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Homework Help: Container with two gases and equation of state

  1. Jun 26, 2013 #1
    1. The problem statement, all variables and given/known data
    We have a cylindrical container filled with a gas: the container has an upper compartment with 2 moles of this gas, and another compartment below with 1 mole of the same gas, separated by a diathermic boundary, and also has an adiabatic mobile plunger over the upper compartment.
    The lower compartment has a kind of grinder (I don't know how to translate it, sorry) which can do work on it. A picture of the container:


    The initial pressure in both compartments and in the outside is [itex]p_0[/itex], and the temperature is [itex]T_0[/itex]. The heat capacity at constant pressure is [itex]C_p = 3R[/itex].
    Now imagine the "grinder" does 500J of work at the lower compartment. What will be the final [itex](p,V,T)[/itex] state of the two gases?

    2. Relevant equations

    The equation of state of the gas is
    p(V-b) = nRT

    Also, the internal energy (as deduced in a previous part of the problem) is

    U = 2nRT

    3. The attempt at a solution

    It is clear that the upper compartment remains at constant pressure and that the temperature at both parts of the tank will be the same along the expansion of the upper gas.
    The work done by the grinder is in form of heat, I think. But i don't know how to proceed. Any clues will be appreciated.

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 26, 2013 #2
    Apply the first law to the two chambers individually. Let Q be the heat transferred from the lower chamber to the upper chamber. The stirrer (grinder) does 500 J of work on the gas in the lower chamber, and the process in the lower chamber takes place at constant volume. The process in the upper chamber takes place at constant pressure, so here Q = ΔH = ΔU + pΔV for that chamber. The final temperatures are the same, but the final pressures are not.
  4. Jun 26, 2013 #3
    Ok. What about this:

    Lower chamber:

    [itex]\Delta U_1 = Q_{stirrer} - Q_{up} + p\Delta V [/itex]

    So using the expression for the internal energy and the constant volume:

    [itex]2R\Delta T = 500 - Q_{up} [/itex]

    Upper chamber:

    [itex]\Delta U_2 = Q_{up} - p\Delta V [/itex]


    [itex]4R\Delta T = Q_{up} - p_0\Delta V [/itex]

    Adding up these two expressions we get:

    [itex]6R\Delta T = 500 - p_0\Delta V [/itex]

    And now, from the equation of state:

    [itex]\dfrac{nRT}{p} + b = V \Rightarrow \Delta V = \dfrac{nR}{p}\Delta T[/itex], so in the upper chamber [itex]\Delta V = \dfrac{2R}{p_0}\Delta T[/itex]

    Replacing it in the last equation we get:

    [itex]6R\Delta T = 500 - p_0\Delta V = 500 - p_0 \dfrac{2R}{p_0}\Delta T = 500 - 2R \Delta T[/itex]

    And finally [itex]8R\Delta T = 500[/itex], from where we can derive the final state of the two chambers.

    Is it correct?
  5. Jun 26, 2013 #4
    Yes. I think this is correct. In the equation for V, I think there should have been an nb rather than a b, but this cancelled out when you wrote the equation for ΔV. You seem to have a very good feel for how to do this type of modelling. Nice work.

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