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First Law of Thermodynamics issue

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data
    In an experiment to simulate conditions within an automobile engine, 0.155 mol of air at a temperature of 780K and a pressure of 3.20x106 Pa is contained in a cylinder of volume 310 cm3. Then 670 J of heat is transferred to the cylinder.

    A) If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas.

    B) If instead the volume of the cylinder is allowed to increase while the pressure remains constant, find the final temperature of the air.

    2. Relevant equations
    Const P W= pdV Q=nCpdT
    Const V W = 0 Q = nCvdT
    dU = nCdT
    Cp = 5/2R
    Cv = 3/2R

    3. The attempt at a solution
    For part A where Volume is constant, it told me that 670 J of heat is added which would be the Q value for dU. so the overall internal energy is 670 J from that I used the equation dU= nCv(Tf-Ti) which with values came out too 670 = (.155)(3/2*8.314)(Tf-780) which equaled 1126 K. which ended up not being the right answer. I do not know what else to try.

    For Part B, I know I need to figure out dU which is Q-W so for this equation it would come out to dU = 670 - pdV or dU = 670 - 3.20E6(Vf-3.1E-4m3). but don't really know where to go from here
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 21, 2014 #2

    • Nitrogen is a diatomic gas therefore Cv=2.5

      For part B, find Vf as you have been finding and then use the ideal gas relation:

      PV=nRT

     
  4. Mar 21, 2014 #3
    the Diatomic gas worked, forgot about that tidbit.

    but for the Part B I don't fully understand what you mean because I have 2 unknown variables V_f and T_f which is why I am a little lost.
     
  5. Mar 21, 2014 #4
    Figured it out thanks!!!!, took a good minute though
     
  6. Oct 30, 2016 #5
    Can you please share the workings of the solution
     
  7. Dec 13, 2016 #6
    For part b, you use the same equation and everything except that Cv is now Cp, which is just Cv + R.
     
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