Explanation Write the expression to calculate the mole number of N 2 ( N N 2 ) . N N 2 = m N 2 M N 2 (I) Here, molar mass of N 2 is M N 2 and mass of N 2 is m N 2 . Write the expression to calculate the mole number of CO 2 ( N CO 2 ) . N CO 2 = m CO 2 M CO 2 (II) Here, molar mass of CO 2 is M CO 2 and mass of CO 2 is m CO 2 . Write the expression to calculate the total number of moles ( N m ) . N m = N CO 2 + N N 2 (III) Write the expression to calculate the apparent molecular weight of the mixture ( M m ) . M m = m m N m (IV) Write the expression to calculate the constant volume specific heat of the mixture ( c v ) . c v = mf N 2 c v N 2 + mf C O 2 c v , C O 2 (V) Here, mole fraction of N 2 , and CO 2 is mf N 2 , and mf C O 2 , specific heat of N 2 is c v , N 2 , and specific heat of CO 2 is c v , CO 2 . Write the expression to calculate the apparent gas constant of the mixture ( R ) . R = R u M m (VI) Here, universal gas constant is R u . Write the expression to calculate the final volume ν 2 . ν 2 = 2 ν 1 (VII) Write the expression for the mass contained in the system ( m ) . m = P 1 ν 1 R T 1 (VIII) Write the expression to calculate the final temperature ( T 2 ) T 2 = P 2 ν 2 m R (IX) Write the expression to calculate the work done during the process. W o u t = P 1 + P 2 2 ( ν 2 − ν 1 ) (X) Conclusion: From Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the values of molar masses for N 2 and CO 2 as 28.2 kg / kmol and 44.0 kg / kmol respectively, From Table A-2a, “Ideal-gas specific heats of various common gases”, obtain the following properties for nitrogen and carbon dioxide. For N 2 , c v = 0.743 kJ / kg ⋅ K For CO 2 , c v = 0.657 kJ / kg ⋅ K Substitute 55 kg for m N 2 and 28 kg / mol for M N 2 in Equation (I). N N 2 = 55 kg 28 kg / mol = 1.964 kmol Substitute 45 kg for m CO 2 and 28 kg / mol for M CO 2 in Equation (II). N CO 2 = 45 kg 28 kg / mol = 1.023 kmol Substitute 1.023 kmol for N CO 2 , and 1.964 kmol for N N 2 in Equation (III). N m = 1.023 kmol + 1.964 kmol . = 2.987 kmol Substitute 100 kg for N m and 2.987 kmol for m m in Equation (IV). M m = 100 kg 2.987 kmol = 33.48 kg / kmol Substitute 0.743 kJ / kg ⋅ K for c v , N 2 , 0.657 kJ / kg ⋅ K for c v , CO 2 , 0.55 for mf N 2 , and 0.45 for mf C O 2 in Equation (V). c v = ( 0.55 ) ( 0.743 kJ / kg ⋅ K ) + ( 0.45 ) ( 0.657 kJ / kg ⋅ K ) = 0.7043 kJ / kg ⋅ K Substitute 8.134 kJ / Kmol ⋅ K for R u and 33

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Chapter 13.3, Problem 95RP

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The total work done and heat transfer for the process

Solution Summary: The author explains the total work done and heat transfer for the process. The expression to calculate the mole number of CO_2.

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