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Adiabatic expansion

  1. May 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A flexible balloon contains 0.375 mol of hydrogen sulfide gas H2S. Initially the balloon of H2S has a volume of 6750 cm^3 and a temperature of 29.0 C. The H2S first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the H2S may be treated as an ideal gas with C_p = 34.60 J/mol*K and gamma = 4/3.

    What is the final volume in meters cubed?
  2. jcsd
  3. May 2, 2008 #2

    Andrew Mason

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    Show us your thoughts on this first.

    This a first Law problem.

    Can you draw a PV diagram of this two stage process? What is the Work done by the gas in the first stage? What is the heat flow into the gas in each stage? What is the total change in internal energy of the gas?

    If you answer those questions, you will be able to answer the question.

  4. Apr 25, 2009 #3
    as a continuation of this problem, I tried to solve it by using the equation:

    v2^(1/3) = [t1*v1^(1/3)]/t2

    Found t2 by:

    so... i got v2^(1/3) = {[299*(7050cm3 *10-9 m3)^1/3] / 598}
    =8.81 * 10-7

    The answer machine says wrong. What did I do wrong??? Thanks in advance!
  5. Apr 25, 2009 #4
    You've been given a gamma, why do you think this is?

    Recheck the ideal gas adiabatic relations. As this is
  6. Apr 25, 2009 #5

    Andrew Mason

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    I may have confused you. This does not require application of the first law. Sorry about that.

    Determine the temperature after the first expansion (PV=nRT). Then apply the adiabatic condition (PV^{\gamma} = nRTV^{\gamma-1} = K)to determine the volume when T is back to the original T.

  7. Apr 25, 2009 #6
    I'm confused. What have I done wrong with my equation.

    By the way, I'm a different person asking the same question. :)

    What is K, the original temperature? And how do you get P, through the p0=t1/v1 equation?
  8. Apr 25, 2009 #7

    Andrew Mason

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    K is a constant. The adiabatic condition is PV^{gamma} = constant.

    As far as your equation is concerned, I am not sure what you are doing. You haven't provided any explanation. You have to break the question into two parts. The first part is simple. You can easily find T after the isobaric expansion. To find the volume after the adiabatic expansion you apply the adiabatic condition.

  9. Apr 25, 2009 #8
    I found help elsewhere. Your way with the K constant is actually quite unnecessary. It's faster to use just V1 and gamma. Thanks for coming back and trying to help I guess.
  10. Apr 26, 2009 #9

    Andrew Mason

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    You appear to be applying the adiabatic condition but you seem to be unaware of it.

    The adiabatic condition is:


    If you substitute P = nRT/V this becomes:

    [/tex]nRTV^{\gamma-1} = K[/tex]

    This means that [/itex]TV^{\gamma-1}[/itex] does not change, so:

    [/tex]T_2V_2^{\gamma-1} = T_3V_3^{\gamma}[/tex]

    T1 = 302K
    T2 = 604K
    [/itex]V2 = 6750 cm^3 = 6.75 x 10^3 (x 10^{-6}) m^3 = 6.75 x 10^{-3} m^3[/itex]
    V3 = 13.50 x 10^{-3}m^3


    [/tex](604)(1.350 x 10^{-2})^{1/3} = (302)V_3^{1/3}[/tex]

    [/tex]V_3^{1/3} = .476 m^3[/tex]

    [/tex]V_3 = .78 m^3 = 7.8 x 10^5 cm^3[/tex]

    You started out ok but your numbers are wrong. Also the temperature of 29C is 273+29 = 302K.

    (Latex seems to be down so I have put a / in front of the tags so you can see them)
  11. Nov 20, 2009 #10
    How can V_3 = .78 m^3
    V_3^(1/3) = .476 m^3

    3sqrt(V_3) = .476 m^3

    V_3 = (.476 m^3)^3 = 0.1078 ?????

    I have trouble finding delta_U

    delta_U = nCv delta_T

    T = 300 K
    n = 0.35 mol
    Cv = 25.95 J/mol*K
    delta_T = 0 I not sure about this one
    The question states
    "Then it expands adiabatically until the temperature returns to its initial value"

    The balloon double it volume isobarically and then adiabatically until the temperature returns to its initial value

    Please help me
    Last edited: Nov 20, 2009
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