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Thermodynamics - Work

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data

    A gas-tight frictionless piston of small thermal conductivity slides in a thermally insulated cylinder, dividing it into two compartments, A and B, each containing nA=n and nB=0.45n moles of ideal monatomic gas (n moles). Initially the temperature of the gas is To in compartment A and 4To in B. Assume that the system is in mechanical equilibrium at all times and that the mass of the piston and the effect of gravity are negligible. Consider the process by which the system reaches thermal equilibrium:

    (a) What is the final temperature, Tf?

    (b) What is the ratio of the volume of A to that of B initially, ri?

    (c) What is the ratio of the volume of A to that of B after thermal equilibrium is reached, rf?

    (d) Calculate the work done on the gas in compartment A, and then do the same for compartment B.

    I managed to answer (a), (b), and (c) correctly, getting values of 1.93To, 0.55, and 2.22 respectively.

    Having a bit of difficulty with (d)

    2. Relevant equations

    dW = -PdV

    W = -∫ Pdv = -∫ (nRT/V)dV

    3. The attempt at a solution

    The answer is to be in terms of nRT

    I computed the work integral and I have,

    W = -nRT ln (Vfa/Via)

    Getting stuck as to what the ratio of Vfa to Via is.

    I played around with the ideal gas law and have,

    (Vfa/Via) = (1.93Pia/Pfa)

    From here I am just running in circles.

    Any help is appreciated!
  2. jcsd
  3. Feb 28, 2015 #2

    Suraj M

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    why?? I would rather use Charle's law .. constant Pressure
    this is because the system is not affected by any external agent so system pressure remains constant, also its a frictionless piston which maintains equal pressure b/w the 2 compartments! so pressure is a constant for each compartment = ##\frac{ P_{total}}{2}##. Can you do the rest?
  4. Feb 28, 2015 #3

    Suraj M

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  5. Feb 28, 2015 #4
    This is not correct. The pressure is not constant for each compartment, and the pressure in each compartment is certainly not half the total pressure.

  6. Feb 28, 2015 #5
    Let P represent the pressure within both chambers (the process is quasistatic and the piston is massless), and let V be the total volume of the two chambers. If you did part b, then you know the ratio of the volumes initially. In terms of V, what are the individual volumes? In terms of V and T0, what is the initial pressure P0.

    You know that the sum of the internal energies of the gases in the two chambers is constant during the process. If, during the transition from the initial to the final state, the temperature of the gas in chamber is TA and the temperature in chamber B it TB, express TA in terms of TB. In terms of TA and TB, what is the ratio of the volumes in the two chambers? In terms of TA and TB and V, what are the individual volumes in the two chambers? In terms of TA, TB, and V, what is the pressure in the two chambers?

  7. Feb 28, 2015 #6

    Suraj M

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    There is a massless, frictionless piston between the 2 compartments, doesn't that maintain the constant pressure between the compartments? Yes but not half of total, i was wrong there!
  8. Mar 1, 2015 #7


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    The system is isolated, so the internal energy is constant. It is ideal (monoatomic gas), so the total energy is U=3/2 nRTA+3/2 0.45 nRTB=const.

    The total volume VA+VB is also constant, and the pressure is equal at both sides.
    p(VA +VB) = nRTA+0.45 nRTB= 2/3 U . The pressure has to be constant during the process.
  9. Mar 1, 2015 #8
    The pressures in the two compartments are equal to one another, but your implication that the pressure remains constant at its initial value during the entire process was far from obvious (at least to me). ehild's analysis in post #7 elegently demonstrated this.

  10. Mar 1, 2015 #9


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    That was a hasty statement ... :devil:
  11. Mar 1, 2015 #10
    Yes. I made this statement before I had solved the problem to completion, but, even if the assumption were made that the pressure was not constant, the analysis would have led to the conclusion that it was. I doubt that the member I had responded to had obtained this result by analyzing the problem, given the remainder of his response. In retrospect, I probably should have said "The pressure is not necessarily constant for each compartment"

  12. Mar 1, 2015 #11


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    You can not read the mind of other people. Suray M guess was right, although not a complete proof.
    Yes, it would have been more appropriate to say that " the pressure was not necessarily constant" during the process.
  13. Mar 1, 2015 #12


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    That is wrong, T changes during the process, you can not pull it out from the integral. Find how Va, Ta, Vb, Tb, and P change during the process.
    Hint: The total volume of the compartment is constant. The total energy of the whole system is also constant. It is an ideal gas. How is the internal energy of the ideal gas related to the temperature?
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