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Need help finding temperature-volume relation in solving Adiabatic problem

  1. Apr 24, 2012 #1
    A monatomic ideal gas (γ = 5/3) is contained within a perfectly insulated cylinder that is fitted with a movable piston. The initial pressure of the gas is 1.31 × 105 Pa. The piston is pushed so as to compress the gas, with the result that the Kelvin temperature doubles. What is the final pressure of the gas?

    I know the equation I will need to solve this with is P_i*(V_i)^γ = P_f*(V_f)^γ
    I'm stuck at figuring out how to plug that T_f = 2*T_i to find the volume. Any pointers?
     
  2. jcsd
  3. Apr 25, 2012 #2

    Andrew Mason

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    Substitute nRT/V for P into the equation.

    AM
     
  4. Apr 25, 2012 #3
    In that case I'm getting an equation that reduces to (V_f/V_i)^γ = V_i/(2*V_f).

    I don't know the initial or final volume or how to solve from here.
     
  5. Apr 26, 2012 #4

    Andrew Mason

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    Substituting nRT/V for P gives:

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

    So if:

    [tex]\frac{T_f}{T_i} = 2[/tex]

    then what is

    [tex]\left(\frac{V_i}{V_f}\right)^{(\gamma-1)} = ?[/tex]

    AM
     
  6. Apr 26, 2012 #5
    I don't know. I would say either 2 or 1/2, but that'd just be me guessing. Or I can say it's (V_i/V_f)^(2/3) - but that doesn't help me either. I don't know how to make sense of this because the problem is none of constant volume, temperature or pressure.
     
  7. Apr 26, 2012 #6

    Andrew Mason

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    You are making it harder than it is. It is just algebra. I gave you the equation!

    Since:

    [tex]\frac{T_f}{T_i} = \left(\frac{V_i}{V_f}\right)^{(\gamma-1)}[/tex]

    and since:

    [tex]\frac{T_f}{T_i} = 2[/tex]

    then

    [tex]\left(\frac{V_i}{V_f}\right)^{(\gamma-1)} = 2[/tex]

    Solve that for Vi/Vf. How do you determine Pf/Pi from Tf/Ti and Vi/Vf?

    AM
     
  8. Apr 27, 2012 #7
    Thank you so far. I took (V_i/V_f)^(2/3) = 2 ==> (V_i/V_f) = 2*SQRT(2).
    I then attempted to apply both Charles and Gay-Lussac's law to come up with:
    P_f/P_i = 2*SQRT(2) ==> P_f = P_i*2*SQRT(2), but this is not correct. I'm sure I'm missing something obvious and simple, but I'm oblivious to it yet.
     
  9. Apr 27, 2012 #8

    Andrew Mason

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    You may need to brush up a bit on algebra involving logarithms.

    [tex]\left(\frac{V_i}{V_f}\right)^{\gamma-1} = 2[/tex]

    Take the log of both sides:

    [tex](\gamma-1)\ln{\left(\frac{V_i}{V_f}\right)} = \ln{2}[/tex]

    Since [itex]\gamma = 5/3[/itex]

    [tex]\ln{\left(\frac{V_i}{V_f}\right)} = 3\ln{2}/2[/tex]

    Take the antilog of both sides:

    [tex]\left(\frac{V_i}{V_f}\right) = e^{1.04} = 2.83[/tex]

    Use PV = nRT to determine the final pressure.

    AM

    (P.S. One could also express the adiabatic condition in terms of T and P and work it out - just avoids the last step)
     
  10. Apr 28, 2012 #9
    Thank you. I knew I brain-farted something obvious. At this point I found it easier to plug the V_i/V_f value back into the P_i*(V_i)^γ = P_f*(V_f)^γ equation and solve for:
    P_f = P_i*(V_i/V_f)^γ.

    Thank you again
     
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