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Molar specific heat of an ideal gas

  1. Dec 10, 2007 #1
    A vertical cylinder with a heavy piston contains air at a temperature of 300 K. The initial pressure is 250 kPa and the initial volume is 0.350 m^3. Take the molar mass of air as 28.9 g/mol and assume that CV = 5R / 2.



    (b) Calculate the mass of the air in the cylinder.
    I know I have to use the volume that they provided me and the molar mass of air to somehow calculate the mass of the air in cylinder. Could someone please give me a hint in the right direction?
     
    Last edited: Dec 10, 2007
  2. jcsd
  3. Dec 10, 2007 #2

    dynamicsolo

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    Do you have sufficient information to find the number of moles, n, in the cylinder?
     
  4. Dec 10, 2007 #3
    I am not sure I do, maybe I could figure out how many moles are in 1 liter? Otherwise, I suppose I could calculate the specific heat into kj/kg*K to get the kg units, but I am not sure that would help all that much.
     
    Last edited: Dec 10, 2007
  5. Dec 10, 2007 #4

    dynamicsolo

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    Won't n = PV/RT?

    What are the other parts of this problem?
     
  6. Dec 10, 2007 #5
    Thanks alot for the help, i figured out the answer to part b of this question along with parts (a) and (c). I am stuck on part (d) now though...

    The questions and answers to the respective problems are listed below:


    (a) Find the specific heat of air at constant volume in units of kJ/kg·K.

    .71 kg/ j*K

    (b) Calculate the mass of the air in the cylinder.

    (c) Suppose the piston is held fixed. Find the energy input required to raise the temperature of the air to 700 K.

    291.7 J

    (d) Assume again the conditions of the initial state and that the heavy piston is free to move. Find the energy input required to raise the temperature to 700 K.

    I would use n* deltaT * 5R/2 to calculate this, but I am not sure how to calculate how much more energy I would need if the "heavy" piston is free to move.
     
  7. Dec 10, 2007 #6

    dynamicsolo

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    That's what you'd use for part (c) also. Since it's a constant volume process, Q = delta-U = (5/2) · nR · delta T. What did you find for n?

    For part (d), letting the piston move freely means that the force applied by the gas pressure balances the weight of the piston. In other words, this is a constant-pressure process. While delta-U is again (5/2) · nR · delta T, you now need to find the work done in this process with the volume going to its new value (which must also be found) in order to determine the heat input.
     
  8. Dec 11, 2007 #7
    n came out to be 35.098 mol which i multiplied by .0289 kg/mol to get the mass in kg. How would i calculate the new volume though?
     
  9. Dec 11, 2007 #8

    dynamicsolo

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    You know the pressure remains at 250 kPa and you have the number of moles and the new temperature. So V' = nRT'/P .

    BTW, while I agree with your result for n, I rather think that the answer to part (c) is somewhat more than 292 J...

    [EDIT: This wouldn't happen to be Problem 17.33 in Serway & Jewett, would it? I just spotted it in the book when I met some of my students today...]
     
    Last edited: Dec 11, 2007
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