Explanation Write the differential of entropy ( d s ) as the functions of temperature, pressure, specific volume and specific heat at constant volume c v . d s = c v d T T + ( ∂ P ∂ T ) v d v (I) Write the differential of entropy ( d s ) as the functions of temperature, pressure, specific volume and specific heat at constant pressure c p . d s = c p d T T − ( ∂ v ∂ T ) P d P (II) Here, the temperature is T , the specific volume is v , the pressure is P , and the symbol ∂ indicates the partial derivative of variables. Write the cyclic relation of P , T , v as follows. ( ∂ v ∂ T ) P ( ∂ P ∂ v ) T ( ∂ T ∂ P ) v = − 1 (III) The specific heat ratio ( k ) is expressed as follows. k = c p c v (IV) Write the formula for isothermal compressibility ( α ) . α = − 1 v ( ∂ v ∂ P ) T (V) Conclusion: When the entropy ( s ) is kept constant, the differential entropy or change in entropy becomes zero. s = constant d s = 0 Here, the rate of change of specific volume with respect to pressure at constant entropy is expressed as follows. ∂ v ∂ P = ( ∂ v ∂ P ) s Substitute 0 for d s in Equation (I) to obtain c v . 0 = c v d T T + ( ∂ P ∂ T ) v d v c v d T T = − ( ∂ P ∂ T ) v d v (VI) Substitute 0 for d s in Equation (II) to obtain c p . 0 = c p d T T − ( ∂ v ∂ T ) P d P c p d T T = ( ∂ v ∂ T ) P d P (VII) Divide the Equation (VII) by Equation (VI)

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Chapter 12.6, Problem 52P

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To show that k=cpcv=−vα(∂v∂P)s.

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To show that k = c p c v = − v α ( ∂ v ∂ P ) s .

Solution Summary: The author explains the differential of entropy (ds) as the functions of temperature, pressure, specific volume and specific heat at constant volume.

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To show that k=cpcv=−vα(∂v∂P)s.

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