Thermodynamics question

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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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Andrew Mason
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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?
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
 
  • #3
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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?
 
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Andrew Mason
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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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