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Please solve the ques in the attachment. ...... . ... ... .

Pawan joshi , 7 Years ago
Grade 12th pass
anser 1 Answers
Arun
In an adiabatic process, No heat is exchanged
dQ=dU+PdV
0=dU+PdV
 d(a+bPV)+PdV=0
bPdV+bVdP+PdV=0
(b+1)PdV+bVdP=0
(b+1)(dV/V) +b(dP/P) = 0
 
Integrating we get
(b+1)logV +b(logP)=constant
 
 
logV(b+1) +logPb=constant
log (V(b+1)xPb)= constant
V(b+1)xPb=constant
divide by 
Vb and p(b-1) on both sides
 
PV((b+1)/b)= constant=PVr
 
r=((b+1)/b)
Last Activity: 7 Years ago
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