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Ideal gas in a cylinder closed by a piston connected to a spring

  1. Nov 24, 2006 #1
    A cylinder is closed by a piston connected to a spring of constant 2.00 X 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00L of gas at a pressure of 1.00 atm and a temperature of 20.0ºC. A) If the piston has a cross-sectional area of 0.0100 m^2 and neglible mass, how high will it rise when the temperature is raised to 250º? b) What is the pressure of the gas at 250º?

    This whole question has me stumped, at first I thought it would be simple ideal gas law question, but then I thought about the force of the spring back on the gas and how that would affect pressure. The way I have it so far is net force on the piston in the direction of the spring if the cylinder is vertical with the end sealed by the piston at the top (y axis) is F = -Fs + [P(gas) X A(piston)] at 20ºC system is in equilibrium so kh = P(gas) X A(piston) ---> h = [P(gas) X A(piston)] /k ---> h=(101300Pa X 0.0100m^2)/(2.00 X 10^3 N/m) =0.507m Does this seem right? I don't even think i'm on the right track here.... any help would be great thanks!
     
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  3. Nov 24, 2006 #2

    daniel_i_l

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    In both the initial state and the final state the piston is in equilibrium. but in the beginning there's the same amount of pressure on the top and on the bottom of the piston (1) and so the spring is relaxed - the force of the spring is 0 (h=0). but when the gas is heated then there's more pressure inside so in order to be in equilibrium the spring has to push harder, this means that the spring has to move up so that it pushes more (f=k*x).
    and don't forget that as the piston goes up the volume gets bigger and the pressure goes down so the pressure is also a function of h.
     
  4. Nov 24, 2006 #3
    Ok thanks, that was another way I looked at doing it but I made some serious mistakes when I laid out my eqs... what I have this time is P=F/A and since spring is in equilibrium with force exerted by the gas... P=kh/A
    Volume = A X h so P1V1/T1 = P2V2/T2 ---> [(101300Pa)(5.00m^3)] / 293K = [(kh/A) X (A X h)] / 543K ---> 1730 = k X h^2 h=0.930m I think this works out or is V2 = 5.00 + (AXh)?? then for the pressure, its simply hk/A = 186000 Pa
     
  5. Nov 24, 2006 #4

    daniel_i_l

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    V2 = 5.00 + (AXh) because in the initial condition h=0 and V=5.
    and also, in the final state P=kh/A + 1 because the pressure in the container also has to push against the pressure outside the container - you know this cause in the initial state there was no force on the spring but there was pressure of 1atm in the cylinder.
     
  6. Nov 24, 2006 #5
    Ok awesome thanks so much!
     
  7. Feb 22, 2008 #6
    please let me see the answer of your question exactly

    hi, it seems we have got the same problem.well, i am new in this area so,i am afraid of that i couldn't understand.could you let me see the answer step by step...such as like...

    a)
    b)


    thanks....
     
  8. Sep 4, 2008 #7
    Thanks for the jump start-- 5 L is not equal to 5 m^3 though. 1000 Liters is 1 m^3
     
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