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Change of temperature of ideal gas

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data

    We have 1m3 of Ar gas M=39.948g/mol, pressure=101300Pa, temperature T=298K. We put the energy Q=1000J in that gas. Whats the temperature after we put 103J. (c=0.5203x103)

    2. Relevant equations

    Im not interested in solution. I tried to solve it on two ways using formula for internal energy of ideal gas [tex]\Delta[/tex]U = [tex]\frac{3}{2}[/tex]nR[tex]\Delta[/tex]T and the formula Q = mc[tex]\Delta[/tex]T

    However solutions are not the same, why?

    3. The attempt at a solution

    pV=nRT => n = pV / RT = 40.89mol
    m = Mn = 1633g = 1.633kg

    [tex]\Delta[/tex]U = [tex]\frac{3}{2}[/tex]nR[tex]\Delta[/tex]T
    thus [tex]\Delta[/tex]T = Q / [tex]\frac{3}{2}[/tex]nR = 1000 / (1.5*40.89*8.314) = 1.961°K

    however
    Q = mc[tex]\Delta[/tex]T
    thus [tex]\Delta[/tex]T = Q / mc = 1000 / (1.633*0.5203*103) = 1.177°K

    what is wrog please?
     
  2. jcsd
  3. May 7, 2009 #2
    You took c to be c_p while you need c_v here.
     
  4. May 7, 2009 #3
    right. Thanks very much!
     
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