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Carnot Engine Between Reservoirs with Varying Temperatures

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Two containers with fixed volume are each filled with n moles of a monatomic ideal gas with constant heat capacity. Initially, the volumes of gas are at temperatures [itex]T_1,i [/itex]and [itex]T_2,i[/itex]. What is the maximum amount of work that can be obtained by operating an engine between the two containers?

    2. Relevant equations
    Let [itex]T_1 > T_2[/itex]

    (1) [itex]\eta = \frac{|dW|}{|dQ|} = 1 - \frac{T_2}{T_1}[/itex]

    (2) [itex]|dQ| = -c_vdT_1[/itex] because |dT1| is negative so |dQ| = -dQ

    (3) [itex]c_v = \frac{3}{2}nR[/itex]


    3. The attempt at a solution
    Substituting (2) into (1) and solving for |dw|gives
    [itex]|dW| = -c_v \Big(1 - \frac{T_2}{T_1}\Big)dT_1[/itex]

    I need to integrate this somehow, but both T1 and T2 are changing so I'm not sure how to do this. I'm assuming there's a relationship between T1 and T2 that I can use that will allow me to integrate? Any help pointing me in the right direction would be greatly appreciated.
     
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  3. Mar 24, 2013 #2

    vela

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    Try coming up with a relationship between T1 and T2. You should be able to calculate how much heat of the dQ taken from the hot reservoir is expelled into the cold reservoir, which you can then use to get the change in T2.
     
    Last edited: Mar 25, 2013
  4. Mar 24, 2013 #3

    TSny

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    Hello doombanana. Welcome to PF!

    Try setting up an equation for the the cooler gas similar to your equation (2). Also, I think you'll need to use the 1st law of thermodynamics to relate dW, dQ1, and dQ2.
     
  5. Mar 24, 2013 #4


    I guess finding this relationship is where I'm having trouble.

    Using Tsny's suggestion, from the first law I found that
    [itex]1- \frac{|dQ_2|}{|dQ_1|}= 1-\frac{c_vdT_2}{-c_vdT_1}=1-\frac{T_2}{T_1}[/itex]
    solving the last two parts of the equation for T2 gives
    [itex]T_2=\frac{T_1c_vdT_2}{-c_vdT_1}[/itex]

    substituting that into my eq for |dW| gives
    [itex]|dW|= -c_v( dT_1 - dT_2)[/itex]

    integrating gives
    [itex]|W| = c_v(T_{1,i}-T_{2,i})[/itex]


    I know this isn't right because |W| ends up being larger than |Q_1|, but I'm not sure where I went wrong, or if this is even remotely close to the correct way to go about this. (I have a feeling it's not.)

    EDIT: when integrating I assumed that the final temperature in both reservoirs was equal to (T1,i + T2,i)/2 which now I'm thinking isn't the case. If so, I'm even more lost than before.:uhh:

    Thank you! I don't know how I've made it this far without creating an account here.
     
    Last edited: Mar 25, 2013
  6. Mar 25, 2013 #5

    TSny

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    OK!
    Check for a sign error inside the parentheses.
    You'll see that with the sign corrected in the previous expression, the integrated expression for W will include the final equilibrium temperature.
    See if you can find the final temperature by integrating your equation
    [itex]T_2=\frac{T_1c_vdT_2}{-c_vdT_1}[/itex]
     
  7. Mar 25, 2013 #6

    TSny

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    Just a note. You can apply the 1st law to the entire process as

    W = |Q1|-|Q2|

    Each Q can be expressed in terms of Cv and the overall temperature changes. This avoids having to first set up an expression for dW and integrating. But the result is the same.

    You're still going to need to determine the final equilibrium temperature in terms of the initial temperatures.
     
  8. Mar 25, 2013 #7
    Sign errors are the bane of my existence, but at least I see where it came from.:smile:

    Got it!

    Oooh, that is a bit of a shortcut and a more intuitive of setting up the problem.

    Thank you very much for your help! You've saved me hours of beating my head against the wall.
     
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