1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook