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Thermodynamics - cyclic pressure/volume process

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data

    A sample of an ideal gas goes through the process shown below. From A to B, the process is adiabatic; from B to C, it is isobaric with 98 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 158 kJ of energy leaving the system by heat. Determine the difference in internal energy, Eint, B-Eint, A.

    (sorry I cant post the link to the actual graph)...
    Point B: P=3atm, V=.09 m^3
    Point C: P=3atm, V=.40
    Point D: P=1atm, V=1.2
    Point A: P=1atm, V=.20

    2. Relevant equations

    [tex]\Delta[/tex]E= Q + W
    For cyclic process, [tex]\Delta[/tex]E=0, Q = -W

    isobaric process, W=-P(V[itex]_{f}[/itex]-V[itex]_{i}[/itex]
    isothermal process, W=nrTln(V[itex]_{i}[/itex]/V[itex]_{f}[/itex])

    3. The attempt at a solution

    I understand that based on conservation of energy, after completing one complete cycle, the overall change in energy will be 0. Basically, this means the net work done equals the area enclosed by the path. I'm confused at how you would solve for the energy at a single point though -- in order to find the difference between E,B and E,A. What equation would I use to calculate such an energy?

    Using the equation Q=-W:

    98 - 158 = -[W[itex]_{B,C}[/itex] + W[itex]_{C,D}[/itex] + W[itex]_{D,A}[/itex] + W[itex]_{A,B}[/itex]]

    W[itex]_{B,C}[/itex] = -P(V[itex]_{f}[/itex]-V[itex]_{i}[/itex] = -94kJ
    W[itex]_{D,A}[/itex] = -P(V[itex]_{f}[/itex]-V[itex]_{i}[/itex] = 101.325

    W[itex]_{C,D}[/itex] and W[itex]_{A,B}[/itex] are isothermal processes which can be solved with W=nrTln(V[itex]_{i}[/itex]/V[itex]_{f}[/itex]), but no information is provided about the temperature.

    At this point, I'm stuck.

    Any help would be greatly appreciated.
  2. jcsd
  3. Apr 28, 2008 #2
    This is also a problem I am having trouble with, the last point I need. Here's the picture relevant to our problem:

  4. Apr 28, 2008 #3

    Andrew Mason

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    Homework Helper

    The process from A to B is not isothermal. It is adiabatic.

    I am not sure what the question is. Are you trying to find the difference in internal energy of the gas between points B and A?

    If so, Find the change in internal energy from B-C, C-D and D-A using the first law:

    [tex]\Delta U_{BC} = Q_{BC} - W_{BC}[/tex]

    i) Work out W from the area under the graph from B-C.
    ii) What is the change in U from C-D? (easy)
    iii) What is it from D-A? (similar to i))

    How is the total of i + ii + iii related to the change in U from A-B?

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