Calculation of ΔH for Ideal Gas Processes

In summary: Therefore, the heat capacity (with consistent units) is Cp=0.193 Btu/lbmF.ChetIn summary, the problem involves calculating the variation of ΔH in Btu for three processes: heating carbon dioxide from 77 to 1000ºF, cooling benzene from its normal boiling point to 77ºF, and heating calcium carbonate from 70 to 1200ºF. The first two processes can be solved using the heat capacity and mass of the substances, while the third process involves using a heat capacity equation and the gas constant R. All calculations can be done using English units without the need for conversion.
  • #1
ruiwp13
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Homework Statement


Calculate the variation of ΔH in Btu for the following processes:

a) Heating of 7Ibmol of carbon dioxide from 77 to 1000ºF at 1atm using property tables for ideal gases.

b) Cooling of 22Ibmol of benzene from it's normal boiling point(80.1ºC) to 77ºF at 1atm.

c) Heating of 10tons of calcium carbonite from 70 to 1200ºF for which:

Cp/R=12.572+(2.637*10^-3)T - (3.120*10^5)T^-2 (T in K)


Homework Equations


ΔH=ΔU+ΔPV
h=u+pv
m=n*M


The Attempt at a Solution



a) and b) are very similar, so for a) I calculated the number of mols (1Ibmol=453.6mol) and I got 3175.2mol, then I got the M of carbon dioxide (44.01kg/kmol) and I multiplied it by 3,1752 obtaining 139.7405kg/kmol. I can get the mass trough m=n*M and on the table I have volume/kmol = 0.0943, but I don't know if I'm heading the right way because I need to know the volume @ 77ºF(25ºC) and @ 1000ºF(537ºC). And how do I find the ΔU in this case? Is it trough the Ibmol and the temperature difference?

in c) I can substitute the T for the temperature they gave me and I can find R. But I don't know after it what I need to do
 
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  • #2
ruiwp13 said:

Homework Statement


Calculate the variation of ΔH in Btu for the following processes:

a) Heating of 7Ibmol of carbon dioxide from 77 to 1000ºF at 1atm using property tables for ideal gases.

b) Cooling of 22Ibmol of benzene from it's normal boiling point(80.1ºC) to 77ºF at 1atm.

c) Heating of 10tons of calcium carbonite from 70 to 1200ºF for which:

Cp/R=12.572+(2.637*10^-3)T - (3.120*10^5)T^-2 (T in K)


Homework Equations


ΔH=ΔU+ΔPV
h=u+pv
m=n*M


The Attempt at a Solution



a) and b) are very similar, so for a) I calculated the number of mols (1Ibmol=453.6mol) and I got 3175.2mol, then I got the M of carbon dioxide (44.01kg/kmol) and I multiplied it by 3,1752 obtaining 139.7405kg/kmol. I can get the mass trough m=n*M and on the table I have volume/kmol = 0.0943, but I don't know if I'm heading the right way because I need to know the volume @ 77ºF(25ºC) and @ 1000ºF(537ºC). And how do I find the ΔU in this case? Is it trough the Ibmol and the temperature difference?

It's very hard to help you unless we know the nature of the data you have available to you. Do you have a table of enthalpies or internal energies, or do you have an expression for the heat capacities for (a) and (b).

Either way, you should be able to do the whole problem in terms of lbm, °F, and BTU without converting to metric.

Chet
 
  • #3
Chestermiller said:
It's very hard to help you unless we know the nature of the data you have available to you. Do you have a table of enthalpies or internal energies, or do you have an expression for the heat capacities for (a) and (b).

Either way, you should be able to do the whole problem in terms of lbm, °F, and BTU without converting to metric.

Chet

For a) and b) I have the property tables of ideal gases, for c) I have that expression. I'm finding it hard because I don't really know which table to use and because I'm having trouble calculating the ΔU. The only thing the problem says is to use the property tables of ideal gases.
 
  • #4
ruiwp13 said:
For a) and b) I have the property tables of ideal gases, for c) I have that expression. I'm finding it hard because I don't really know which table to use and because I'm having trouble calculating the ΔU. The only thing the problem says is to use the property tables of ideal gases.

