Question

# The relation between internal energy U, pressure P and volume V of a gas in an adiabatic process is given as U=a+bPV where a and b are constants. What is the value of adiabatic constant (γ)?abb+1ba+1aba

Solution

## The correct option is B b+1bFrom first law of thermodynamics, dQ=dU+dW ...(i) we can write small amount of work done as dW=PdV For an adiabatic process heat supplied is zero i.e dQ=0 From Eq(i), ⇒dQ=dU+PdV ⇒0=dU+PdV substituting value of U  ⇒d(a+bPV)+PdV=0 ...(ii) since a & b are constants, da=0, db=0 Applying chain rule for finding d(bPV) and putting in Eq (ii)  ⇒bPdV+bVdP+PdV=0 or (b+1)PdV+bVdP=0 Dividing both sides by PV, ⇒(b+1)dVV+bdPP=0 ...(iii) Integrating Eq.(iii) both sides, ⇒(b+1)logV+blogP=constant ⇒logVb+1+logPb=constant Applying logm+logn=log(mn) ⇒log(Vb+1Pb)=constant ⇒Vb+1Pb=constant Taking power 1b both sides, ⇒PVb+1b=constant1b ⇒PVb+1b=constant ...(iv) Comparing Eq. (iv) with PVγ=constant ∴γ=b+1b

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