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Adiabatic expansion in a thermodynamic system

  1. May 10, 2010 #1
    1. The problem statement, all variables and given/known data
    4 liters of an ideal diatomic gas are compressed in a cilinder. In a closed process, the following steps are taken :

    1) Initial state :
    [tex]p_1 = 1 atm = 1.013*10^5 N/m^2 [/tex]
    and :
    [tex]T_1 = 300 K[/tex]

    2) Isochoric proces resulting in :
    [tex]p_2 = 3.p_1[/tex]

    3) Adiabatic expansion resulting in:
    [tex]p_3 = p_1[/tex]

    4) Isobar compression resulting in
    [tex] V_4 = V_1 [/tex]

    Note : State 4 = State 1

    Questions :
    a) What is the volume of the gas at the end of the adiabatic proces ? (solution : 8.77*10^-3 )
    b) What is the temperature of just before the adiabatic expansion? (solution : 902 K)
    c) What is the work performed by the gas in this cycle ? (solution : 335.0 J)

    2. Relevant equations

    ideal gas :

    [tex] pV = nRT [/tex]
    [tex] \frac{p_xV_x}{T_x} = \frac{p_y V_y} {T_y} [/tex]
    [tex]\Delta U = n Cv \Delta T[/tex]

    ideal gas + adiabatic expansion :
    [tex]pV^{\lambda} = c[/tex]

    Note : lambda is Youngs module, c is a constant value.

    diatomic :
    [tex]C_v = 5/2 R[/tex]
    [tex]\lambda = 1.4[/tex]



    3. The attempt at a solution
    b) I'm starting with this one since it seemed easier :
    [tex]\frac{p_1 V_1}{T_1} = \frac{p_2 V_2} { T_2 }[/tex]
    [tex]\Rightarrow T_2 = \frac{p_2 V_2 T_1}{p_1 V_1}[/tex]
    [tex]\Rightarrow T_2 = 900 [/tex]
    My guess is that my teacher approached this through using the
    [tex]pV = nRT[/tex] equation twice, introducing rounding errors ?

    a) I've tried to approach this in several ways, but none seem to give me the correct solution... My current approach :

    Since we know it's an ideal gas that undergoes an adiabatic expansion, we could use the formula given above :
    [tex] p_2V_2^\lambda = constant = p_3V_3^\lambda [/tex]
    [tex] \Rightarrow V_3 = \log_\lambda ( \frac{p_2}{p_3} * V_2^\lambda)[/tex]
    [tex] \Rightarrow V_3 = \log_{1.4} ( 3 (4*10^{-3})^{1.4} ) [/tex]
    [tex] \Rightarrow V_3 = -19.70870781 [/tex]

    That is wrong in many ways, but I don't know which assumption I made is wrong ...

    c) Don't know yet, I'm guessing something like :

    [tex] \sum W = W_{1,2} + W_{2,3} + W_{3,1}[/tex]
    [tex] \sum W = 0 + W_{2,3} + p_3 (V_1 - V_3) [/tex]

    But I'll have to integrate over an unknown volume to get the value of [tex]W_{2,3}[/tex] ...
    If I know the value of [tex]V_3[/tex] I think I can calculate the value of [tex]T_3[/tex] using [tex]p_3V_3=nRT_3[/tex]. Once I know [tex]T_3[/tex] I can use [tex]\Delta U = n Cv \Delta T_{2,3} = - W_{2,3}[/tex].
     
    Last edited: May 11, 2010
  2. jcsd
  3. May 10, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    b) the final temperature has to be 900 K.

    a) Use the adiabatic condition [itex]PV^\gamma = K[/itex] to give:

    [tex]\left(\frac{V_f}{V_i}\right)^\gamma = \frac{P_i}{P_f}[/tex]

    c) The work done from 1-2 and 3-4 is easy. To determine the work done from 2-3 use the first law: dQ = dU + dW to determine the work done (what is Q for this adiabatic expansion?). Hint: you just have to know the change in temperature - use:

    [itex]T_2V_2^{\gamma -1} = T_3V_3^{\gamma -1}[/itex] to find the temperature at 3.)

    AM
     
  4. May 11, 2010 #3
    I am using that formula for (a) in my attempted solution, but te result is not correct. The values I use are:

    [tex]V_f = unknown[/tex]
    [tex]V_i = 4*10^{-3}[/tex]
    [tex]P_i = 3 * P_f[/tex]
    [tex]P_f = 1.013 * 10^{5}[/tex]
    [tex]\lambda = 1.40[/tex]

    Am i using wrong values ?

    Attempt 2 :

    [tex] V_f = V_i * \log_{1.4}(p_i / p_f)[/tex]
    [tex] \Rightarrow V_f = 4*10^{-3} * \log_{1.4}(3)[/tex]
    [tex] \Rightarrow V_f = 13 * 10^{-3} [/tex]
     
    Last edited: May 11, 2010
  5. May 11, 2010 #4

    Andrew Mason

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    Science Advisor
    Homework Helper

    [tex]x^\gamma = (e^{\ln{x})^\gamma} = e^{\gamma\ln{x}[/tex]


    AM
     
  6. May 11, 2010 #5

    Andrew Mason

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    Science Advisor
    Homework Helper

    As a follow-up to my last post, in case you found it too cryptic: your figures are correct (although you should state the units, particularly in your answer). The problem is with algebra.

    AM
     
  7. May 11, 2010 #6
    Thank you ! Using this and your previous posts I was able to find the answer for both (a) and (c) now. (how do i mark this thread as solved ?)
     
    Last edited: May 11, 2010
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