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Entropy Change for an Ideal Gas

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    10 moles of an ideal gas Cv = 20.8 J/mol at T0 = 300 K and P0 =0.3 MPa occupy the left half of an insulated vessel. At time t=0 a 1 kW electrical heating element is turned on. after 30 s, the partition dividing the vessel ruptures and the heating element is turned off. Calculate a) the final temperature Tf and pressure Pf of the gas in the vessel and b) the entropy change of the gas during the process.

    2. Relevant equations
    PV= nRT
    du = TdS - PdV
    delta U = Q+ W

    3. The attempt at a solution
    1 kw = 1 kJ/s
    1 kJ/s x 30 s = 30 kJ

    delta U = Q+ W -----> Q= 0 because system is insulated therefore delta U = W
    30 kJ is the total amount of internal energy added to system
    -----------------------------------------------------------------------------------------

    My friend and I were attempting the solution this way. However, our textbook has set W=0 for some reason, and said that deltaU = n*Cv*delta(T) =Q.

    We don't understand how W= 0 unless we assume that the gas is expanding against a vacuum which wasn't stated explicitly. .....Also, how is Q not equal to zero is the system is insulated ???
     
  2. jcsd
  3. Feb 28, 2015 #2
    The implication of the problem statement is that vacuum is present in the other half of the vessel initially. Even though the vessel is insulated, they expect you to accept that heat is added to the vessel contents from the heater. Maybe the heater is inside the vessel, and, in that way, the heater can be considered as part of the surroundings which transfer heat Q to the system. The overall vessel is rigid, so it does no work on the surroundings. Based on this, you end up with the equation given in your book. Even if the gas were not expanding against a vacuum, the contents of the rigid container would not be doing any work on the surroundings. So, in any event, W = 0.

    Chet
     
  4. Feb 28, 2015 #3
    I understand what you are saying. However, I am still confused as to what you are considering the system to be: the vessel or the gas? If the system is the gas, then it makes sense that Q has a value(not zero) and that W = 0(but only if it is expanding against a vacuum). However, if the system is the vessel, which is insulated, then Q must definitely be zero and it makes sense that work is zero too because the vessel is rigid.....

    What I don't get about thermodynamics or maybe it is just this textbook that the professor chose, is how we are expected to make these assumptions while being given insufficient or even misleading information...... It's not not that I don't understand the concepts or how to apply them, but this textbook expects us to pull certain assumptions out of the air unrealistically.

    In reality, if I ever had to encounter a scenario, such as the one in the problem, I would not have to guess if the gas were expanding against a vacuum or if the heater is inside the vessel. I don't see why can't these things just can't be stated explicitly, without making it hard for the student.
     
  5. Mar 1, 2015 #4
    The problem statement is imprecise and ambiguous to say the least.

    There is always the decision you need to make of what is the most convenient and optimum choice for the system. In this case, I reckoned that the easiest choice to work with would be the gas. The surroundings for the gas are the container, and the gas does no work on its rigid walls of the container.
    It isn't just your textbook. Most textbooks on thermodynamics are very imprecise about how they teach the concepts, and very ambiguous about how they word the problems. It's very hard to word a problem that someone won't find a little ambiguous. Just try not to be too frustrated and don't blame yourself. Maybe, in a little while with a little more practice, you'll be about to read between the lines.

    Chet
     
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