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Molecular Specific Heat of an Ideal Gas: Computations

  1. Aug 3, 2017 #1
    1. The problem statement, all variables and given/known data

    A cylinder with a heavy ram/piston contains air at T = 300 K. Pi = 2.00 * 105 Pa, Vi = 0.350 m3, Mr = 28.9 g/mol & Cv = 5R/2

    (a) What's the Molecular Specific Heat of an Ideal Gas, with a constant volume, computed at J/KgC ? (Cv)
    (b) What's the mass of the air inside the cylinder?
    (c) If the piston stands still, how much energy is required for Tf = 700 K?
    (d) If the piston is free to move, how much energy is required for Tf = 700 K?

    2. Relevant equations

    ΔE = nCvΔT
    PV = nRT
    Cv = R + Cp

    3. The attempt at a solution

    (a) No clue what to do here.

    (b) PiVi = nRTi <=> .... <=> n = 28.065 mol = m/Mr = m/28.9(g/mol) <=> ... <=> m = 0.811 kg

    (c) ΔE = nCvΔT = 28.065 mol * (5/2 * 8,314 J/molK) * (700 - 300) K <=> ΔE = 233 kJ

    (d) No clue what to do here either.

    Any help is appreciated!
     
  2. jcsd
  3. Aug 3, 2017 #2
    In part (a) if you know the molar heat capacity (J/mole-K) and you know the molecular weight (gm/mole), what is the heat capacity in J/kg-K?
     
  4. Aug 3, 2017 #3
    Kind of a correction here, I made a mistake. (a) is about the Molar Heat Capacity in J/kgC, not the Molecular Specific Heat. So in otherwords c, not Cv or Cp. I'm putting this in bold since I can't edit the OP (kinda messed up here, sorry). The book doesn't explain the differences that well.

    Anyway, I couldn't find anything to connect Cv & Cp in my book, but I looked around and found that C = M * c.

    So, now we know that the Molecular Specific Heat of an Ideal Gas and the Molar Heat Capacity are connected through this formula: C = M * c

    In this case we have a constant volume and M = 28.9 g/mol. We are given that Cv = 5/2*R = 2,5 * 8,314 J/molK

    20,785 J/molK = 0,0289 kg/mol * c <=> c = 719.7 J/kgK. So now I need to somehow turn the K into a C.
     
  5. Aug 3, 2017 #4
    The molar heat capacity refers to J/molC, not J/kgC. I, who have had lots of experience with thermodynamics, am having trouble interpreting the terminology you are using.

    The equation C = Mc is correct, but I'm surprised you had to refer to your book for this. Have you learned to work with cancellation of units?

    You are aware that a change of 1 K is the same as a change of 1 C?
     
  6. Aug 4, 2017 #5
    Yeah, I made a mistake in the OP.

    Well, english isn't my native, which means that my books are in another tongue. So when I want to ask something I have to search online for the correct words in order to translate the exercises, and because I do not know the proper terminology I get a bit lost sometimes.

    The theory wasn't that clear on the distinction between the two, so I'm still a bit iffy on it.

    You mean stuff like 1 kg = 103 g & 1 atm = 101325 Pa, right?

    Yes, that's in my book (in the "end notes" section). Does that mean that I can simply "replace" K with C in my c = 719.7 J/kgK result?
     
  7. Aug 4, 2017 #6
    Yes.
     
  8. Aug 5, 2017 #7
    Okay, gotcha.

    As for (d), this is what I came up with:

    Q = mcΔT = nMrcΔT = ... = 233 kJ

    P = constant, so: PfVf = nRTf <=> Vf = 8.065 mol * 8,314 J/molK * 700 K / 2*105 Pa = 0.820 m3

    W = -P∫VfVidV = ... = - 94000 J

    Now, the Work is negative, which means that since the gas' volume expands, it's energy from the gas that is applied to the piston. Either way, (d) asks for the total energy that is required for T to go from 300 K to 700 K, so I assume it means all the energies, regardless of the "direction". So:

    ΔΕ = Q + |W| = 327 kJ

    This is the book's answer, but I'm not sure whethermy method is correct. The exercise just says that "the initial data is valid, and the piston can move freely".
     
  9. Aug 5, 2017 #8
    Which version of the first law are you using, $$\Delta E = Q - W$$where W is the work done by the gas on the piston or $$\Delta E=Q+W$$where W is the work done by the piston on the gas?

    Also, in the last part (d), you are not trying to determine ##\Delta E##. You are trying to determine Q.
     
  10. Aug 5, 2017 #9
    I'm going with the second. I know about the two versions, but it's the one I've been taught since HS. Anyway, with my theory we have: W > 0 means that it's work done on the gas, W < 0 means it's work done by the gas.

    Oh yeah, no wonder I had to do "mental gymnastics" to get the correct result. My bad. It should be:

    ΔE = nCvΔT = ... = 233 kJ
    W = -94000 J (as shown in my above post-the n should be 28.065 up there, my keyboard's a bit busted)

    So: ΔE = Q + W <=> 233 kJ = Q - 94 kJ <=> Q = 327 kJ

    PS: Thanks a ton for the help!
     
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