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Adiabatic Gas expansion

  1. Sep 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi, I did a lab experiment where I took a 5L vessel made of some material that isolates gas inside and thus behaves like an isolated system (adiabatic). I then pumped gas from ~90kPa to up to 150 kPa... recorded the temperature, then let the gas 'expand' by opening the piston and recorded the change in Temperature (usually of about 0.4 °C) with an about ~20-30 kPa change in internal pressure.

    I have the following data.

    T1 P1 T2 P2
    Trial 1 27.1 142.5 26.7 95.7
    Trial 2 28.1 143.5 27.7 83.9
    trial 3 29.2 147.16 28.7 108.00


    I need to calculate the amount of work done by this gas.


    2. Relevant equations

    ΔU = q + w

    adiabatic

    ΔU = w

    w = pΔV

    3. The attempt at a solution

    I dont know how to begin... I think i approximated the volume to be constant (Despite some gas being released.

    But that doesnt make sense because then no expansion, means no work....

    How then do i calculate the work done adiabatically? The temperature dropped, and so the pressure... any help?
     
  2. jcsd
  3. Sep 5, 2014 #2
    I have come up with something.

    Since w = p (V2 - V1) i am thinking....

    w = p ( nRT2/P2 - nRT1/P1) ... assuming n = approximately constant...

    would this work?

    thanks
     
  4. Sep 5, 2014 #3
    What does "opening the piston" mean?

    Chet
     
  5. Sep 5, 2014 #4
    Sorry. The vessel had a stopper that you could open and let the gas expand... That's what i meant..
     
  6. Sep 5, 2014 #5

    Matterwave

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    If the expansion is free then no work is being done because the gas is not pushing on a piston or anything like that. If the expansion is adiabatic as well, the ideal gas law tells you that the temperature of the gas will remain constant since U=Q+W and Q=0 due to adiabaticity and W=0 due to free expansion, and an ideal gas's temperature is dependent only on U (U=3/2NkT). This is the adiabatic expansion of a free gas
     
  7. Sep 5, 2014 #6
    isnt the gas pushing against the atmosphere?
     
  8. Sep 5, 2014 #7

    Matterwave

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    I thought that the gas expanded into a vacuum. You are just letting the gas out? In that case, it will quickly just come to equilibrium with the atmosphere...

    Both processes are non-reversible, and non-quasistatic.
     
  9. Sep 5, 2014 #8
    Some questions:

    Are those absolute pressures or gage pressures?

    Is the gas allowed to expand freely against the atmosphere, or does it pass through some kind of constraint like a valve?

    Why isn't the final pressure equal to atmospheric pressure?

    Chet
     
  10. Sep 5, 2014 #9
    Absolute pressure.

    The gas is allowed to expand against the atmosphere, no valve involved.

    Because the stopper was opened for a split second and thus some of the compressed air pushed against atmosphere (left the vessel) while still some remained slightly compressed.



    I think I didnt explain it correctly so let me do it once more.

    The vessel had a wide opening which a rubber stopper (like a flask stopper but bigger and with holes) was placed in order to seal the vessel. A tube that pumps air was connected to the stopper (through one of the holes) into the vessel. Also, a temperature probe was placed through the stopper (through another hole) into the vessel.. Finally, a tube connected to the pressure reader was also connected to the stopper into the vessel. The sealed vessel then is filled with AIR until a certain arbitrary (usually close to 150 kPa) pressure was reached. My friend was in charge of holding the stopper in place and not allow it to pop out due to the high pressure (risky move hehehe). Then after the temperature increased and stabilized at some point, he let the stopper go for a split second (allowing the compressed air to expand against the atmosphere), and quickly closed it again. Not all the air left the vessel which is why the last reading is still higher than atmosphere.

    Attaching an image.

    hope this helps.

    EDIT: I just noticed.. I dont have n !!! the number of moles of 'air' ... so i cant use that formula.... ARG!! i dont know what to do :(

    EDIT2: What about using dw = Cv (dT)... we assume constant volume?
     

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    Last edited: Sep 5, 2014
  11. Sep 6, 2014 #10
    So when your friend opened the stopper for a split second, air was escaping around the stopper. This is similar to a valve, although it was more open than most valves. If the air were released gradually, the expansion would have been reversible, but if it were released to the atmosphere, you would have had irreversible expansion against a constant pressure. Your situation was somewhere in between. Because the pressure change was relatively small, you will probably get the same answer for the temperature change doing it either way. So do it both ways and see what you get.

    Also, you can get the number if moles in the container at the higher pressure by using the ideal gas law.

    Chet
     
  12. Sep 6, 2014 #11
    Incidentally, your closed system is going to be the air that still remains I'm the flask after the stopper is closed. This air has undergone an expansion. To get the temperature change for this air, you don't need to know the number of moles.

