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Ideal Gas Expansion State Properties & Exergy Balance

  1. Jan 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Two well-insulated rigid tanks of equal volume, tank A and tank B, are connected via a valve. Tank A is initially empty. Tank B has 2 kg of Argon at 350 K and 5000 kPa. The valve is opened and the Argon fills both tanks. State 2 is the final equilibrium state. The temperature and pressure of the room in which the tanks sit are 300 K and 100 kPa, respectively. Perform a closed system analysis.
    a. Determine the volume of tank B.
    b. Determine the final temperature.
    c. Determine the final pressure.
    d. Determine the entropy produced in the process.
    e. Determine the exergy (or availability) destroyed in the process.
    i. Using an exergy balance.
    ii. Using the entropy produced.​
    f. Determine the total exergy in tank B initially.

    2. Relevant equations
    Ideal Gas law:
    PV = ZmRT​
    First Law of Thermo
    E2-E1 = Q - W + ∑m(h+v2/2+gz)​
    Second Law of Thermo

    3. The attempt at a solution

    PART A)
    using tabulated values for Tc and Pc of Argon, I found Z (compressibility factor) to be 0.9945~1, so I'm treating the Argon as behaving like an ideal gas.
    given TB1 and PB1, specific volume is found to be vB1=0.00137 m3/kg
    VB = vB1*mB1
    VB = 0.02914 m3

    PART B)

    My approach to find the temperatures would be to use conservation of internal energy between initial and final states.

    U1 = U2
    U1 = UA1 + UB1
    U1 = uA1*mA1 + uB1*mA1
    where tank A is initially evacuated, so mA1 = 0, therefor
    U1 = uB1*mB1

    and then for U2

    U2 = uA2*mA2 + uB2*mB2
    assuming from conservation of mass: mB1 = mA2 + mB2
    and that since PA2=PB2 and TA2 = TB2, mA2 = mB2 = mB1/2​
    U2 = (mB1/2)*(uA2 + uB2)
    assuming uA2 = uB2 = .5*u2
    U2 = (mB1/2)*(u2)
    U2 = .5*mB1*u2

    now equating U1 and U2:

    uB1*mB1 = .5*mB1*u2
    uB1 = .5*U2 = uB2 = uA2

    so since I have uB1, TB1 and have equated the final specific internal energies to uB1, I can find the final temperature and move on. Yes?

    PART C)
    In order to find the pressure I would equate the specific volume of the tanks, resulting in
    .5*vB1 = vB2 = vA2

    next using final specific volume and final temperature to determine the final pressure.

    Is this heading in the right direction?
  2. jcsd
  3. Jan 19, 2016 #2
    No. This assumes that the two tanks equilibrate thermally, such that uA2=uB2=uB1, and TA2=TB2=TB1. It also implies that pA2=pB2=pB1/2. However, the tanks are insulated, and so, the gases in the two tanks do not thermally equilibrate.

    What is your assessment of what is happening to the gas in insulated tank B during the time that gas is seeping through the valve (while the system is in the process of equilibrating)?

  4. Jan 19, 2016 #3
    The gas in tank B is expanding, or decreasing in Pressure (so also in Temperature).
  5. Jan 19, 2016 #4
    So, are you saying that the gas remaining in the tank at any instant of time has been experiencing something very close to an adiabatic reversible expansion (particularly if the seepage rate through the valve is very slow)?
  6. Jan 19, 2016 #5
    That's what it sounds like to me. So the 2nd law is where this is headed...
  7. Jan 19, 2016 #6
    No. We are going to express everything in terms of the final pressure P, and then solve for P under the constraint that the change in internal energy of the overall system is zero. So, lets get started.

    Given that the initial temperature and pressure in tank B is 350 K and 5000 kPa, in terms of P, what is the final temperature T in tank B?
    Given the final volume in tank B is 0.02914 cubic meters, in terms of P, what is the final mass of argon in tank B?
  8. Jan 20, 2016 #7
    for T2:
    So if I'm not using the second law, the other equation I see T and P in is the ideal gas law. Can I use this even though I can't check the validity of the second state behaving as an ideal gas?

    is it appropriate here to use V2 = 2*V1, since I've been told the tanks are equal in size? Then it would become:

    for m2:
    the ideal gas law also seems appropriate here because it has P, V and m in it. Yeah?

    Last edited: Jan 20, 2016
  9. Jan 20, 2016 #8
    None of this is correct. Our focus is on the gas that remains in tank B during and after the system equilibrates. This gas is experiencing and adiabatic reversible expansion. Do you know the relationship between temperature and pressure for an ideal gas experiencing an adiabatic reversible expansion? If not, what about the relationship between molar volume and pressure for an ideal gas experiencing and adiabatic reversible expansion?

  10. Jan 20, 2016 #9

    So I followed similar approach to yours and I believe we get the same answer
    my approach:

    assumption 1: Argon is an ideal gas
    mAB = mB (COM)

    1st law:
    E2- E1 =Q -W
    since it is well insulated Q = 0 (Adiabatic)
    and because it is rigid body I assumed there is boundary work so W = 0

    From this I went on and
    equated E2-E1=0
    so U2-U1 +change in KE and PE =0 (assumption neglect PE and KE)

    mAB * uAB = mA*uA + mBuB
    since mA is 0 and mAB = mB
    uAB = uB
    so TempB = T2

    for final pressure
    we know that v2 = 2*vB
    so P2 = (m*R_bar*TempB)/(2*vB)

    which is approximately half PressureB

    I got stuck in the last part of the question which is finding the total energy in tank B initially!!! :(
  11. Jan 21, 2016 #10
    As I said earlier, this is only correct if the two tanks can reach thermal equilibrium with one another, but, according to the problem statement, they cannot.
  12. Jan 23, 2016 #11
    For a nice analysis of the change in tank B between the initial and final states of the system, see Example 6.10 in Fundamentals of Engineering Thermodynamics by Moran, Shapiro, Boettner, and Bailey.
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