms the calculation of 1240 K as correct but notes that if the air were heated to achieve 200 atm, the temperature would be 273*200 K, which seems excessively high.
  • There is a discussion about using the formula P1/T1 = P2/T2 for determining pressure after allowing the adiabatically compressed air to cool to 273K, resulting in a calculated pressure of 44 atm.
  • In a separate thread, participants discuss the thermodynamics of gas expansion, noting that during reversible expansion, the temperature drops, while in free expansion, it remains constant.
  • One participant raises a question about the work done by gas expanding through a needle valve and whether it does work against the outside air, linking this to temperature changes and the Joule-Thomson effect.
  • Another participant clarifies that real gases may behave differently than ideal gases, particularly in terms of temperature changes during expansion and the role of intermolecular forces.

Areas of Agreement / Disagreement

Participants generally agree on the application of certain thermodynamic principles, but there are competing views regarding the behavior of gases under different conditions, particularly between ideal and real gases. The discussion remains unresolved in terms of the implications of heating air to 200 atm and the effects of gas expansion.

Contextual Notes

Limitations include assumptions about ideal gas behavior versus real gas behavior, the dependence on specific conditions of the gas, and the unresolved nature of temperature changes during various expansion scenarios.

imsmooth
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I am trying to understand what will happen to the temperature under these conditions (starting temp T1 = 273K).

I compress 1 atm air adiabatically to 200 atm. Using the formula T2/T1 = (P2/P1)^([g-1]/g)

where g = 1.4 for air

I get a temperature of about 1240 K. Is this correct?

Now if I "heated" the air so the pressure went up to 200 atm, would the temperature be 273*200K? That is very hot. I'm now using the formula P1/T1 = P2/T2. This seems too hot, but is this right?

Finally, for the question at the top, if I take the adiabatically compressed air and let the temperature fall to 273K, would I use the P1/T1 = P2/T2 formula? This seems to give a pressure of 44 atm.

Is my reasoning correct?
 
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imsmooth said:
I am trying to understand what will happen to the temperature under these conditions (starting temp T1 = 273K).

I compress 1 atm air adiabatically to 200 atm. Using the formula T2/T1 = (P2/P1)^([g-1]/g)

where g = 1.4 for air

I get a temperature of about 1240 K. Is this correct?
Correct.

Now if I "heated" the air so the pressure went up to 200 atm, would the temperature be 273*200K? That is very hot. I'm now using the formula P1/T1 = P2/T2. This seems too hot, but is this right?
That would be right if air was an ideal gas at such a temperature. But it would not be. For an ideal gas, if volume is constant, P1/T1 = P2/T2 so if the pressure increases by a factor of 200 the temperature does as well.

Finally, for the question at the top, if I take the adiabatically compressed air and let the temperature fall to 273K, would I use the P1/T1 = P2/T2 formula? This seems to give a pressure of 44 atm.
Is my reasoning correct?
That is correct. If volume is constant then P1/T1 = P2/T2. 200*(273/1240) = 44 atm.

AM
 
Next question on thermodynamics:

If a gas expands reversibly and does work the temperature drops. During free expansion (W = 0, Q = 0, so Ui = Uf) the temperature remains the same.

If the gas expands through a needle valve (high pressure going to low pressure, throttling) during a refrigeration cycle, is the work obtained from this gas expanding against the gas on the low pressure side of the valve?

In case the wording is confusing, if I puncture a pressurized CO2 cartride, does the expanding CO2 do work by pusing against the outside air? By doing work, the kinetic energy drops along with the temperature? If the CO2 canister is punctured within a larger container that has a vacuum, and the walls are totally insulated, then there is no change in temperature?
 
Last edited:
imsmooth said:
Next question on thermodynamics:

If a gas expands reversibly and does work the temperature drops. During free expansion (W = 0, Q = 0, so Ui = Uf) the temperature remains the same.
If you are talking about an ideal gas, this is correct. But the internal potential energy of a real gas may increase with expansion, so its temperature (internal translational kinetic energy) must decrease so its total internal energy remains unchanged.
If the gas expands through a needle valve (high pressure going to low pressure, throttling) during a refrigeration cycle, is the work obtained from this gas expanding against the gas on the low pressure side of the valve?
If it is an ideal gas and a free expansion, there will be no decrease in temperature. So real gases are used. These real gases cool when they expand. This is due to the intermolecular bonds. Increasing the separation of molecules requires energy and this must come from the internal energy of the compressed gas. This is the Joule-Thomson effect.

In case the wording is confusing, if I puncture a pressurized CO2 cartride, does the expanding CO2 do work by pusing against the outside air? By doing work, the kinetic energy drops along with the temperature? If the CO2 canister is punctured within a larger container that has a vacuum, and the walls are totally insulated, then there is no change in temperature?
CO2 is not an ideal gas. So the Joule-Thomson effect results in cooling in the free expansion.

AM
 

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