Explanation The given equation of state of helium 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 enthalpy ( d h ) given in Equation (12-29). d h = c P d T + [ v − 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 and R T P + a for v in Equation (III). d h = c P d T + [ R T P + a − T ( R P ) ] d P = c P d T + ( R T P + a − R T P ) d P = c P d T + ( a ) d P = c P d T + a d P (IV) Integrate the Equation (IV) with two states 1 and 2 gives as follows. h 2 − h 1 = c P ( T 2 − T 1 ) + a ( P 2 − P 1 ) (V) Here, the change in enthalpy is h 2 − h 1 and the subscripts 1 and 2 indicates the two states of helium. Write the formula for change in enthalpy of ideal gas model. d h = c P d T When integrated for two states, h 2 − h 1 = c P ( T 2 − T 1 ) (VI) Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”. The specific heat of helium at constant pressure ( c P ) is 5

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Author: Yunus A. Cengel Dr., Michael A. Boles

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Chapter 12.6, Problem 38P

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Chapter 12.6, Problem 37P

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

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

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