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Adiabatic piston/cylinder problem

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data
    1, 2, and 3 use the following information: a cylinder is closed at both ends and has adiabatic walls. It is divided into two compartments by a movable, frictionless, adiabatic piston. Initially the pressure, volume, and temperature on both sides are equal and given by P=10.0 atm, V=100 Liters and T=300k. The gas is perfect(ideal) with the MOLAR heat capacity of Cv=3R/2(Joules/mol.K)). By means of a heating coil, which is switched on at time t=0, heat is supplied to the compartment on the left, causing the piston to move and the pressures and volumes of the gas in the two compartments to change. A constant current of 10.0 Amps flows through the heating coil; the potential difference across the coil is 100.0 Volts. Assume the power of the heater is also at which heat is delivered to the gas on the left. The heater is turned on for 303.0 seconds, and the system is allowed to come to a new equilibrium. For this new equilibrium state, calculate:

    1.The pressure in atmospheres of the gas in the right hand compartment.
    2. The volume of the right hand compartment.
    3. The final temperature of the left hand compartment



    2. Relevant equations

    PV=nRT
    dW=PdV
    du=dq+dW
    P1V1=P2V2

    Power=I*Voltage
    Power=work/time

    3. The attempt at a solution
    I've found the work to be 303 kJ. I also know that the final pressure on both sides will equal each other. I know that the piston will expand to the right, increasing the volume of the left, while decreasing the volume on the right. I know that because this is an adiabatic closed system that the change in internal energy will be zero.

    I'm fairly sure that the temperature will go up on the right hand side but I'm not 100% sure.

    I worked out a very roundabout way of solving this problem, which involved doing the questions out of order. My professor assured me there is an easier and more efficient way of solving this problem by doing them in order, which is why he presented it in this order.

    My problem is that since I don't know any of the final conditions, how can I determine what any of them are in this order? For instance I know I could integrate PdV=dW, but I need the final volume to have my bounds of integration.

    So I know I'm missing something simple here. Can I do something with the fact that the total volume of this cylinder is 200L? Or is there a way to determine pressure using the work and time? I've searched high and low for some way to do this and I really have no idea.
     
  2. jcsd
  3. Feb 9, 2013 #2

    Andrew Mason

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    Homework Helper

    The right side will undergo adiabatic, quasi-static compression. So use the adiabatic condition for temperature and pressure for the right side:

    [tex]PV^{\gamma} = P(nRT/P)^{\gamma} ⇔ P^{1-\gamma}T^{\gamma} = K [/tex]

    That will give you an expression for the final right T in terms of P (which as you point out has to be the same for both sides).

    Also, look at the work on the surroundings. What does that tell you about how the total heat flow is related to the total change in internal energy? That will tell you how the two temperatures are related

    I think you should be able to work everything out from that.

    AM
     
  4. Feb 9, 2013 #3
    Part 1:
    The combined system does no work, so the amount of heat added to the left chamber equals the change in internal energy for the combined system. The change in the sum of the temperatures in the two chambers is equal to the combined internal energy change divided by Cv. The new volumes in each chamber are given by:

    [tex]V_L=V_0\frac{T_L}{T_0}\frac{p_0}{p_f}[/tex]
    [tex]V_R=V_0\frac{T_R}{T_0}\frac{p_0}{p_f}[/tex]
    If you add these two equations together, you get:
    [tex]V_L+V_R=2V_0=V_0\frac{(T_L+T_R)}{T_0}\frac{p_0}{p_f}[/tex]
    But
    [tex](T_L+T_R)=2T_0+Q/C_v[/tex]
    So combine these last two equations together to get p_f (which is the same for both chambers).
     
  5. Feb 9, 2013 #4

    Andrew Mason

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    Science Advisor
    Homework Helper

    How does your solution depend on the fact that the right side undergoes adiabatic compression?

    You also have to have n in there somewhere, don't you? as in:

    [tex](T_L+T_R)=2T_0+Q/nC_v[/tex]

    AM
     
    Last edited: Feb 9, 2013
  6. Feb 10, 2013 #5
    Good question. The sum of the temperatures in the two cylinders is determined by the above equation, as is the final pressure. This is irrespective of how much heat is added to the individual chambers. The individual temperatures in the two chambers depend on the fact that the right side undergoes adiabatic reversible compression, subject to the constraint that the sum of the temperatures in the two chambers satisfy the above equation. In your previous post, you already presented the equation for the final temperature on the right side, given the final pressure:

    [tex]\frac{T_{Rf}}{T_0}=(\frac{p_f}{p_0})^{(1-1/\gamma)}[/tex]

    Thanks for correcting the temperature equation by including the factor of the number of moles.

    Chet
     
  7. Feb 10, 2013 #6
    First of all I want to thank you guys, I obviously have a lack of understanding of some of this material, but I'm still running into slight issues.

    The final pressure I'm getting is terribly small(0.026atm), which just seems off, and when I throw it into the other equations I'm getting strange answers.

    Here's what I'm doing:

    2(100L)=100L((2(300K)+(-303kJ/((40.6mol)(3/2)(8.314J/mol.K))/300k)*10.0atm/Pf

    The Joules and moles both cancel leaving me with Kelvins on the inside which then also cancel out. Leaving me with L=L*atm. Divide through and I'm left with atm=Pf. So my dimensions are working out.

    Is my heat wrong perhaps?

    I used W=-I^2Rt to get -303000 Joules. Which plugging it into P=IV and Pt=W gives me the same answer. So I'm not sure where I'm going wrong.

    Also, I know it sounds silly, but could you explain where all of these equations came from? I know the adiabatic gas law but there's a missing link between these final equations and the beginning ones, that I would not mind getting a better understanding of.

    Thank you
     
  8. Feb 10, 2013 #7
    I'm just confused isn't Q=0 in an adiabatic process? So is the circuit an outside heat source then?
     
  9. Feb 10, 2013 #8
    Disregard previous posts, I had an extra negative sign in there. I've managed to work out the answers, get reasonable solutions and found that they check out overall.

    Thank you guys so much, this problem has been hell for me all week!
     
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