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Ideal Gas Law

  1. May 2, 2006 #1
    The volume of a quantity of the ideal gas was kept constant in a experiment. The final temp. was 600 kelvins (K) and the final pressure was 300 netwons per square m (Meter). What was the original pressure if the original temperature was 1000 K?

    This is so mind twisting.

    I got it wrong three times
  2. jcsd
  3. May 2, 2006 #2


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    What was your approach on this question? Also please be sure to re-read this thread part #1.

    If you show us what you have done, we can help steer you in a successful direction.

    What kind of relationship (equations) do you know that relate the variables in your question?
    Last edited: May 2, 2006
  4. May 3, 2006 #3
    K, I read it.

    I was doing the problem wrong. But I think i understand it now.
    My equation is

    1. 1000 x 300 = 300,000
    2. 300,000 / 600 = 500


    This all works out right?
  5. May 3, 2006 #4


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    I don't understand.. What are you trying to accomplish with your calculations? Do you know the equation that shows the relationship between your variables (final temperature, pressure, & volume; original temperature, pressure & volume)?
  6. May 4, 2006 #5
    I'm trying to accomplish what the original pressure was if the original temperature was 1000 K.
    1000 [Original Temperature] x 300 [Final Pressure] = 300,000

    300,000 [New Solution] / 600 [Final Temperature] = 500 [Original Pressure]

    Answer = 500

    I asked my advisor if i was correct, and he said yes.

    But, I want to know if you think my steps were the correct way of getting the answer?
  7. May 4, 2006 #6


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    Yes your solution is okay..

    It helps to understand the relationship of your variables to solve any problem. That way you are not limited by step-by-step instructions shown in your text or given by the instructor.

    In your case one form of the general expression for The Ideal Gas Law, where i is initial and f is final, P-pressure, V-volume and T-temperature.

    [tex] \frac {P_i V_i}{T_i} = \frac {P_f V_f}{T_f} [/tex]

    ref (see Ideal Gas Law with Constraints)

    Since your volume is constant [itex] V_i = V_f [/itex]

    You are asked what is the original pressure. Solving for [itex]P_i[/itex]
    (using basic algebra)

    [tex] P_i = \frac {P_f T_i}{T_f} [/tex]
    Once you realize that, the solution becomes trivial (just plugging in the numbers and keeping your units consistent). Speaking of units, I hope remembered to give your solution in [tex] \frac {N}{m^2} [/tex]
    Last edited: May 4, 2006
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