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

Question

Grade 12th passThermal Physics

Pawan joshi

7 Years agoGrade 12th pass

Answer 1

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)