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Ideal Gas - Thermal Equilibrium Problem

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow gas to leak into the other side. Initially, one side of the piston contains 1 cubic meter of Nitrogen (N2) at 500 kPa and 80 degrees centigrade while the other side contains 1 cubic meter of Helium (He) gas at 500 kPa and 25 degrees centigrade. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move?
    N2
    V = 1 m3
    P = 500kPa
    T = 80 C

    He
    V = 1 m3
    P = 500kPa
    T = 25 C

    2. Relevant equations
    PV = mRT
    Q = m(T2-T1)Cp


    3. The attempt at a solution
    I used the specific heat at constant pressure to solve the first part of the problem (Cp)
    Cp N2 = 1.039
    Cp He = 5.1926

    Using the ideal gas law i found the mass (kg) of each gas
    m = (PV) / (RT)
    m N2 = 4.77 kg
    m He = 0.807 kg

    then using the energy from one side of the cylinder equal to the other to find the final temperature of the system
    Q N2 = Q He
    (4.77)(T2-80)(1.039) = (0.808)(T2-25)(5.1926)

    I get T2 = 384K or 111 C This answer seems too high to be correct (it is higher than the initial temp of both gases)

    If someone could point out were i'm going wrong that would be great.
     
  2. jcsd
  3. Sep 30, 2009 #2

    kuruman

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    This equation is incorrectly written. You should write

    (4.77)(T2-80)(1.039) + (0.808)(T2-25)(5.1926) = 0

    One of the two terms is positive and the other is negative so their sum is zero. You could also write

    (4.77)(80-T2)(1.039) = (0.808)(T2-25)(5.1926)

    but not what you have.
     
  4. Sep 30, 2009 #3

    ehild

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    n =(PV) / (RT) is the correct ideal gas law. (n is the number of moles)

    ehild
     
  5. Sep 30, 2009 #4

    Andrew Mason

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    This is essentially a first law problem but it is tricky. You cannot use Cp. You have to use Cv. If you work it out using the following, you will see why.

    First of all find the relative number of moles of He to N2.

    Then apply the first law to the heat flow into/out of the respective sides (you have noticed that they must sum to 0). Write this out as an expression (this will contain two internal energy terms and two work terms. Write out the internal energy change of each gas in terms of n, Cv and T. Just use W1 and W2 for the work terms.)

    What is the relationship between the work done by/on the respective gases in the process of reaching equilibrium? What does that tell you about the relationship between the changes in internal energy of each side?

    From that you should be able to work out the final temperature Tf , and from that the final pressure.

    AM
     
  6. Oct 1, 2009 #5
    Thank you Kuruman, i figured it was some stupid mistake like that. That give me a much more reasonable answer of T2 = 54.78 degrees centigrade.

    there are many forms of the ideal gas law, it depends on what units your R is in. Since this problem uses mass, i chose to use Individual Gas constants with units of (KJ/kg*K) or (kPa*m3/kg*K). In this case, PV = mRT (m = mass) is correct.

    This is a two part problem. The first part asks you to use Cp and the second part asks you to use Cv. I have not attempted the second part yet because my answer from the first part seemed wrong (because it was wrong).

    Thank you all for your help
     
  7. Oct 1, 2009 #6

    kuruman

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    The answer may be reasonable because of the way you set up the equations, however I think you should seriously consider what Andrew Mason posted. I am afraid my reply was a bit hasty. I thought his would be a constant pressure process because the pressure in one side matches the pressure in the other, but this does not mean that the pressure cannot simultaneously increase (or decrease) in each side as heat is transferred and the partition moves. What remains constant is the total internal energy of the two gases and that is tied to the temperature and Cv.
     
  8. Oct 1, 2009 #7
    I don't have my book in front of me right now, but let me see if i can at least set this up right.

    I have DeltaU = 0 (for a closed system)
    DeltaU = (Q + W)N2 - (Q + W)He = 0
    Q = m(Tf - T1)Cv
    W = -P(Vf - V1)

    doesn't this give me too many unknowns? Also, why am i using Cv? if the piston is allowed to move, volume cannot remain constant, correct?
     
  9. Oct 1, 2009 #8

    kuruman

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    No, it's not constant volume, but the sum of internal energy changes is

    ΔE1 + ΔE2 = ΔQ1 + ΔQ2 + W1 + W2

    Do you see why the right side of the equation is zero? If so, then you should see why you should use Cv. As Andrew Mason said, it is a tricky application of the first law.
     
  10. Oct 1, 2009 #9
    Right. its a closed system, so it does not interact with the surrounding environment. so the volume of the system is constant (2 cubic meters). but each gas is allowed to expand or contract within the system.
    but unless im mistaken, there are still two unknowns and only one equation (final temperature and final volume). What am i missing?
     
  11. Oct 1, 2009 #10

    Andrew Mason

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    There is only one unknown: final temperature. All you have to determine are the changes in internal energy of each side. The heat flows and work terms sum to 0 so the internal energy changes must also.

    [tex]\Delta U_1 = - \Delta U_2[/tex]

    What is the expression for internal energy change?

    AM
     
  12. Oct 1, 2009 #11
    DeltaU = mCv(Tf-T1) Cv is in units of kJ/kg*K

    Then for part b if i consider the case were the piston does not move, i have W = 0 (no change in volume) and QN2 = QHe (again Q cancels out) and I'm left with the exact equation i had before, and get the same answer: Tf = 57.2
     
  13. Oct 2, 2009 #12

    Andrew Mason

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    Right. Since in both cases it is a closed system in which no net work is done and no heat is added or removed, the total internal energy of the system in both cases must be the same. If both sides end up at the same temperature in each case, that final temperature must be the same in each case. If it was not, the first law would be broken.

    AM
     
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