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Ideal Gas Problem

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A certain amount of gas at 298.15 K and at a pressure of 0.800 atm is contained in a glass vessel. suppose that the vessel can withstand a pressure of 2.00 atm. How high can you raise the termperature of the gas without bursting the vessel.

    2. Relevant equations

    [tex]PV=nRT[/tex]

    Since the amount of gas remains constant:

    [tex]R=\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

    3. The attempt at a solution

    Obviously I don't have enough information to fill this in directly. Ideal gas problems are usually very simple, but this one stumped me. I don't see how I can solve for [tex]T_2[/tex]
     
  2. jcsd
  3. Jan 7, 2010 #2

    CompuChip

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    Homework Helper

    You are almost there.
    What can you say about the relation between V1 and V2?
     
  4. Jan 7, 2010 #3
    Thanks for your reply. I noticed that they must be equal right? I kind of forgot the basic definition of a gas, lol. The volume must be constant if the gas has filled the container.
     
  5. Jan 7, 2010 #4

    Borek

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    Staff: Mentor

    Yep, it is just pV=const.

    Beware:

    [tex]R=\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

    This is incorrect. It would be correct for exactly 1 mole of gas.

    --
    methods
     
  6. Jan 7, 2010 #5
  7. Jan 7, 2010 #6

    Borek

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    Staff: Mentor

    No, n is number of moles. It is usually constant throughout the problem if the number of moles of gas doesn't change, but it is not constant as R is.

    Ideal gas equation is

    PV=nRT

    That means

    [tex]R = \frac{PV}{nT}[/tex]

    or

    [tex]nR = \frac{PV}{T}[/tex]

    --
    methods
     
  8. Jan 7, 2010 #7
    Yeah, I know that n isn't a natural constant. I meant that it's constant in this particular problem since it doesn't change. The number of moles before and after are equal.

    [tex]\frac{V_1P_1}{n_1T_1}=\frac{V_2P_2}{n_2T_2}[/tex]

    [tex]n_1=n_2=n[/tex]

    [tex]\frac{V_1P_1}{nT_1}=\frac{V_2P_2}{nT_2}[/tex]

    [tex]\frac{V_1P_1}{T_1}=\frac{V_2P_2}{T_2}[/tex]
     
  9. Jan 7, 2010 #8

    Borek

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    Staff: Mentor

    OK, but still

    [tex]R=\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

    is incorrect in general (holds only for one mole of gas). It should be

    [tex]nR=\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

    --
     
  10. Jan 7, 2010 #9

    CompuChip

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    So in this case, you could continue by writing V1 = V2 = V, so
    [tex]\frac{n R}{V} = \frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
    where you can call the left hand side R', or C, or k (not kB :) ), since it is a constant in the current problem.

    You could calculate the constant if you knew n and V and looked up R. However, the equality in that formula which you are interested in, is of course the second one:
    [tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
     
  11. Jan 7, 2010 #10

    Oh, ok, I see what you're saying. I wasn't thinking when I wrote [tex]R=\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]. I agree that this is incorrect. I thought that you were claiming that [tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex] was incorrect.

    Thanks
     
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