Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Chapter 16.6, Problem 94RP

a)

To determine

The amount of heat required for the process.

a)

Expert Solution
Check Mark

Answer to Problem 94RP

The amount of heat required for the process is 324,033kJ.

Explanation of Solution

Write the energy balance equation for the reported process.

EinEout=ΔEsystem (I)

Here, input energy is Ein, output energy is Eout, and the change in system energy is ΔEsystem.

Write the expression to obtain the amount of heat required for the process (Qin).

Qin=N(u¯2u¯1) (II)

Here, number of moles is N, internal energy of the system at state 1 is u¯1, and internal energy of the system at state 2 is u¯2.

Write the expression to obtain the internal energy of the system at state 1 (u¯1).

u¯1=h1RuT1 (III)

Here, enthalpy of the system at state 1 is h1, universal gas constant is Ru, and temperature of the system at state 1 is T1.

Write the expression to obtain the internal energy of the system at state 2 (u¯2).

u¯2=h2RuT2 (IV)

Here, enthalpy of the system at state 2 is h2, and temperature of the system at state 2 is T2.

Write the expression to obtain the change in enthalpy of the system (h2h1).

h2h1=12cpdT (V)

Conclusion:

Substitute (h1RuT1) for u¯1, and (h2RuT2) for u¯2 in Equation (II).

Qin=N(u¯2u¯1)=N(h2RuT2(h1RuT1))=N(h2h1Ru(T2T1)) (VI)

Refer Table A-2c, “Ideal-gas specific heats of various common gases”, obtain the specific heat relation as a+bT+cT2+dT3.

Substitute a+bT+cT2+dT3 for cp in Equation (V) and then integrate.

h2h1=12cpdT=12(a+bT+cT2+dT3)dT=a[T]12+b[T22]12+c[T33]12+d[T44]12=a[T2T1]+b2[T22T12]+c3[T23T13]+d4[T24T14] (VII)

Here, constants are a, b, c and d.

Refer Table A-2c, “Ideal-gas specific heats of various common gases”, obtain the values of constants a, b, c and d for methane as 19.89, 5.024×102, 1.269×105, and 11.01×109 respectively.

Substitute 19.89 for a, 5.024×102 for b, 1.269×105 for c, 11.01×109 for d, 1,000 K for T2, and 298 K for T1 in Equation (VII).

h2h1=a[T2T1]+b2[T22T12]+c3[T23T13]+d4[T24T14]={19.89(1,000K298K)+(5.024×1022)((1,000K)2(298K)2)+(1.269×1053)((1,000K)3(298K)3)+(11.01×1094)((1,000K)4(298K)4)}=13,962.78+22,889+4,1182,730=38,239kJ/kmol

Substitute 38,239kJ/kmol for (h2h1), 8.314kJ/kmolK for Ru, 1,000 K for T2, and 298 K for T1 in Equation (VI).

Qin=N(h2h1Ru(T2T1))=(10mol)(38,239kJ/kmol(8.314kJ/kmolK)(1,000K298K))=324,033kJ

Thus, the amount of heat required for the process is 324,033kJ.

b)

To determine

The amount of heat required for the process.

b)

Expert Solution
Check Mark

Answer to Problem 94RP

The amount of heat required for the process is 245,200kJ.

Explanation of Solution

Write the stoichiometric reaction for the dissociation process.

CH4C+2H2

From the stoichiometric reaction, infer that the stoichiometric coefficient for methane (vCH4) is 1, for hydrogen (vH2) is 2, and for carbon (vC) is 1.

Write the expression to obtain the actual reaction for the dissociation process.

CH4xCH4+yC+zH2 (VIII)

From the actual reaction, infer that the equilibrium composition contains x amount of methane (NCH4), y amount of carbon (NC), and z amount of hydrogen (NH2).

Write the expression to obtain the total number of moles (Ntotal).

