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Homework Help: Heat added

  1. Dec 1, 2006 #1
    1. The problem statement, all variables and given/known data
    A monatomic, ideal gas is in a sealed container (the number of gas molecules is always constant: n = 2 moles); the initial pressure is Pi = 1.01 x 10^5 Pa and the initial volume is Vi = 0.0224 m^3.

    * First, the volume of the gas is decreased at a constant pressure (at Pi = 1.01 x 10^5 Pa) to a final volume of Vf = 0.0155 m^3.
    * Second, the pressure of the gas is increased at a constant volume (at Vf = 0.0155 m3) to a final pressure of Pf = 1.35 x 10^5 Pa.

    How much heat was added to (give as a positive number) or removed from (give as a negative number) the system? (The gas constant R = 8.31 J/mole-K.)


    2. Relevant equations
    PV = nRT


    3. The attempt at a solution

    I guess I am confused as to how I am supposed to solve this problem without knowing the heat capacity or the specific heat of the substance.

    I have calculated Ti=94.14 and Tf=125.83 - deltaT=31.692 C - why is this incorrect?
     
  2. jcsd
  3. Dec 1, 2006 #2
    The heat capacity at constant volume of an ideal gas is: [tex] c_{v}NR [/tex].

    [tex] c_{v} = \frac{3}{2} [/tex] for a monatomic gas, and [tex] \frac{5}{2} [/tex] for a diatomic gas.

    The heat capacity at constant pressure of an ideal gas is:

    [tex] (c_{v}+1})NR [/tex]
     
  4. Dec 3, 2006 #3
    Problem Solved.

    Please look at case one of the pressure vs volume graph:

    http://img99.imageshack.us/img99/2964/followuphz6.gif [Broken]

    I thought to get the total work done by the system I would take the work done by decreasing volume and add the work done by increasing pressure.

    To get the total work done all I had to do was use the work done by the system decreasing volume. (-696.9J)

    Why didn't I have to add on the work done by the increase in pressure? I calculated that to be 527J, where does this energy disappear to?

    thanks
     
    Last edited by a moderator: May 2, 2017
  5. Dec 3, 2006 #4
    The work done is the area under the graph of pressure versus volume. When you increase the volume, the gas does positive work.

    [tex] W = P\Delta V [/tex]

    For non-constant pressure the work is:

    [tex] W = \int_{V_{1}}^{V_{2}} P\; dV [/tex]
     
    Last edited: Dec 3, 2006
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