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Work done by an expanding gas

  1. Jan 27, 2014 #1
    Today in an engineering thermodynamics lecture, the professor gave an example of a gas doing work. We had a cylinder full of helium at a pressure of something like 200kPa absolute and the valve was opened so that the gas would flow out against the atmospheric pressure until the pressures were equal. Also the cylinder was assumed to be in thermal equilibrium with its surroundings so the temperature of the gas was equal to the temperature of the ambient air. However, the way he calculated the work perturbed me. He said that this was an isobaric process because the gas was expanding against a constant atmospheric pressure. I was under the assumption that an isobaric process means that the working fluid stays at constant pressure throughout the process which is not the case in this expansion. And in this case, the gas pressure is dropping as it leaves the cylinder.

    The professor then proceeded to calculate the work as W = Patm*ΔV. But I don't think that is right and that simple.

    Am I correct, or is the professor? Can someone please return me to sanity?
     
  2. jcsd
  3. Jan 27, 2014 #2

    Jano L.

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    Gold Member

    You are right, the helium expands and its pressure decreases from 200kPa to atmospheric 100 kPa. Helium gas does not undergo isobaric process in the common sense of the word (atmosphere does).

    Your professor is right this time - the work helium does on the atmosphere is indeed (approximately) W = Patm*ΔV, where ΔV is the volume of the helium gas outside the cylinder just after it escapes from it. After a while, the helium is heated by the atmosphere and expands even more and does further work, but this work is neglected in the above.
     
  4. Jan 27, 2014 #3
    Thanks for the response Jano,

    If the process is not isobaric, then why is the work not calculated using an integral and instead W = Patm*ΔV. I'm fairly certain he took the system as a control mass/closed system.
     
  5. Jan 27, 2014 #4

    Jano L.

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    You can write the work as integral, but because the pressure of the atmosphere can be assumed constant during the process, the result is just ##P_{atm} \Delta V##.
     
  6. Jan 27, 2014 #5
    Okay, so in this case, why is the pressure of the atmosphere the pressure used to calculate work and not the pressure of the working fluid?
     
  7. Jan 27, 2014 #6
    What I mean to say is, why, since this is not an isobaric process, the work is calculated using a pressure that is assumed not to change?

    edit: Looking at the atmosphere as the working fluid I agree that the work is defined P_atm*deltaV.
     
    Last edited: Jan 27, 2014
  8. Jan 27, 2014 #7
    It depends on what you define as your system. If you define your system as just the gas that remains in the cylinder after equilibration, then that gas has done work on expelling the gas from the cylinder, and the pressure at the interface with the gas that it expelled was not at constant pressure.

    If you define your system by surrounding all the helium that was originally inside the cylinder with an imaginary moving boundary, then, throughout this process, different parts of the helium were at different pressures. However, at the imaginary boundary with the surrounding atmospheric air, the pressure was constant (atmospheric). In the first law, you calculate the work done on the surroundings by calculating the integral of the pressure at the interface with the surroundings integrated over the volume change. (See my Blog on my PF home page.) So, in the case of this system, your professor was correct.


    Chet
     
  9. Oct 22, 2014 #8

    Odd

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    Hello

    Is it possible then to calculate the same work looking only at the work of the expanding gas inside the cylinder ?

    I assume one would have to use d(PV) and then a equation of state for the process.

    Odd
     
  10. Oct 22, 2014 #9
    You would just solve it as an isothermal reversible expansion. The real irreversibility occurs within the valve, where the pressure drops from that inside the cylinder to 1 atm.

    Chet
     
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