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An ideal gas is heated from 500 to 1500 K. Find the change

Fundamentals of Thermodynamcs | 8th Edition | ISBN: 9781118131992 | Authors: Claus Borgnakke, Richard E. Sonntag

Problem 127HP Chapter 3

Fundamentals of Thermodynamcs | 8th Edition

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Fundamentals of Thermodynamcs | 8th Edition | ISBN: 9781118131992 | Authors: Claus Borgnakke, Richard E. Sonntag

Fundamentals of Thermodynamcs | 8th Edition

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Problem 127HP

An ideal gas is heated from 500 to 1500 K. Find the change in enthalpy using constant specific heat from Table A.5 (room temperature value) and discuss the accuracy of the result if the gas is

a. Argon

b. Oxygen

c. Carbon dioxide

TABLE A.5

Properties of Various Ideal Gases at 25°C, 100 kPa* (SI Units)

Gas

ChemicalFormula

Molecular Mass(kg/kmol)

R(kJ/kg-K)

ρ (kg/m3)

Cp0(kJ/kg-K)

Cv0(kJ/kg-K)

Steam

H2O

18.015

0.4615

0.0231

1.872

1.410

1.327

Acetylene

C2H2

26.038

0.3193

1.05

1.699

1.380

1.231

Air

28.97

0.287

1.169

1.004

0.717

1.400

Ammonia

NH3

17.031

0.4882

0.694

2.130

1.642

1.297

Argon

Ar

39.948

0.2081

1.613

0.520

0.312

1.667

Butane

C4H10

58.124

0.1430

2.407

1.716

1.573

1.091

Carton dioxide

CO2

44.01

0.1889

1.775

0.842

0.653

1.289

Carton monoxide

CO

28.01

0.2968

1.13

1.041

0.744

1.399

Ethane

C2H6

30.07

0.2765

1.222

1.766

1.490

1.186

Ethanol

C2H5OH

46.069

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Step-by-Step Solution:

Solution 127HP

Part (a)

Step 1 of 7:

Our aim is to find the change in enthalpy of the Argon (ideal gas condition) and to discuss the accuracy of the change in enthalpy.

The initial temperature T1 = 500 K

The final temperature T2 = 1500 K

The heat capacity of Argon Cp0 = 0.52 kJ/kg.K

The change in temperature ΔT = T2 - T1 = 1000 K

Step 2 of 7:

The change in enthalpy is expressed as

Putting the Values

The change in the enthalpy of Argon is 520 kJ/kg.

The equation for specific heat is given by (Table A.6)

Where C0 = 0.52 kJ/kg.K and rest of the values (C0 - C3) are zero for argon. The change in enthalpy matches with the standard value as the equation given in the table A.6 does not affect the specific heat value.

Part (b)

Step 3 of 7:

Our aim is to find the change in enthalpy of the Oxygen and to discuss the accuracy of the change in enthalpy.

The initial temperature T1 = 500 K

The final temperature T2 = 1500 K

The heat capacity of Oxygen Cp0 = 0.992 kJ/kg.K

The change in temperature ΔT = T2 - T1 = 1000 K

Step 4 of 7:

The change in enthalpy is expressed as

Putting the Values

The change in the enthalpy of Oxygen is 992 kJ/kg.

Step 5 of 7

Chapter 3, Problem 127HP is Solved
Step 6 of 7

Textbook: Fundamentals of Thermodynamcs
Edition: 8th
Author: Claus Borgnakke, Richard E. Sonntag
ISBN: 9781118131992

Since the solution to 127HP from 3 chapter was answered, more than 228 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Fundamentals of Thermodynamcs , edition: 8th. Fundamentals of Thermodynamcs was written by Sieva Kozinsky and is associated to the ISBN: 9781118131992. The answer to “An ideal gas is heated from 500 to 1500 K. Find the change in enthalpy using constant specific heat from Table A.5 (room temperature value) and discuss the accuracy of the result if the gas isa. Argon________________b. Oxygen________________c. Carbon dioxideTABLE A.5Properties of Various Ideal Gases at 25°C, 100 kPa* (SI Units)GasChemicalFormulaMolecular Mass(kg/kmol)R(kJ/kg-K)? (kg/m3)Cp0(kJ/kg-K)Cv0(kJ/kg-K) SteamH2O18.0150.46150.02311.8721.4101.327AcetyleneC2H226.0380.31931.051.6991.3801.231Air—28.970.2871.1691.0040.7171.400AmmoniaNH317.0310.48820.6942.1301.6421.297ArgonAr39.9480.20811.6130.5200.3121.667ButaneC4H1058.1240.14302.4071.7161.5731.091Carton dioxideCO244.010.18891.7750.8420.6531.289Carton monoxideCO28.010.29681.131.0410.7441.399EthaneC2H630.070.27651.2221.7661.4901.186EthanolC2H5OH46.0690.18051.8831.4271.2461.145EthyleneC2H428.0540.29641.1381.5481.2521.237HeliumHe4.0032.07710.16155.1933.1161.667HydrogenH22.0164.12430.081314.20910.0851.409MethaneCH416.0430.51830.6482.2541.7361.299MethanolCH3OH32.0420.25951.311.4051.1461.227NeonNe20.1830.41200.8141.030.6181.667Nitric oxideNO30.0060.27711.210.9930.7161.387NitrogenN228.0130.29681.131.0420.7451.400Nitrous oxideN2O44.0130.18891.7750.8790.6901.274n-OctaneC8H18114.230.072790.0921.7111.6381.044OxygenO231.9990.25981.2920.9220.6621.393PropaneC3H844.0940.18861.8081.6791.4901.126R-12CCI2F2120.9140.068764.980.6160.5471.126R-22CHCIF286.4690.096163.540.6580.5621.171R-32CF2H252.0240.15982.1250.8220.6621.242R-125CHF2CF3120.0220.069274.9180.7910.7221.097R-134aCF3CH2F102.030.081494.200.8520.7711.106R-410a—72.5850.114552.9670.8090.6941.165Sulfur dioxideSO264.0590.12982.6180.6240.4941.263Sulfur trioxideSO380.0530.103863.2720.6350.5311.196” is broken down into a number of easy to follow steps, and 60 words. The full step-by-step solution to problem: 127HP from chapter: 3 was answered by Sieva Kozinsky, our top Engineering and Tech solution expert on 08/03/17, 05:05AM. This full solution covers the following key subjects: gas, dioxide, oxide, oxygen, argon. This expansive textbook survival guide covers 7 chapters, and 1462 solutions.

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