Explanation The given equation of state of air is as follows. P ( v − a ) = R T (I) Here, the pressure is P , the specific volume is v , the universal gas constant is R , the temperature is T and the constant is a . Rewrite the Equation (I) as follows. v = R T P + a (II) Partially differentiate the Equation (II) with respect to the temperature ( T ) by keeping the pressure ( P ) as constant. ( ∂ v ∂ T ) v = R ( 1 ) P + 0 = R P Write the formula for change in entropy ( d s ) given in Equation (12-40). d s = c P d T T − ( ∂ v ∂ T ) P d P (III) Here, the specific heat at constant pressure is c P , the change in temperature is d T , the change in specific pressure is d P and the symbol ∂ indicates the partial derivative of variables. Substitute R P for ( ∂ v ∂ T ) P in Equation (III). d s = c P d T T − ( R P ) d P = c P d T T − R d P P (IV) Integrate the Equation (IV) with two states 1 and 2 gives as follows. s 2 − s 1 = c P ln ( T 2 T 1 ) − R ln ( P 2 P 1 ) (V) Here, the change in entropy is s 2 − s 1 and the subscripts 1 and 2 indicates the two states of air. Write the formula for change in entropy of ideal gas model. d s = c P d T T − R d P P When integrated for two states, s 2 − s 1 = c P ln ( T 2 T 1 ) − R ln ( P 2 P 1 ) (VI) Refer Table A-2 (b), “Ideal-gas specific heats of various common gases”. Obtain the specific heat of air at constant pressure ( c P ) corresponding to the average temperature of 300 ° C + 20 ° C 2 = 160 ° C ( 433 K ) - using interpolation method. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (VII) Show the temperature and specific heat values from the Table A-2 (b) as in below table

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Chapter 12.6, Problem 36P

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Chapter 12.6, Problem 35P

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The change in entropy of air, in kJ/kg and compare the result to the value obtained by using ideal gas equation of state.

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