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Homework Help: Finding the compressibility factor (Z)

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the volume, in m^3, occupied by 20 kg of hydrogen (H2) at 1170 kPa, 2220°C.

    2. Relevant equations

    Z=pv/rt, Pr=P/Pc, Tr=T/Tc, and for hydrogen M = 2.016 (kg/kmol) Tc = 33.2 (K) Pc = 13.0 bar Zc=pc*vc/(RTc)

    3. The attempt at a solution
    I know if I find Z then the problem is a done deal. I'm just confused about how to use Pr,Tr, and the table to find Z. I found Pr=.9 and Tr=1.59 but is there a formula for Z?

    Thanks in advance for the help (I'm not a fan of Thermo so far and my Profs at my old college spoiled me with great teaching)
  2. jcsd
  3. Sep 15, 2012 #2


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    Staff: Mentor

    Isn't it just an ideal gas question?
  4. Sep 15, 2012 #3
    Sadly no, Z is the factor that relates PV/RT of Hydrogen to the PV/RT of and ideal gas. Hydrogen in this problem, because of the high temperatures and pressure, is not an ideal gas.
  5. Sep 16, 2012 #4
    Something is sadly amiss. You have said that you found Tr = 1.59, after also saying that T = 2220+273 K and Tc = 33.2K. But when I calculate using the equation that you have supplied

    Tr = T/Tc

    I get Tr = 2493/33.2 , which is roughly 78, not 1.59! Your value for pr seems about right, although you will need to carry at least 1 more significant figure, and preferably 2 more to get a result with any precision.

    The other point is that I am not seeing an equation for whichever approach you are using to approximate the ideal gas equation, if that is not good enough. Van der Waals equation? Virial theorem calculation? or what?
  6. Sep 16, 2012 #5
    Ok, I feel really bad about this, the T value is -220 °C, which makes the Tr come out to 53.0K/33.2k=1.59. Pr=.900 with the correct number of sig. figs. After reading the question twice more, looking for something about how to approximate Z, i finally found the attached chart hiding in the back of the book. This makes it look like Z should be .95, although it seems like a very inaccurate way to find it. If this is true, v=(8.314m3Pa/(k*mol)*53K*.95/(11.7x10^3 Pa)=.0358m3/mol
    Using dimensional analysis, .0358m3/mol / .00202 kg/mol * 20kg = 354 m3, and it looks like I could have done this problem without wasting everyone's time. Sorry about that.

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