Change in internal energy due to adiabatic expansion of ideal gas

  • Thread starter sudipmaity
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  • #1
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what is the change In internal energy of an ideal gas when it is expanded Adiabatically from (p °,v°) to ( p,v )?
The relevant equations :
dQ= dU+ pdv ; pV^ r = K (constant). r = Cp / Cv.
Attempted solution :
During adiabatic process dQ=0 ; p= k v ^ -r
Th f r : du= -k v^ -rdv
Integrating the above relation : Uf-U i = -k [ ( v ^ -r + 1 ) / -r + 1 ] v to v° .
= -k [ {(v ^ 1-r ) / (1-r)} -{(v° ^ 1-r ) /( 1-r) } ]
= 1 / (1-r)[ (k v ^ -r) v-( k v° ^-r ) v° ] = 1/ (1-r) [ pv-p° v° ] .
IS there any mistake ? Am i absolutely right ?? .
 

Answers and Replies

  • #2
TSny
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Integrating the above relation : Uf-U i = -k [ ( v ^ -r + 1 ) / -r + 1 ] v to v° .
= -k [ {(v ^ 1-r ) / (1-r)} -{(v° ^ 1-r ) /( 1-r) } ]
= 1 / (1-r)[ (k v ^ -r) v-( k v° ^-r ) v° ] = 1/ (1-r) [ pv-p° v° ] .
IS there any mistake ? Am i absolutely right ?? .

I believe you dropped the negative sign in front of the k when writing the last line. Otherwise, it looks good to me.

Using r = CP/Cv and CP-Cv = R , you can rewrite 1/(1-r) in terms of R and Cv if you wish.
 
  • #3
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My bad .what I intended to write was 1/(r-1)( pv-p° v° )
 
  • #4
TSny
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Good.

There's another approach to this problem. Internal energy U is a state variable. So, the change in U between the state (Po,Vo) and (P,V) doesn't depend on whether or not the process was adiabatic. Any process connecting those two states gives the same ΔU.

For an ideal gas, U = nCvT. Using the ideal gas law, PV = nRT, this can be written as U = (CV/R)PV. So, ΔU = (CV/R)(PV - PoVo) and you can show that CV/R = 1/(r-1).
 

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