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Homework Help: Isothermal Question

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A 2.00-mol sample of helium gas initially at 300K and 0.400 atm is compressed isothermally to 1.20atm. Assume the helium behaves as an ideal gas, find (a) the final volume of the gas, (b) the work done on the gas, and (c) the energy temperature?


    2. Relevant equations
    PV = nRT

    [tex]W = nRTln(\frac{Pi}{Pf})[/tex]

    [tex]\delta{U} = Q - W[/tex]


    3. The attempt at a solution
    This is what i have done can anyone tell me what is wrong, because it is not the same as the book's answer.
    a)
    [tex]Vf = \frac{2 * 8.314*10^3 * 300}{1.20}[/tex]

    [tex]Vf = 4157000m^3[/tex]


    b)
    [tex]W = 2 * 8.314 * 10^3 * 300 * ln(\frac{1.20}{0.400})[/tex]
    W = 5480317.541

    c)
    [tex]\delta{U} = -5480317.541[/tex] meaning this is compression

    The correct answers for the following are:

    a) 0.0410 [tex]m^3[/tex]
    b) 5.48 kJ
    c) -5.48 kJ

    P.S
     
  2. jcsd
  3. May 11, 2010 #2

    Mapes

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    It's the classic error: not checking your units.
     
  4. May 11, 2010 #3
    could you tell me specifically which values have the wrong units?
     
  5. May 11, 2010 #4
    Check your units for "R"
     
  6. May 11, 2010 #5
    how do you convert mol into kmol??
     
  7. May 11, 2010 #6
    ok would 2mol be 0.200 kmol because 1 mol is 1000mols?
     
  8. May 11, 2010 #7
    yes because 1000 moles= 1 kmol
     
  9. May 12, 2010 #8
    after i changed the units i still didn't get the correct answer:

    This is what i changed:

    [tex]Vf = \frac{.200 * 8.314*10^3 * 300}{1.20}[/tex]
    [tex]Vf = 415700m^3[/tex]
     
  10. May 12, 2010 #9

    Andrew Mason

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    This is an enormous volume and obviously wrong!

    What are your units for pressure?

    What is the initial volume? Since T is unchanged, what is the final volume if the pressure is 3 x greater?

    AM
     
  11. May 12, 2010 #10
    It's .002 Kmole, change 1.2 atm to units of pressure in N/m^2 and 1 J = 1 N*m
     
  12. May 12, 2010 #11

    Andrew Mason

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    I am not sure why you are using Kmol. It is much easier to use moles and R = 8.314 (J/mol - K) and pressure as J/m^3 or N/m^2

    AM
     
  13. May 12, 2010 #12
    My reference uses R= 8.314 kJ/kmole-K same as 8.314 J/mole-K pick your poison but keep the units consistent
     
  14. May 15, 2010 #13
    i changed the units of atm to Pa and i got the correct answer.
     
  15. May 15, 2010 #14

    Andrew Mason

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    I trust that you used .002Kmol not .2Kmol. as you have shown.

    AM
     
  16. May 15, 2010 #15
    oh yeh oops typo.
     
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