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Interstellar Cloud collapse

  1. Aug 16, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    An interstellar cloud, made up of an ideal gas, collapses with its radius decreasing as $$R = 10^{13} \left(\frac{-t}{216}\right)^{2/3} \text{m}$$ with ##t## measured in years. The time ##t## is taken to be zero at zero radius so that ##t## is always negative.

    The cloud collapses isothermally at 10K until its radius reaches 1013m. It then becomes opaque so that from then on, the collapse takes place adiabatically and reversibly. How many years does it take for the temperature to rise by 800K measured from the time the cloud reaches a radius of 1013m.

    2. Relevant equations
    The one in question, First Law of Thermodynamics, Adiabatic expansion

    3. The attempt at a solution
    The question has not specified how they define 'radius', but I assumed a spherical cloud. Considering the end of the isothermal phase, start of the adiabatic phase, and the end of the adiabatic phase, the following holds: $$\frac{T_i}{P_i^{1-1/\gamma}} = \frac{T_f}{P_f^{1-1/\gamma}},$$ Reexpressing gives: $$T_i^{1/\gamma} V_i^{1-1/\gamma} = T_f^{1/\gamma} V_f^{1-1/\gamma},$$ where ##T_i = 10, V_i = 4/3 \pi (10^{13})^3, T_f = 810K, V_f = 4/3 \pi R^3## When I solve for t, I obtain the incorrect answer.

    Perhaps my assumption of the spherical shape of the cloud is incorrect?
     
  2. jcsd
  3. Aug 16, 2013 #2
    What answer did you obtain, and what is the correct answer?
     
  4. Aug 16, 2013 #3

    CAF123

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    Hi voko,
    I obtained approx t = 57 years which means t ≈ 21 years (57 - 36) in answering the given question. (36 yrs the amount of time for R = 1013 ). The correct result is t = 208 yrs.
     
  5. Aug 16, 2013 #4
    What do you use for ##gamma## and why?
     
  6. Aug 16, 2013 #5

    CAF123

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    Oops, ##\gamma## was given in the question to be 5/3. Sorry about that.
     
  7. Aug 16, 2013 #6

    D H

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    The correct answer is indeed 208.

    How did you get 36 years as the time it takes to collapse to R=1013, and how did you get 57 years for the time it takes to collapse to the stage where T=810K? Both of those numbers are wrong. They don't even have the correct sign.
     
  8. Aug 16, 2013 #7

    D H

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    In fact, looking at it as the time it takes to collapse to R=1013 is wrong. t=0 is the time at which R=0.

    Start with R=1013 meters. Given that ##R(t)=10^{13}\left(\frac {-t} {216} \right)^{2/3}\,\text{m}##, what is the value of t that yields R=1013 meters?
     
  9. Aug 17, 2013 #8

    CAF123

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    That would be t ≈ 10.9 yrs. I made an error previously.
     
  10. Aug 17, 2013 #9
    How is that possible if ##R(t)=10^{13}\left(\frac {-t} {216} \right)^{2/3} = 10^{13}##?
     
  11. Aug 17, 2013 #10

    CAF123

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    It's not , I made another clumsy error - rather t = -216 yrs.
     
  12. Aug 17, 2013 #11

    D H

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    Much better. Now how about T=810K? What do you get for the radius, and hence for time?
     
  13. Aug 17, 2013 #12
    Do you get ## t_f / t_i = (T_i / T_f)^{3/4} ##?
     
  14. Aug 17, 2013 #13
    Hmm. That ratio gives me the correct answer, and, frankly, I do not see why it is incorrect.
     
  15. Aug 17, 2013 #14

    D H

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    It is correct. I wouldn't solve it that way, though. Personal preference.
     
  16. Aug 17, 2013 #15

    CAF123

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    My simplified expression was: $$\left(\frac{T_i}{T_f}\right)^{1/\gamma} = \left(\frac{-t}{216}\right)^{4/5},$$ which gives t = -8 yrs. Relative to the state of R=0, 216 yrs ago the cloud was at R = 1013m and temperature T = 10K. 8 years ago, it was at a temperature of 810K and R ≈ 1012m. So it takes (216 - 8) = 208 yrs to reach the latter state from the former state.
    I obtained a different expression. How did you get yours? Edit: Actually, they are the same.

    I have some questions about the underlying physics of this process that I would like to ask later.

    Thanks.
     
  17. Aug 18, 2013 #16

    CAF123

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    Just want to check a few conceptual things:
    In the isothermal phase, dU = 0 so that dQ = -dW. So the heat that enters the cloud (which is regarded as the system) is then released as work done by the cloud.

    In the adiabatic phase, dU = dW, so the work done on the cloud by environment (in compressing it) is increasing its temperature.

    Is this fine?

    Also, why is it that ##\gamma = 5/3##?
     
  18. Aug 18, 2013 #17
    Is it the cloud that does the work, or is the work being performed on it, and by what?

    The heat capacity ratio is ## \gamma = 5/3 ## for monatomic ideal gas. This is related to the degrees of freedom of a gas molecule, which is 3 for monatomic gases.
     
  19. Aug 18, 2013 #18

    CAF123

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    Both the cloud and the environment do work because of NIII, yes? Wsystem on surr = - Wsurr on system. I suppose an external agent in the environment that could do work on the cloud would be varying pressure.
     
  20. Aug 18, 2013 #19
    That is certainly true. However, in thermodynamics we usually distinguish between work on and work by based on the sign of work.

    What external agent is at play here? The cloud is in the interstellar space, which we could consider perfectly void for this problem.
     
  21. Aug 18, 2013 #20

    CAF123

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    Hmm, I am not sure what a possible source could be then to explain the compression of the cloud. Or maybe you could argue that the compression is caused by internal effects and there is no need for any external agent to do the compressing.
     
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