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Thermodynamics and cycle

  1. Feb 21, 2017 #1
    1. The problem statement, all variables and given/known data
    2.00 moles of gas is held in a cylinder with a piston and is initially held at 0.300atm and has an initial volume of 0.200 m^3. The molar heat capacity of the gas at constant volume is 24.94 J mol^−1 K^−1
    . The gas is then brought from this initial state (State A) through the following processes:
    From state A to B:
    Gas is allowed to expand isothermally.
    From state B to C:
    The temperature of the gas drops by 100 K while it is being held at constant volume.
    From state C to A:
    The volume of the gas is then compressed in an adiabatic process back to its initial state.
    (a) What is the initial temperature of the gas in state A?
    (b) What is the ratio of the molar heat capacity at constant pressure (C P to the molar heat capacity at constant volume (C V ) of the gas?
    (c) What is the volume of the gas at state C? Hence, sketch a P −V
    curve depicting the processes, indicating the pressure and volume at each point.
    (d) In which of the processes is heat being transferred to the system and in which process is the heat being expelled from the system? Hence, calculate the network done by the system.
    (e) Assume that process B to C is instead stated as “The temperature of the gas rises by 100 K while it is being held at constant volume.” Is it possible then to return the gas to its initial state via an adiabatic process? Why or why not

    2. Relevant equations
    q = n cv ΔT
    Δu = Q - w
    p v^ϒ = constant for adiabatic process
    pv = nrt
    cp = r + cv

    3. The attempt at a solution
    i solved part a and b but got stuck at c i have shown my answer for a and b

    a)T = pv/nr = 364K
    b)cp /cv = 1+ r/cv = 1.33
    c)so for c how am i supposed to know the volume at c without knowing the pressure at c please help thanks!!
     
  2. jcsd
  3. Feb 21, 2017 #2

    kuruman

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    Gold Member

    Can you find the volume at b? It is the same as at c.

    On edit: How are ΔUbc and ΔUca related? Can you find an equivalent expression of pVγ = const. that involves temperature instead of pressure?
     
    Last edited: Feb 21, 2017
  4. Feb 21, 2017 #3
    well,
    Δubc = q
    where q = n cv Δt
    Δt = -100k
    and
    Δuca = -w
    and
    adiabatic process can also be written as t Vγ-1
    i had realised this but how does this help
     
  5. Feb 21, 2017 #4

    kuruman

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    TcVcγ-1=TaVaγ-1
    How many unknowns in the above equation?
     
  6. Feb 21, 2017 #5
    The trick in this problem is to focus on the change from state C to state A. Presumably, they meant for the process to be reversible. So, instead of trying to guess the pressure and volume in state C, run this process step in reverse. You know the conditions in state A, and you know the final temperature in state C. So start at state A, and allow the gas to expand adiabatically and reversibly until the temperature is 100 C lower. This will get you to the conditions at state C.
     
  7. Feb 22, 2017 #6
    oh my god i cant believe the equation TVγ-1 was staring at me all along and i could not figure it our my careless mistake sorry for wasting all of your time i got so involved that i forgot i know V1 thanks anyways!!!
     
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