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Homework Help: Differential equation for concentration of gas in the atmosphere

  1. Apr 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi everyone, I'm currently studying an online course on climate science and am a bit overwhelmed by the calculus. I have studied calculus to second year of college but that was a while ago and I'm very rusty.

    A few weeks ago I was a question to find the how long it would take for the concentration of a gas in the atmosphere to reduce by a given percentage. The answer was given, though steps were missing and I can't duplicate the result.

    E = emission rate [Tg/yr]
    C = the concentration of gas [Tg]
    [itex]τ[/itex] = lifetime of gas in atmospher [yr]

    2. Relevant equations

    (eq1)[itex]dC/dt = E - C/τ[/itex], where E = 0 (ie. emissions are stopped)

    the solution to this differential equation is given as;

    (eq2)[itex]C = C_0 exp(-t/τ)[/itex], and

    (eq3)[itex]t = -τ log(C/C_0)[/itex]

    3. The attempt at a solution

    [itex]dC/dt = - C/τ[/itex]

    [itex]∫ -τ/C dC = ∫ dt [/itex]

    [itex]-τ∫ 1/C dC = ∫ dt [/itex]

    [itex]-τ log(C) = ∫ dt [/itex]

    I can't figure out how to get eq2 from eq1. If anyone can fill me in on the intermediate steps from eq1 to eq2 to eq3, I would be very grateful.

  2. jcsd
  3. Apr 20, 2014 #2
    ##\int t## is ##t## plus some constant, which we write as ##- \log C_0## for convenience. That gives you eq. 3. Transition between eq. 2 and eq. 3 is not even calculus, it involves simple algebra and the relationship between ##\exp## and ##\log##.
  4. Apr 20, 2014 #3
    Thanks voko

    The main part I'm struggling to understand is the steps between eq1 and eq2. Looks like an integration on both sides, but I can't reproduce the steps to get there.
  5. Apr 20, 2014 #4
    From eq1, you go to eq3, like you (almost) did, then you go to eq2 as I said in #2.
  6. Apr 20, 2014 #5
    ok, thank you
  7. Apr 20, 2014 #6


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    Okay, with E= 0, this is [itex]dC/dt= -C/\tau[/itex] and, separating,
    [itex]dC/C= -dt/\tau[/itex]. Integrate both sides to get
    [itex]ln(C)= -t/\tau+ K[/itex] where K is the constant of integration.

    Taking the exponential of both sides, [itex]e^{ln(C)}= C= e^{-t/\tau+ K}= e^Ke^{-t/tau}[/itex]. Let [itex]C_0= e^K[/itex] and you have the solution below.

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