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Thermodynamics piston problem

  1. Oct 7, 2006 #1

    quasar987

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    A cylindrical container 80 cm long is separated in two compartments by a thin piston, originally held 30cm away from the left side of the piston. The left compartment is filled with 1 mol of an helium gaz at a pressure of 5 atm; the right compartment is filled with Argon at a pressure of 1 atm. The gaz can be considered as ideal. The cylinder is submerged in 1 litre of water (thermal contact) and the system is initially at a uniform temperature of 298 K. The calorific capacities of the cylinder and the piston can be neglected. When the piston is set free, a new equilibrium condition is ultimately attained with the piston in a new position.

    a) What is the new water temperature?

    b) What is the new position of the piston?

    c) What is the change in the total entropy of the system?


    I can do b). I know how to do c) knowing a), but how do I do a)? All I know is that at equilibrium, the temperature of both gazes and of the water will be the same.

    As the piston moves, work is done by the He gaz and on the Ar one, so their energy will change, which will lead of a change in their temperature, which will lead to an exchange of heat btw the 3 mediums. But this interaction is too complex for me to "monitor". How do I proceed?

    Thanx!
     
    Last edited: Oct 7, 2006
  2. jcsd
  3. Oct 9, 2006 #2

    quasar987

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    help?


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  4. Oct 9, 2006 #3

    Astronuc

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    One way to solve this is to ask, what is changing, when the He at 5 atm is doing work on the 1 atm of Ar. How does the enthalpy of each gas change? Can one determine an equation for the enthalpy of the gases and water at equilibrium. Whatever energy is transfered from the He must go to the Ar and H2O.

    [itex]\Delta H = mc_P\Delta T[/itex].
     
  5. Oct 9, 2006 #4

    quasar987

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    We have not yet encountered the notion of enthalpy. Do you have another idea or can you reformula that idea without the notion of enthalpy?
     
  6. Oct 9, 2006 #5

    Astronuc

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    Has your class covered polytropic processes?
     
  7. Oct 9, 2006 #6

    quasar987

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    Yes, but I'm curious to see how that applies here, since neither T, p or V are held constant.
     
  8. Oct 12, 2006 #7

    siddharth

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    Have you tried taking the system as both the gases? Here's how I think you can approach this.
    • The net expansion work will be 0(?) and you'll only have to equate the change in internal energy to the heat exchange (by the first law).
    • Since the gases are ideal, the internal energy change will be functions of temp only. So, if you know the final temp of each gas, you'll be able to calculate the temp of water.
    • To find the final temp of each gas, you can use the ideal gas equation. The values of [itex] P_i , V_i , Y_i [/itex] are given in the problem. If you solve for part (b), you can find the value of [itex] V_f[/itex]. You need to calculate the value of [itex]P_f[/itex] and [itex]T_f[/itex]
    • Because everything is in equilibirium finally, the zeroth law says the temperatures must be same everywhere. So, you'll have 2 equations to calculate [itex]T_f , P_f [/itex]. (One ideal gas equation for each substance)
    I also think the expansion in this case is irreverisble. So, the equation of the process can't be plotted.

    To calculate the change in entropy, consider a reversible process between the inital and final states for each substance.
     
    Last edited: Oct 12, 2006
  9. Oct 12, 2006 #8

    quasar987

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    I solved it, but thanks. Here's a brief outline of the solution:

    The whole system is isolated, so [itex]\Delta E=0[/itex]. Aditionally, since all 3 sub-systems start at the same temperature and end at the same temperature, the variation in their respective temperature will be the same. No work can be done on the water, so it's varitation in energy is purely a result of heat flow: [itex]\Delta E_{H_2O}=Q_{H_2O}[/itex]. For the two ideal monoatomic gases, the energy is linked directly to the temperature by [itex]\Delta E=\frac{3}{2}\nu R\Delta T[/itex]. Putting all this in the energy conservation law gives [itex]Q_{H_2O}=-\frac{3}{2}R(\nu_{He}+\nu_{Ar})\Delta T[/itex]. This equation is in contradiction with the law of heat flow as soon as [itex]\Delta T \neq 0[/itex], because remember that [itex]\Delta T = \Delta T_{H_2O}[/itex]. For instance, suppose that [itex]\Delta T_{H_2O}<0[/itex]. Then [itex]Q_{H_2O}>0[/itex] which means that the water absorbed heat. But heat flows from the hotest substance to the coolest until they are both at the same temperature, so heat absorbed cannot be positive and variation in temp. negative. So [itex]\Delta T = 0[/itex].
     
    Last edited: Oct 12, 2006
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