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Homework Help: Thermodynamics - This time it really is adiabatic!

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data

    14 grams of Nitrogen at 20 degrees Celsius in 1 atmosphere of pressure is compressed to a pressure of 20 atmospheres. What is its final temperature and what is the work done on the gas if the process is adiabatic? Obtain the compression ratio (i.e. Vmax/Vmin). How much heat comes in or leaves the gas? What is the ratio of Pmax/Pmin? Show this change on a P-V graph.

    2. Relevant equations

    PV = NkT

    PinitialVinitial[tex]\gamma[/tex] = PfinalVfinal[tex]\gamma[/tex]

    VinitialTinitial[tex]3/2[/tex] = VfinalTfinal[tex]3/2[/tex]

    W = -[((PinitialVinitial[tex]\gamma[/tex])/(1-[tex]\gamma[/tex]))(Vfinal(1-[tex]\gamma[/tex]) - Vinitial(1-[tex]\gamma[/tex]))

    3. The attempt at a solution

    14 grams of Nitrogen = 1 atom

    Tinitial = 20 degrees Celsius = 293 Kelvin

    Tfinal = ?

    Pinitial = 1 atm = 1.013 * 105 Pa (1 Pa = 1 N/m2)


    PinitialVinitial = NkTinitial

    (1.013 * 105 N/m2)Vinitial = 1(1.381 * 10-23 Nm/K)(293K)

    Vinitial [tex]\approx[/tex] 4 * 10-26


    PinitialVinitial[tex]\gamma[/tex] = PfinalVfinal[tex]\gamma[/tex]

    (1.013 * 105 N/m2)(4 * 10-26) = (20 atm)(1.013 * 105 N/m2)Vfinal[tex]\gamma[/tex]

    Vfinal = (4 * 10-26 m3) [tex]\sqrt[5]{[1.013 * 105 N/m2) / ((20 atm)(1.013 * 105 N/m2))]3}[/tex]

    where [tex]\gamma[/tex] = 5/3

    Vfinal [tex]\approx[/tex] 6.6 * 10-29 m3

    Okay... I'm not done yet, but I am going to submit this to check out how my latex reference skills are working out. This is tedious.
  2. jcsd
  3. Oct 5, 2008 #2
    Oh holy crap, it is a disaster... well, the important parts can be made out. it would be nice to have a little chalk board we could draw on. Maybe I'll continue the problem after a while.. that is enough for now. Hopefully I will get better and faster at inputting this stuff.
  4. Oct 5, 2008 #3
    I thought of a comment... under the fifth-root-radical it is suppose to be (# of Pascals in an atmosphere)/(20 atmospheres)(# of Pascals in an atmosphere) all cubed... because it is Pinitial / Pfinal. I did it with the Pascals because that helps me to work with my units. I think it should be the same outcome if I just use atmospheres because it seems like the # of Pascals in an atmosphere would cancel out anyway. Maybe I am putting in the calculator wrong, but I get different values when I do it with Pascals or atmospheres. It is a difference of like two orders of magnitude. That is turning out the be the most troublesome part of this problem for me. If anyone could figure out what Vfinal should be, that would be nice. :smile:
  5. Oct 6, 2008 #4
    I went with 6.6*10-27 instead of 10-29 because the latter made way to much of a temperature increase. Other than that, I think I can move on from this point. I ended up getting that the change in heat coming in or leaving the system was zero. Does that make sense? I mean, it is adiabatic, so the change in heat has to be zero during the process, but there is adiabatic cooling. Should I worry about that here?
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