1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Adiabatic Expansion Pressure Temperature Relation

  1. Jan 26, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem is in the context of convection in the troposphere

    Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation

    [tex]\frac{dT}{dP} = \frac{2}{f+2} \frac{T}{P}[/tex]
    2. Relevant equations
    Ideal gas law PV = nRT
    adiabatic relations ##VT^{f/2}= constant##, ##V^{\gamma}P = constant##
    where ##\gamma = \frac{f+2}{f}##

    3. The attempt at a solution

    The first thing I tried was taking the derivative of each of the adiabatic relations w.r.t. the T or P in each of them. Since they are equal to constants, their derivatives are equal to zero and thus each other. I did not take a partial derivative because V is not constant. Temperature, pressure and volume of the air mass are changing as it moves through the atmosphere.

    I am rusty with my derivatives right now, and was plagued with simple errors as I went through taking them and doing the arithmetic. Like right now I see a mistake that I made that I didn't notice I had and yeah, I gave up on this route because it was too messy.

    Then I tried the idea of dividing the one adiabatic relation by the other, and since the ratio of two constants is a constant, I could then take the derivative of what I got and work from there.

    $$PV^\gamma = A$$
    $$VT^{f/2} = B$$
    $$\frac{PV^\gamma}{VT^{f/2}} = A/B$$
    $$PV^{\gamma - 1}T^{-f/2} = constant$$

    From there I uh wrote the "implicit derivative" of each. I'm not entirely sure if this is legal, or if I did it right but

    $$V^{\gamma - 1}T^{-f/2}dP + PT^{-f/2}V^{\gamma - 2}(\gamma - 2)dV + PV^{\gamma - 1}T^{-f/2 - 1}(-f/2 - 1)dT = 0$$

    which is admittedly still messy but whatever. I moved all the Vs to the one side of the equation to leave the Ps and Ts on the other

    $$\frac{dP}{P} = (f/2 + 1) \frac{dT}{T} - (\gamma - 2)\frac{dV}{V}$$

    From an example in the book I was shown that ##\frac{f}{2}\frac{dT}{T} = -\frac{dV}{V}##. I'm assuming this is true for my problem because the example was about adiabatic compression and so P, V and T were dependent on each other like in the problem I'm doing.

    And so $$\frac{dP}{P} = \frac{dT}{T}[(\gamma -2)(f/2) + (f/2 + 1)]$$

    so now I'm gonna simplify and turn the ##\gamma## into ##(f+2)/f##


    and holy crap if I didn't redo this a million times but I keep ending up with

    $$\frac{dP}{P} = \frac{f+4}{2}\frac{dT}{T}$$

    Which is exactly the answer (with some maneuvering) except there's a FOUR where a TWO should be!

    Ok I found that I was writing a 1 where the 2 should be in ##\gamma - 2##... which makes the above turn out to be $$\frac{dP}{P} = 2\frac{dT}{T}$$

    I can only pore over my handiwork so many times looking for tiny errors until my mind starts to turn off, but I can't help but think I did something stupid. Either that or perhaps that substitution is fallacious? I tried changing the sign of that part: ##\frac{f}{2}\frac{dT}{T} = \frac{dV}{V}## (maybe because the work is in the opposite direction for expansion) but it did not help either.

    And ok at this point I'm just confusing myself with all my tiny changes and am gonna go take a break. I am very appreciative of any help, thank you.
     
  2. jcsd
  3. Jan 26, 2016 #2
    Start with PVγ=constant, and substitute V from the ideal gas law into the equation. What do you get?
     
  4. Jan 26, 2016 #3
    I get $$\frac{dT}{dP} = \frac{\gamma}{\gamma - 1} \frac{T}{P^{\gamma + 1}}$$

    Which evaluates to $$\frac{dT}{dP} = \frac{f+2}{2} \frac{T}{P^{\frac{2f+2}{f}}}$$ and I really don't know how to proceed from there.
     
  5. Jan 26, 2016 #4
    Oh wow wait I think I got it, I took the natural log of the initial substitution and I think that's exactly what I needed to do. I can nicely separate P and T from each other using log properties, and no weird exponents.

    I could take the log, of course, because a ln(constant) is still some constant.
     
  6. Jan 26, 2016 #5
    That's not what I get. I get:
    $$T=CP^{\frac{(\gamma-1)}{\gamma}}$$
    where C is a constant.
     
  7. Jan 26, 2016 #6
    Thanks, that did it!
     
  8. Jul 25, 2017 #7
    For future people.

    Use
    \begin{equation}
    VT^{ \frac {f} {2}} = C
    \end{equation}
    and
    \begin{equation}
    V= \frac {NKT} {P}
    \end{equation}
    Enter Equation (2) into Equation (1)
    \begin{equation}
    ( \frac {NKT} {P})T^{\frac {f} {2}}=C
    \end{equation}
    Combine T's and rearrange to be a function of T
    \begin{equation}
    T^ {\frac {f+2} {2}}= \frac {CP} {NK}
    \end{equation}
    Recognize that
    \begin{equation}
    \frac {C} {NK} = constant
    \end{equation}

    Enter that information
    \begin{equation}
    T^{ \frac {f+2}{2} }= CP
    \end{equation}
    Multiply by exponential \begin{equation} ( \frac {2} {f+2} ) \end{equation} to isolate a single T. (C raised to any power is still a constant).
    \begin{equation}
    T=CP^{ \frac {2} {f+2}}
    \end{equation}
    Now take the derivative of T with respect to P.
    \begin{equation}
    \frac {dT} {dP} = \frac {2C} {f+2} P^{\frac {2} {f+2} -1}
    \end{equation}
    separate P-1 from P2/f+2
    \begin{equation}
    \frac {dT} {dP} = \frac {2C} {f+2} P^{\frac {2} {f+2}}P^{-1}
    \end{equation}
    Rearrange (
    \begin{equation}
    \frac {T} {C} = P^{\frac {2} {f+2}}
    \end{equation}
    Input above equation into the differential for matching P.
    \begin{equation}
    \frac {dT} {dP} = \frac {2C} {f+2} (\frac {T} {C})P^{-1}
    \end{equation}
    Constants are equal so they can cancel out. Leaving you with
    \begin{equation}
    \frac {dT} {dP} = \frac {2} {f+2} \frac {T} {P}
    \end{equation}
     
  9. Jul 25, 2017 #8
    You realize that this thread is over a year old, right? Do you feel that the answer we previously obtained was incorrect?
     
  10. Jul 25, 2017 #9
    In this problem you aren't searching for an answer. You are trying to figure out the necessary steps for derivation. So I was just putting this here for future people searching for it.
     
  11. Jul 25, 2017 #10
    Well, OK. Since the OP has not been seen in over a year, and there have been no other responses until yours, I think this thread has pretty much run its course. I am hereby closing it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Adiabatic Expansion Pressure Temperature Relation
  1. Adiabatic expansion (Replies: 1)

Loading...