Determine the internal energy change \triangle u of hydrogen, in

Gregory Emery 2021-12-15 Answered
Determine the internal energy change u of hydrogen, in kJ/kg, as it is heated from 200 to 800 K, using (a) the empirical specific heat equation as a function of temperature (Table A-2c), (b) the cv value at the average temperature (Table A-2b), and (c) the cv value at room temperature (Table A-2a).
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Expert Answer

vrangett
Answered 2021-12-16 Author has 36 answers
Step 1
Given:
- Initial temperature T1=200K
- Final temperature T2=800K
Required
- Determine the internal energy change of hydrogen using
a) The empirical specific heat equation
b) The cv value at average temperature
c) The cv value at room temperature
Step 2
Solution
Part a
- Using the empirical relation of cp(T) from table (A-2C) and relating it to cv(T)
cv(T)=cpRu=(aRu)+bT+cT2+dT3
Where (a=29.11,b=0.1916×102,c=0.4003×105,d=0.8704×109)
- The internal energy change could be defined as the following
u=12cv(T)dT=12[(aRu)+bT+cT2+dT3]dT
u=(aRu)(T2T1)+12b(T22T12)+13c(T23T13)+14d(T24T14)
u=(29.118.314)(800200)+12×(0.1916×102)×(80022002)+13×(0.4003×105)×(80032003)+14×(0.8704×109)×(80042004)=12487KJKmol

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Marcus Herman
Answered 2021-12-17 Author has 41 answers
Step 1
a) In this problem we need to determine the internal energy change u by using three different methods.
The first method is with the empirical specific heat equation. The equation gives us cp and we need the cv for the calculation.
cp=a+bT+cT2+dT3
cv=cpRu
cv=aRu+bT+cT2+dT3
The values for the constants a,b,c and d we find in the table A-2. We will also need the constant Ru
a=29.11
b=0.1916×102
c=0.4003×105
d=0.8704×109
Ru=8.31447kJkmolK
Step 2
To calculate the internal energy change per mole u we need to integrate the cv from the initial T1=200K to the final T2=800K temperature.
u=T1T2cv(T)dT
u=T1T2(aRu+bT+cT2+dT3)dT
u=T1T2(aRu+bT+cT2+dT3)dT
u=(aRu)T1T2dT+bT1T2TdT+cT1T2T2dT+dT1T2T3dT
nick1337
Answered 2021-12-27 Author has 575 answers

a) From Table A-2 C
cv=(aR)+bT+cT2+dT3
where: a=29.11
b=0.1916×102
c=0.4003×105
d=0.8704×109
Substituting:
u=(29.118.314)+(0.1916×102)(800200)+(0.4003×105)(80022002)+(0.8704×109)(80032003)
u=12487 kJ/kmol
u=6194 kJ/kg
b) From Table B-2
At 500 K, (average Temperature)
cv=10.839 kJ/kgK
u=cv(T2T1)
u=6233 kJ/kg
c) Table A-2a
cv=10.183 kJ/kgK
u=cv(T2T1)
u=6110 kJ/kg

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