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Homework Help: Finding ΔS and P(final)?

  1. Dec 6, 2017 #1
    So, there is a problem given, where diatomic ideal gas N2 is involved. Both blocks have same volume: 1m3. The molar mass of gas is: 28 g/mol. Initial temperature is 300K.

    I need to find ΔS and PF.

    First, I did find P1 by using formula: PV = nRT
    so, P1 = 1*8.3145*300/1 = 2494.35 Pa
    and P2 = 2*8.3145*300/1 = 4988.7 Pa

    Finally: I have total moles = 3, and total V = 2 m3.
    What should I consider next?
  2. jcsd
  3. Dec 6, 2017 #2
    What does the first law of thermodynamics tell you about the change in internal energy of this system? Based on that, what is the final temperature of the system?
  4. Dec 6, 2017 #3
    The first law of the thermodynamics says that the change in internal energy = heat added to the system - work done by the system.
    However, I can't connect the dots with the final temperature based on this definition. Any hints please?
  5. Dec 6, 2017 #4
    As I said, in this particular problem, what is the total change in internal energy of the system comprised of the two gases? How much heat is added to the system? How much work do the gases do on their rigid container?
  6. Dec 6, 2017 #5
    I am so confused right now.
    I have the value of ϒ, which is 1.4. I can also use formula: pVϒ = K, but I don't know which p should I use. As, I come up with values for pressure. Further, work done could be found out by K(Vf1-ϒ - Vi1-ϒ)/(1-ϒ). Still I don't know how to find the value of K.
  7. Dec 6, 2017 #6
    What is the change in volume of the rigid container holding the gases? Based on this, how much work does the system comprised of the two gases do on their surroundings (I.e., on the rigid container)?
  8. Dec 6, 2017 #7
    Volume is constant.
  9. Dec 6, 2017 #8
    And the work the gases do on the rigid container comprising their surroundings?
  10. Dec 6, 2017 #9
    I can't find out work done, since I don't have the value of K. I can figure out K, but I just don't know whether I use p2 or p1 in the formula.
  11. Dec 6, 2017 #10
    The gases do zero work on their surroundings (I.e., the rigid container) because, treating the container as a black box, ##\Delta V=0##. And the heat entering is also zero, since the container is insulated. So the change in internal energy= ???
    Last edited: Dec 6, 2017
  12. Dec 7, 2017 #11
    Is the partition forced to move gradually, or is it free to move on its own (unconstrained)? Is the partition insulated, so that no heat can pass from one side to the other, or is it conductive to heat?
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