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Change of temperature of ideal gas

  • Thread starter hubber26
  • Start date
  • #1
5
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Homework Statement



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)

Homework 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?

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?
 

Answers and Replies

  • #2
1,838
7
You took c to be c_p while you need c_v here.
 
  • #3
5
0
right. Thanks very much!
 

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