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Homework Help: Clyinder problem

  1. Jan 24, 2008 #1
    Hello I am having trouble with this problem I was wondering if anyone could help me out here goes,

    During some actual expansion and compression process in piston-cylinder devices, the gases are observed to satisfy the relationship PV=C, where n and C are constants. Calculate the work done in KJ when gas expands from 150kPa and 0.03m to a final volume of 0.2m for the case of n = 1.3.

    Ok I understand that P1 = 150kpa and V1 = 0.03m and V2 = 0.2m when I try to solve for P2 using P1V1=P2V2 I get 22.5kPa but it doesn't make sense because its expanding and I'm not sure how to incorporate n into the problem, also I am confused as to finding the work done??? can anyone hint me in the right direction? thanks.
  2. jcsd
  3. Jan 24, 2008 #2


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    Does the problem indicate if it is an adiabatic process?

    Did you mean to writne [tex]PV^n= C [/tex] instead?

  4. Jan 25, 2008 #3
    no it does not state that it is an adiabatic, I considered that formula but I'm not sure as to how I would use it because there is a beginning volume and an ending volume. I'm still pretty confused it has been a while since Thermo I this is just a review for Thermo II. I appreciate the help thanks!
  5. Jan 25, 2008 #4


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    You really need to know the process by which the final state of the gas is obtained, however since you don't know that, lets assume it is isothermal.

    The general equation for work done by an ideal gas is...

    [tex] W = \int_{V_i}^{V_f} PdV [/tex]

    The equation of state for an ideal gas is...

    [tex]PV = nRT[/tex], which once rearranged gives [tex]P = \frac{nRT}{V} [/tex].

    So the work becomes...

    [tex] W = \int_{V_i}^{V_f} \frac{nRT}{V}dV [/tex]

    Since this is isothermal [tex]nRT[/tex] is constant, which then gives...

    [tex] W = nRT \int_{V_i}^{V_f} \frac{dV}{V} [/tex]

    After integrating and using a log identity you have...

    [tex] W = nRT \cdot \ln \frac{V_f}{V_i} [/tex]

    If it is not isothermal then of course the work will be different. Let me know if it is not isothermal.

    Hope that helps...

  6. Jan 25, 2008 #5
    it doesnt state that is isothermal ho it is written above is exactly how the problem is written, but I will attempt it that way thank you.
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