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Help with Ideal gas equation of state

  1. Mar 19, 2012 #1
    I am reading through my thermodynamics book. Going over a air-standard otto cycle example problem. For reference the example problem is as follows:

    The temperature at the beginning of the compression process of an air-standard Otto cycle with a compression
    ratio of 8 is 5408R, the pressure is 1 atm, and the cylinder volume is 0.02 ft3. The maximum temperature during
    the cycle is 36008R. Determine (a) the temperature and pressure at the end of each process of the cycle, (b) the
    thermal efficiency, and (c) the mean effective pressure, in atm.


    After finding the values for u1 and vr1 from a table and using the air-standard relation of vr2=(V2/V1)(vr1) which I followed. Then using that to interpolate a value for T2 and u2, the solution then states that since process 2-3 occurs at constant volume we can find p2 by the following equation:

    p2 = p1((T2/T1)(V1/V2))

    I'm trying to understand where this is coming from because I don't see this on any ideal gas tables. I'm sure I'm missing some simple relation but if anyone can help explain where this relation has come from and why we can use it I would appreciate it.
     
  2. jcsd
  3. Mar 19, 2012 #2

    jack action

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    [itex]P_2 V_2 = m_2 RT_2[/itex] and [itex]P_1 V_1 = m_1 RT_1[/itex]

    We know that:

    [itex]m_2 = m_1[/itex]

    So:

    [itex]\frac{P_2 V_2}{T_2} = m_2 R[/itex] and [itex]\frac{P_1 V_1}{T_1} = m_1 R[/itex]

    Or:

    [itex]\frac{P_2 V_2}{T_2} = \frac{P_1 V_1}{T_1}[/itex]

    [itex]P_2 = P_1 \frac{ T_2}{T_1}\frac{V_1}{V_2}[/itex]
     
  4. Mar 19, 2012 #3
    Wow that easy huh?.. I feel stupid now. I guess I didn't fully understand how to use the ideal gas law all these years. Thanks a ton
     
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