Ntotal=NCH4+NC+NH2 (IX)

Here, number of moles of CH4 is NCH4, number of moles of C is NC, and number of moles of H2 is NH2.

Write the expression to obtain the equilibrium constant (Kp) for the dissociation process.

Kp=NCvCNH2vH2NCH4vCH4(PNtotal)(vC+vH2vCH4) (X)

Here, pressure is P.

Write the expression to obtain the mole fraction of Methane (yCH4).

yCH4=NCH4Ntotal (XI)

Write the expression to obtain the mole fraction of carbon (yC).

yC=NCNtotal (XII)

Write the expression to obtain the mole fraction of hydrogen (yH2).

yH2=NH2Ntotal (XIII)

Write the expression to obtain the amount of heat required for the process (Qin).

Qin=N(yCH4cV, CH4T2+yH2cV, H2T2+yCcV, CT2)NcV, CH4T1 (XIV)

Here, specific heat of methane is cV, CH4, specific heat of hydrogen is cV, H2, and specific heat of carbon is cV, C.

Conclusion:

Write the carbon balance equation from Equation (VIII).

1=x+yy=1x (XV)

Write the hydrogen balance equation from Equation (VIII).

4=4x+2zz=22x (XVI)

Substitute x for NCH4, y for NC, z for NH2, 1x for y, and 22x for z in Equation (IX).

Ntotal=x+y+z=x+1x+22x=32x (XVII)

Substitute e2.328 for Kp, x for NCH4, y for NC, z for NH2, 1x for y, 22x for z.

32x for Ntotal,1 atm for P, 2 for vH2, and 1 for both vC and vCH4 in Equation (X).

e2.328=(1x)(22x)2x(1atm32x)(1+21)0.0975x=(1x)(22x)2(1(32x)2)0.0975x[9+4x212x]=(1x)(4+4x28x)0.8775x+0.39x31.17x2=(4+4x28x4x4x3+8x2)

3.61x313.17x2+12.8775x4=0x=0.641

Substitute 0.641 for x in equation (XV).

y=1x=10.641=0.359

Substitute 0.641 for x in equation (XVI).

z=22x=22(0.641)=0.718

Substitute 0.641 for x in equation (XVII).

Ntotal=32x=32(0.641)=1.718

Substitute 0.641 for x, 0.359 for y, and 0.718 for z in Equation (VIII).

CH4xCH4+yC+zH2CH40.641CH4+0.359C+0.718H2

Substitute 0.641 for x, and 1.718 for Ntotal in Equation (XI).

yCH4=0.6411.718=0.37

Substitute 0.359 for x, and 1.718 for Ntotal in Equation (XII).

yC=0.3591.718=0.2

Substitute 0.718 for x, and 1.718 for Ntotal in Equation (XIII).

yH2=0.7181.718=0.41

Substitute 10 kmol for N, 0.37 for yCH4, 63.3kJ/kmolK for cV, CH4, 0.41 for yH2, 21.7kJ/kmolK for cV, H2, 0.2 for yC, 0.711kJ/kmolK for cV, C, 1,000 K for T2, and 298 K for T1 in Equation (XIV).

Qin={(10kmol)((0.37)(63.3kJ/kmolK)(1,000K)+(0.41)(21.7kJ/kmolK)(1,000K)+(0.2)(0.711kJ/kmolK)(1,000K))(10kmol)(27.8kJ/kmolK)(298K)}=245,200kJ

