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

[imsmooth]

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?

 

[Andrew Mason]

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

 

[imsmooth]

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?

 

[Andrew Mason]

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

 

[sardar]

I would like to clarify a few things about the scenario described. First, adiabatic compression means that no heat is exchanged with the surroundings, so the temperature of the air will increase due to the work done on it during compression. This is described by the equation T2/T1 = (P2/P1)^([g-1]/g), where T2 is the final temperature, T1 is the initial temperature, P2 is the final pressure, and P1 is the initial pressure.

Using this equation and the given values, the final temperature would indeed be approximately 1240 K. This is a significant increase from the initial temperature of 273 K.

Next, it is important to note that heating the air to increase the pressure is not the same as adiabatic compression. In this case, heat is being added to the system and the temperature will increase accordingly. The equation P1/T1 = P2/T2 can be used to calculate the final temperature, but it is not directly related to the adiabatic compression scenario.

Finally, if you let the compressed air cool to 273 K, the temperature will decrease and the pressure will also decrease. The equation P1/T1 = P2/T2 can be used to calculate the final pressure, but again, it is not directly related to the adiabatic compression scenario.

In summary, the temperature of air will increase during adiabatic compression, and this increase can be calculated using the appropriate equation. Heating the air or allowing it to cool to a certain temperature will result in different final temperatures and pressures, and these calculations would require different equations.