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Derive the equation 1.40 from the equation 1.39

Short Answer

Expert verified

The Equation 1.40 from the equation 1.39 is derived

as Vγ P = Constant.

Step by step solution

01

Step 1. Given data.

Equation 1.39 from the book as below

VTf/2=³¦´Ç²Ô²õ³Ù²¹²Ô³Ù.⋯⋯⋯⋯⋯⋯(1.39).

02

Step 2. Derive Equation 1.40from1.39.

Relationship between Volume and temperature is given as

VTf2=Constant.................................(1)

Expression for the ideal gas equation

PV=NkTsimplifyT=PVNk

Apply the power f2on both sides, we get

Tf2=PVNkf2..............................(2)

Multiplying equation(2)by V on both sides and simplify


VTf2=VPVNkf2VTf2=V1+f2PNkf2

Using equation(1).

VTf2=V1+f2PNkf2=constantV1+f2Pf2=(Nk)f2·constant.................................(3)

Here N.k,fare the constants.

(Nk)f2·constant=constant

V1+f2Pf2=constant

03

Step 3. Adiabatic exponent.

Apply the power 2f

V2f1+f2Pf2·2f=constantV1+2fP=constant

1+2f=f+2f

Therefore,

Vf+2fP=constant

Defining the adiabatic exponent γ=f+2fthe relation for a quasi-static adiabatic process

VγP=constant.

Hence the equation 1.40 is obtained from equation 1.39.

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