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Thermodynamics Piston-Cylinder Question

  1. Jul 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Air contained in a piston–cylinder assembly, initially at 2 bar, 200 K, and a volume of 1 L, undergoes a process to a final state where the pressure is 8 bar and the volume is 2 L. During the process, the pressure–volume relationship is linear. Assuming the ideal gas model for the air, determine the work and heat transfer, each in kJ.

    2. Relevant equations
    pv = RT (ideal gas, v = specific volume)
    W = pdV
    ΔE = Q - W
    h = u(T) + RT

    3. The attempt at a solution
    pv = RT
    => v = RT/p
    v = [ (8.314 kJ/kmol*k)/(28.97 kg/kmol) * (200 K) ] / (2 bar) * (105 N/m2 / bar) (1 kJ / 103 N*m) = 0.2689 m3/kg

    v = V/m
    => m = V/v
    m = 0.001 m3/ 0.2869 m3/kg = 0.003486 kg

    That's my calculations for the mass of the initial state, m1. Since the assembly is a closed system, the initial mass m1 should be equal to the final mass m2. I'm assuming the mass of the air would be used to find the change in internal energy. But, since the problem does not explicitly state it, am I allowed to assumed there are no effects in kinetic/potential energy?

    Also, as for the work and heat transfer calculations, I'm assuming that W = pdV would come into play, but the statement regarding the linear pressure-volume relationship is throwing me off. Does that mean the same thing as pV = constant?

    I'm fairly lost right now, so if anyone can steer me in the right direction it would be greatly appreciated.
     
  2. jcsd
  3. Jul 18, 2017 #2
    Yes.
    No. If P is the pressure in bars and v is the volume in liters, what is the equation for the straight line passing through the points (1,2) and (2,8)? In Joules, what is the amount of work done by the gas on the surroundings?
    You know the initial temperature and the initial- and final pressures and volumes. From the ideal gas law, what is the final temperature? Knowing the number of moles and the initial and final temperatures, what is the change in internal energy?
     
  4. Jul 19, 2017 #3
    Chester,

    With your hint, I was able to find a final temperature of 1600, and corresponding internal energy values of u1 = 142.56 kJ/kg, and u2 = 1298.30 kJ/kg, according to the tables in the back of my textbook. I calculated the corresponding change in internal energy to be 4.0289 kJ.

    The heat transfer calculation will be a cakewalk once I find the work value, however, I'm still a bit confused on how to do just that. Pressure isn't constant, so W=∫pdV can't just become p (V2 - V1), and temperature isn't constant either so the integral can't become W = pV ln(V2 / V1).

    Am I even remotely on the right track?
     
  5. Jul 27, 2017 #4
    I didn't read the whole text but I just solved it. Open the attachment.

    First, you can draw a P-V diagram if its relation is linear so it is easy to find the boundry work by calculating the area of trepazoid. It's 0.5 kJ. And then, from the energy balance $$ Q_(Net) - W_(Net) = \Delta U $$ We are looking for Q net. We can do this by two ways. First we can arrange the eq. like $$ Q_Net = \Delta U $$ since W net + Delta U = the change in the entalphy. I did the calculation on the top right side of the pic. It is 5.43 kJ The other way is since we know the W net, we can find the internal energy change and sum up these two. To be able to that, you can determine the Cv(avg), find the final temp. by P1.V1 / T1 = P2.V2/T2 eq, that is about 1600 K. And from the 1600+200/2 = 900K you can take the Cv(avg) @ 900K and you know the rest. I hope I could help you.

    Regards
     

    Attached Files:

    • q1.PNG
      q1.PNG
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    Last edited: Jul 27, 2017
  6. Jul 27, 2017 #5
    By the way, I realized that I wrote wrong, the second equation should be $$ Q_Net = \Delta H $$
     
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