Thus, the amount of heat required for the process is 245,200kJ.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
The products from the combustion of a stoichiometric mixture of CO and O2 are at a pressure of latm and a certain temperature. The products analysis shows that 35% of each kmol if CO, is dissociated. Determine the equilibrium constant for this temperature, and hence find the percentage dissociation when the products are at the same temperature but compressed to 10 atmospheres.
At 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane, C5H10. What is the mole fraction of cyclopentane in the vapor that is in equilibrium with this solution? The vapor pressure of the pure liquids at 25°C are 451 torr for pentane and 321 torr for cyclopentane. Assume that the solution is an ideal solution.    (a) 0.284       (b) 0.551       (c) 0.716       (d) 0.643       (e) 0.357
The standard enthalpy of formation of solid barium oxide, BaO, is –553.5 kJ/mol, andthe standard enthalpy of formation of barium peroxide, BaO2, is –634.3 kJ/mol.(a) Calculate the standard enthalpy change for the following reaction. Is the reactionexothermic or endothermic?2 BaO2(s) → 2 BaO(s) + O2(g)(b) Draw an energy level diagram that shows the relationship between the enthalpychange of the decomposition of BaO2, to BaO and O2, and the enthalpies of formationof BaO2(s) and BaO2(s)

Chapter 16 Solutions

Thermodynamics: An Engineering Approach

Ch. 16.6 - Prob. 11PCh. 16.6 - Prob. 12PCh. 16.6 - Prob. 13PCh. 16.6 - Prob. 14PCh. 16.6 - Prob. 15PCh. 16.6 - Prob. 16PCh. 16.6 - Prob. 17PCh. 16.6 - Prob. 18PCh. 16.6 - Prob. 19PCh. 16.6 - Prob. 20PCh. 16.6 - Prob. 21PCh. 16.6 - Prob. 22PCh. 16.6 - Prob. 23PCh. 16.6 - Determine the equilibrium constant KP for the...Ch. 16.6 - Prob. 26PCh. 16.6 - Prob. 27PCh. 16.6 - Carbon monoxide is burned with 100 percent excess...Ch. 16.6 - Prob. 30PCh. 16.6 - Prob. 31PCh. 16.6 - Estimate KP for the following equilibrium reaction...Ch. 16.6 - Prob. 33PCh. 16.6 - A mixture of 3 mol of N2, 1 mol of O2, and 0.1 mol...Ch. 16.6 - Prob. 35PCh. 16.6 - Prob. 36PCh. 16.6 - Prob. 37PCh. 16.6 - Prob. 38PCh. 16.6 - Prob. 40PCh. 16.6 - What is the equilibrium criterion for systems that...Ch. 16.6 - Prob. 43PCh. 16.6 - Prob. 44PCh. 16.6 - Prob. 45PCh. 16.6 - Prob. 47PCh. 16.6 - Prob. 48PCh. 16.6 - Prob. 51PCh. 16.6 - Prob. 52PCh. 16.6 - Prob. 53PCh. 16.6 - Prob. 54PCh. 16.6 - Prob. 55PCh. 16.6 - Prob. 56PCh. 16.6 - Prob. 58PCh. 16.6 - Prob. 59PCh. 16.6 - Prob. 60PCh. 16.6 - Prob. 61PCh. 16.6 - Using the Henrys constant data for a gas dissolved...Ch. 16.6 - Prob. 63PCh. 16.6 - Prob. 64PCh. 16.6 - Prob. 65PCh. 16.6 - Prob. 66PCh. 16.6 - A liquid-vapor mixture of refrigerant-134a is at...Ch. 16.6 - Prob. 68PCh. 16.6 - Prob. 69PCh. 16.6 - An oxygennitrogen mixture consists of 30 kg of...Ch. 16.6 - Prob. 71PCh. 16.6 - Prob. 72PCh. 16.6 - Prob. 73PCh. 16.6 - Prob. 74PCh. 16.6 - Prob. 75PCh. 16.6 - Prob. 76PCh. 16.6 - An ammoniawater absorption refrigeration unit...Ch. 16.6 - Prob. 78PCh. 16.6 - Prob. 79PCh. 16.6 - Prob. 80PCh. 16.6 - One lbmol of refrigerant-134a is mixed with 1...Ch. 16.6 - Prob. 82RPCh. 16.6 - Prob. 83RPCh. 16.6 - Prob. 84RPCh. 16.6 - Prob. 85RPCh. 16.6 - Prob. 88RPCh. 16.6 - Prob. 89RPCh. 16.6 - Prob. 90RPCh. 16.6 - Prob. 91RPCh. 16.6 - Prob. 92RPCh. 16.6 - A constant-volume tank contains a mixture of 1 mol...Ch. 16.6 - Prob. 94RPCh. 16.6 - Prob. 95RPCh. 16.6 - Prob. 96RPCh. 16.6 - Prob. 97RPCh. 16.6 - Prob. 99RPCh. 16.6 - Consider a glass of water in a room at 25C and 100...Ch. 16.6 - Prob. 101RPCh. 16.6 - Prob. 102RPCh. 16.6 - Prob. 105RPCh. 16.6 - Prob. 106RPCh. 16.6 - Prob. 107RPCh. 16.6 - Prob. 108RPCh. 16.6 - Prob. 109FEPCh. 16.6 - Prob. 110FEPCh. 16.6 - Prob. 111FEPCh. 16.6 - Prob. 112FEPCh. 16.6 - Prob. 113FEPCh. 16.6 - Prob. 114FEPCh. 16.6 - Propane C3H8 is burned with air, and the...Ch. 16.6 - Prob. 116FEPCh. 16.6 - Prob. 117FEPCh. 16.6 - The solubility of nitrogen gas in rubber at 25C is...
Knowledge Booster
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
    • SEE MORE QUESTIONS
    Recommended textbooks for you
  • Elements Of Electromagnetics
    Mechanical Engineering
    ISBN:9780190698614
    Author:Sadiku, Matthew N. O.
    Publisher:Oxford University Press
    Mechanics of Materials (10th Edition)
    Mechanical Engineering
    ISBN:9780134319650
    Author:Russell C. Hibbeler
    Publisher:PEARSON
    Thermodynamics: An Engineering Approach
    Mechanical Engineering
    ISBN:9781259822674
    Author:Yunus A. Cengel Dr., Michael A. Boles
    Publisher:McGraw-Hill Education
  • Control Systems Engineering
    Mechanical Engineering
    ISBN:9781118170519
    Author:Norman S. Nise
    Publisher:WILEY
    Mechanics of Materials (MindTap Course List)
    Mechanical Engineering
    ISBN:9781337093347
    Author:Barry J. Goodno, James M. Gere
    Publisher:Cengage Learning
    Engineering Mechanics: Statics
    Mechanical Engineering
    ISBN:9781118807330
    Author:James L. Meriam, L. G. Kraige, J. N. Bolton
    Publisher:WILEY
  • Elements Of Electromagnetics
    Mechanical Engineering
    ISBN:9780190698614
    Author:Sadiku, Matthew N. O.
    Publisher:Oxford University Press
    Mechanics of Materials (10th Edition)
    Mechanical Engineering
    ISBN:9780134319650
    Author:Russell C. Hibbeler
    Publisher:PEARSON
    Thermodynamics: An Engineering Approach
    Mechanical Engineering
    ISBN:9781259822674
    Author:Yunus A. Cengel Dr., Michael A. Boles
    Publisher:McGraw-Hill Education
    Control Systems Engineering
    Mechanical Engineering
    ISBN:9781118170519
    Author:Norman S. Nise
    Publisher:WILEY
    Mechanics of Materials (MindTap Course List)
    Mechanical Engineering
    ISBN:9781337093347
    Author:Barry J. Goodno, James M. Gere
    Publisher:Cengage Learning
    Engineering Mechanics: Statics
    Mechanical Engineering
    ISBN:9781118807330
    Author:James L. Meriam, L. G. Kraige, J. N. Bolton
    Publisher:WILEY
    Extent of Reaction; Author: LearnChemE;https://www.youtube.com/watch?v=__stMf3OLP4;License: Standard Youtube License