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Isobaric process

  • Thread starter Amith2006
  • Start date
  • #1
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Homework Statement



Consider 1 gram of an ideal gas undergoing isobaric process. Suppose del(H) be the amount of heat given to it. Then,
del(H) = dU + del(W)
del(H) = 1 x C(v)dT + (P.dV)/J
But del(H) = C(p)dT
P.dV = r.dT
C(p)dT = C(v)dT + (r.dT)/J
C(p) - C(v) = r/J


Homework Equations





The Attempt at a Solution



In the above derivation, when volume is changing, how can they take dU = mC(v)dT?
Here m = mass of gas,r = gas constant,J = Mechanical equivalent of heat
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
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In the above derivation, when volume is changing, how can they take dU = mC(v)dT?
Here m = mass of gas,r = gas constant,J = Mechanical equivalent of heat
U is a function of temperature only. It does not depend on volume or pressure (although those will affect temperature, of course). dU is always = mC(v)dT

AM
 
  • #3
427
2
That was a nice piece of information.Thanks.
 

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