    Chet
     
  13. Sep 6, 2014 #12
    well, i know the temperature change because the probe was measuring the temperature inside of the vessel at all times... what is the difference between a reversible and irreversible expansion... sorry but i took p.chem long ago and i forgot already some stuff...

    EDIT: Also, in a reversible expansion... the external pressure is equal to the internal pressure... why would I approach this experiment as being reversible when clearly the external pressure is ~100 kPa and the internal was 50kPA higher? (just read this on a book)
     
    Last edited: Sep 6, 2014
  14. Sep 6, 2014 #13
    Also, we did not intend to reach any specific final or initial pressure reading (due to the difficulty of the experiment itself, holding the stopper of the flask was very difficult to do and thus we needed to be quick). Due to this, our readings vary a lot and thus the degree of expansion is different and not necessarily the same... do I still calculate error? I mean, when we perform an experiment in which an error is calculated, we usually have fixed values we expect to obtain (for instance, burning 5 g of X chemical releases 5 cal of heat <- we expect to obtain this number using the same initial conditions in many different trials). In this case, we started from totally different pressures and ended at some different final pressure after releasing the stopper. It is not that we tried to start at say at 150kPA and by letting go the stopper we expected to reach 100kPa... like i said the difficulty of the experiment itself made it difficult to reach any specific target and thus for each trial I calculated each work done by the expansion without comparing to other trials since clearly they were performed at different initials and final pressures. do i still have to calculate the error? thank you

    By the way we have 20 trials, I only included the first 3.
     
  15. Sep 6, 2014 #14
    You are using the term "external pressure" incorrectly. It is not the pressure external to the chamber. It is the pressure acting on the interface between your closed system and its surroundings. In this case, your closed system is the gas that eventually remains in the vessel after part of the original gas has escaped through the stopper. The external pressure on this closed system is the pressure of the gas that it is pushing ahead of itself (which constitutes its surroundings) out through the clearance between the cork and the wall. There is a pressure drop across the cork, and the pressure on the gas that will eventually remain in the chamber is higher than the pressure of the outside atmosphere throughout the entire expansion. In my judgement, this expansion will be close to reversible.
     
  16. Sep 6, 2014 #15
    My understanding of what you are trying to do is to compare the observed temperature drops with theoretical estimates of the temperature drops that you make based on thermodynamic principles. Do you remember the equation for determining the temperature change for a gas undergoing a reversible adiabatic expansion from one pressure to an lower pressure? Look that up, and apply it to all 20 cases, and see how the results compare. Maybe it will match up pretty well with the observed temperature drops. If not, you can start to think about ways to improve the theoretical model.

    Chet
     
  17. Sep 6, 2014 #16
    I am having difficulties trying to understand the external pressure... You are saying that the pressure of the gas 'escaping' is now the external pressure of the gas that remains inside?

    How then do I calculate the work done by this expansion?

    w = nRT ln(v2/v1) ? but T isnt constant.. im very confused. thanks for your help though


    Also, to answer your previous question...isnt it V2/V1 = (T2/T1)^gama ?
     
  18. Sep 6, 2014 #17
    Yes. I'm saying that the pressure of the gas 'escaping' is the external pressure on the gas that will be remaining inside after the expansion is complete. Think of there being an imaginary boundary that, throughout the expansion, surrounds the gas that will eventually remain in the chamber after the expansion is complete.
    This is the equation for the reversible work in an isothermal process. Your process is not isothermal. It's adiabatic. For an adiabatic reversible expansion, you have:
    [tex]dU = nC_vdT=PdV=nRT\frac{dV}{V}[/tex]
    where n is the number of moles of gas that will be remaining in the cylinder after the expansion is complete. Note that n cancels out between the left hand side and the right hand side of the equation.

    No. Integrate the above equation, and tell us what you get.

    Chet
     
  19. Sep 6, 2014 #18
    Thank you. That is indeed very difficult to visualize... i can see the difference between this and piston that just allows the gas to expand rather than expand and escape.

    integrating gives me

    ΔU = nCv*ΔT = nRTln(v2/v1)

    is this correct?

    then what? do i use the left side of the equation ? or the right side?
     
  20. Sep 7, 2014 #19
    Neither. You integrated incorrectly.

    Try dividing both sides by nT first. Then, from the ideal gas law, ΔlnV=ΔlnT-ΔlnP, so combine this with the result of your integration (i.e., after you integrate correctly this time).

    Also, I am again asking, "are those absolute pressures or gage pressures that you measured?"

    Chet
     
  21. Sep 7, 2014 #20
    ΔU = Cv * ΔlnT = R * Δ(lnV) = R (ΔlnT - ΔlnP)

    do I keep going?

    Cv ΔlnT = R*ΔlnT - R*ΔlnP -> ΔlnT ( Cv - R) = - R*ΔlnP ?


    still lost... sorry :(

    (could you share a link where I can read about this topic, maybe i can save you some time if you could just direct me to the right topic)
     
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