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Question about deriving Gay-Lussac's Law

  1. Feb 16, 2009 #1
    There's a flaw in my logic, and I cannot find it.

    Ok, all you are allowed to use is Charles' Law (V/T = K) and Boyle's Law (PV = K) and you have to derive an equation relating pressure and temperature.

    V = kT
    V = k/P
    kT = k/P
    PT = 1
    Therefore P1T1 = P2T2

    But, this is the exact opposite of what Gay-Lussac's Law states :frown:

  2. jcsd
  3. Feb 17, 2009 #2


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    Note that V/T = K is only valid for constant pressure, and P V = C is only valid for constant temperature (I think K and C are in general different constants, so let me use different symbols for them).

    If you want to vary the pressure, for example, you actually need to write V / T = K(P).
    For the Gay-Lussac Law, you need a constant volume (note that the constant quantity is always the one "missing" from the equation, of course that's just because we're sweeping it into the constant on the right hand side).

    Though all three (Charles, Boyle and Gay-Lussac) can be easily derived from the ideal gas law,
    [tex]\frac{P V}{T} = \text{constant}[/tex]
    I don't immediately see how you can solve your question... I keep getting identities like
    [tex]\frac{P}{T} = \frac{ K(P) C(T) }{ V^2} [/tex]
    where K(P) and C(T) are only constants for constant pressure and temperature, respectively.
  4. Feb 17, 2009 #3


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    Compuchip is correct, the K's in the two equations are different, and each K can contain information about whatever quantity is being held constant.

    I'm wondering if it's helpful to rewrite the equations as
    V = K1T
    V = K2/P
    From there, can the OP come up with an equation with all three variables (and perhaps a different constant K3), eg.
    V = K3 x ???
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