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Homework Help: AP Question involving Ideal Gas Law - Thermal Physics - Pressure

  1. Jan 22, 2009 #1
    [SOLVED]AP Question involving Ideal Gas Law - Thermal Physics - Pressure

    I am having some trouble with an old ap problem(PART C ONLY), see below:

    1. The problem statement, all variables and given/known data
    1996 AP PHYSICS B Free Response Question:
    The inside of the cylindrical can has a cross sectional area of 0.005 m^3 and length of 0.15 m. The can is filled with an ideal gas and covered with a loose cap. The gas is heated to 363 K and some is allowed to escape from the can so that the remaining gas reaches atmospheric pressure (1 x 10^5 Pa). The cap is now tightened, and the gas is cooled to 298 K.

    a. What is the pressure of the cooled gas?
    b. Determine the upward force exerted on the cap by the cooled gas inside the can.
    c. If the cap develops a leak, how many moles of air would enter the can as it reaches a final equilibrium at 298 K and atmospheric pressure? (assume that air is an ideal gas)

    2. Relevant equations

    3. The attempt at a solution

    a) i figured this one out..
    P2 = ~82093 Pascals

    b) got this one too..
    P = F/A
    F = PA
    F = ~410.5 Newtons

    c) cannot get this one..
    the volume of the gas that will enter into the leak is equal to the volume of the gas that escaped when the can was heated, correct? someone has said that this part will involve the volume of an ideal gas at STP, which is 22.4 Liters.
    I am not sure what to do..

    V2= 9.136 x 10^-4 m^3

    V2-V1 = V(escaped)
    V(escaped) = 1.92 x 10^-4 m^3

    n= ((1x10^5)(1.92x10^-4)) / ((8.31)(298))
    n= 0.0078 moles ??

    Any help, explanations, and/or answers will be greatly appreciated.
    Thank you!
    Last edited: Jan 23, 2009
  2. jcsd
  3. Jan 22, 2009 #2


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    For c) note that everything is an ideal gas.

    You calculated the underpressure inside the can. That would be 82093 Pa.

    But that is an underpressure.

    Consider then what volume it will occupy at 100,000 Pa. or 1 atm.

    Armed with that you know how much air must have entered to equalize the pressure to 100,000.

    Knowing the volume of air entered at 298K allows you to determine the number of moles that entered.
  4. Jan 22, 2009 #3
    hmm not sure i understand.

    do you mean:

    P1V1=P2V2 with P1 = 82093 and P2 = 1atm?
    V2 = ((82093)(0.005*0.15)) / (1x10^5)
    V2 = 6.157x10^-4 m^3

    then do the n = PV/RT with V = V2 = 6.157x10^-4 and P = 1atm and T =298 ?
    n= 0.0248 moles

    still doesn't seem right.. although i am sure i didn't follow your advice, since i didn't really get it
  5. Jan 22, 2009 #4


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    Not quite.

    At the under pressure how much might you have needed to reduce the volume in order to equalize the pressure back to 1 atm. You got that part OK.

    (1atm)*V = .82atm*(.005)(.15)
    That gives the 6.1*10-4

    Subtract that from the overall volume 7.5*10-4 - 6.1*10-4 = 1.4*10-4 = ΔV

    This change in volume is what needs to be supplied in outside air then right?
  6. Jan 22, 2009 #5
    Oh ok i got that part now(ΔV). thanks.
    so basically what thats asking is what change in volume will occur when the pressure increases from 82093 to 100,000, right?
    I forgot to subtract the overall volume, not sure why.. i did on my first attempt.

    Anyway, so now would I still use n = PV/RT ? since n = moles..
    T = 298 and P = 1atm?

    Note: So temperature doesn't change, since it has already been cooled to 298k, and the equilibrium point is also 298k, correct?
    Last edited: Jan 22, 2009
  7. Jan 22, 2009 #6


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    Yes. That was my approach.
  8. Jan 23, 2009 #7
    Thanks a lot for the help, LowlyPion! :D
    I've got the answer..

    .00542 moles of air enters the can

    Your method did work..
    V2 = ((82093)(0.005*0.15)) / (1x10^5)
    V2 = 6.157x10^-4 m^3

    ΔV = V(can) - V2
    ΔV = (0.005*0.15) - 6.157*10-4
    ΔV = 1.343x10^-4

    n=PV/RT = ((100000)(1.343x10^-4)) / ((8.31)(298))
    n = 0.00542 moles

    There are also other ways of doing the problem.
    A user(codygo) in the physics channel on freenode irc pointed some out for me..

    For those interested:

    Alternative solution:

    PV/nRT (initial) = PV/nRT (final)
    since T is the same (298) and so is V (0.005*0.15), they can be eliminated
    the formula now becomes


    to find ni use ni=PV/RT = ((82093)(0.005*.15)) / ((8.31)(298)) = 0.02486
    (Must use combination of either the cooled values, or the heated values - So 100,000Pa and 363K - OR 82093Pa and 298K -i used the cooled ones above - either combo will work..)

    nf = Pf*ni/Pi
    nf = ((100,000)(0.02486)) / (82093)
    nf = 0.030286

    Amount of air entered = nf-ni
    nf-ni = (0.030286)-(0.02486) = 0.00542 moles

    Thanks again,

    UPDATE: i just realized the mistake i was making in my first attempt at part c..i was using the wrong temperatures..i had T1 and T2 reversed!
    It should have been:

    V2= 6.157 x 10^-4 m^3

    Vi - V2 = V(escaped)
    V(escaped) = (0.005*0.15) - (6.157 x 10^-4)
    V(escaped) = 1.343x10^-4

    n= ((1x10^5)(1.343x10^-4)) / ((8.31)(298))
    n= 0.00542 moles


    So, like i said, there are many ways of doing this problem..
    Last edited: Jan 23, 2009
  9. Jan 23, 2009 #8


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    And think of what you learned along the way.

    Good luck.
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