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Polytropic process vs perfect gas eq

  1. Dec 25, 2013 #1
    The polytropic law states:

    (1) P1V1n = P2V2n

    The perfect gas equation states:

    PV = mRT --> P1V1/T1 = P2/V2/T2

    If T1 = T2 then
    (2) P1V1 = P2V2

    So, how can equation 1 and 2 both be true for the same gas? If the gas follows a polytropic process, where n ≠ 1, how can 2 be correct when there is no temperature change?
  2. jcsd
  3. Dec 25, 2013 #2
    Merry Christmas imsmooth,

    The polytropic law describes a process that a gas would follow from state 1 to state 2 during a compression or expansion. The value of n can be anything from 0 to infiniti for a set process.

    For your question if T1=T2, then this process is descibed by the polytropic expression with the value of n = 1. This means that during the process of compression or expansion PV = a constant = mRT ( since m, R, T are all fixed values for only this process where n = 1 ). Where the temperature does not change, the process is called an isothermal process and the state of the gas follows a constant temperature profile called isotherms.

    For any other value of n, there are other descriptions of the process, during which for an ideal gas, the equation PV=mRT will hold true, and as P or V are altered so will the value of T alter.

    This has a brief summary:

    all the best.
  4. Dec 25, 2013 #3
    I appreciate the answer, but I know this.

    On page 116 of Rayner Joel's Engineering Thermodynamics, both equations are used to derive another set of equations. One equation is setting n = 1; the other is just leaving it as n. This does not make sense as n should be the same for both equations for deriving the third.

    Even using your reference on page 8, PV = mRT. mR is a constant. Thus, P1V1 = T = P2V2. Here, n = 1. This is rearranged to have V1 = mRT/P2 and subsituted into PV^n

    How can n = 1 for PV = nRT, but it is just n for PV^n?
  5. Dec 25, 2013 #4
    For polytropic processes with n≠1, the temperature of the gas changes during the process. This does not mean that the ideal gas law doesn't also apply to these processes. In such cases, P1V1/T1 = P2V2/T2.
  6. Dec 25, 2013 #5
    That makes sense. Thanks.
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