OK. In part (a), you know that, for an ideal gas, dH=mCpdT, so you don't need to work through the internal energy. What is the parameter used in your property tables, Cp or H?

In part (b), you can again use the heat capacity (in this case for liquid benzene) to calculate the change in enthalpy. Do you have heat capacity data for benzene liquid?

In part (c), you have the heat capacity equation given for CaCO3 solid. So, you do the same thing again. Also note that R = 1.987 cal/gmoleC =1.987 Btu/lbmole-F, so you already know the heat capacity in the correct units for Btu's.

Chet
 
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  • #5
Chestermiller said:
OK. In part (a), you know that, for an ideal gas, dH=mCpdT, so you don't need to work through the internal energy. What is the parameter used in your property tables, Cp or H?

In part (b), you can again use the heat capacity (in this case for liquid benzene) to calculate the change in enthalpy. Do you have heat capacity data for benzene liquid?

In part (c), you have the heat capacity equation given for CaCO3 solid. So, you do the same thing again. Also note that R = 1.987 cal/gmC =1.987 Btu/lbmF, so you already know the heat capacity in the correct units for Btu's.

Chet

For a) I have the specific heat at 300K and 800K. H=mCpT where m is the mass, Cp the specific heat and T the temperature in F? For the mass I have the molar mass(139.7405) and number of moles(3175,2) so I just need to multiply them to get the mass right?

Best Regards
 
  • #6
ruiwp13 said:
For a) I have the specific heat at 300K and 800K. H=mCpT where m is the mass, Cp the specific heat and T the temperature in F? For the mass I have the molar mass(139.7405) and number of moles(3175,2) so I just need to multiply them to get the mass right?

Best Regards
Let me understand correctly. You have the specific heat of CO2 at 300K and 800K? H is not given by the equation you wrote. It should be ΔH=m∫CpdT. What units do you have Cp in? There is no need to convert to metric quantities. Try working entirely with the english units. There are 44 lbm of CO2 in 1 lbmole.

Chet
 
  • #7
Chestermiller said:
Let me understand correctly. You have the specific heat of CO2 at 300K and 800K? H is not given by the equation you wrote. It should be ΔH=m∫CpdT. What units do you have Cp in? There is no need to convert to metric quantities. Try working entirely with the english units. There are 44 lbm of CO2 in 1 lbmole.

Chet

I have the specific heat in (kJ/kgK). I got it from this website http://www.engineeringtoolbox.com/carbon-dioxide-d_974.html.
But with that how do I do the m∫CpdT part? The integral is between the two temperatures but I have two Cp's according to that table.

I also found this table, could be more appropriate: http://thermo.sdsu.edu/testcenter/testhome/Test/solve/basics/tables/tablesIG/igCO2.html
 
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  • #8
ruiwp13 said:
I have the specific heat in (kJ/kgK). I got it from this website http://www.engineeringtoolbox.com/carbon-dioxide-d_974.html.
But with that how do I do the m∫CpdT part? The integral is between the two temperatures but I have two Cp's according to that table.

I also found this table, could be more appropriate: http://thermo.sdsu.edu/testcenter/testhome/Test/solve/basics/tables/tablesIG/igCO2.html

The table I have has the h values
 
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  • #9
ruiwp13 said:
The table I have has the h values
For the Cp data, you would first fit a straight line to the data and then integrate (the equation). If you are using the table, you would linearly interpolate between values in the table. The results you get from integrating the heat capacity should closely match the results you get from interpolating in the table.

Chet
 
  • #10
Chestermiller said:
For the Cp data, you would first fit a straight line to the data and then integrate (the equation). If you are using the table, you would linearly interpolate between values in the table. The results you get from integrating the heat capacity should closely match the results you get from interpolating in the table.

Chet

With the h values according to the second table that are 0 for 298K and 23330 for 810K, to get the ΔH I just need to do (h2-h1)*m right?
 
  • #11
ruiwp13 said:
With the h values according to the second table that are 0 for 298K and 23330 for 810K, to get the ΔH I just need to do (h2-h1)*m right?
Yes, that's fine (811K, but who's counting?). Chet.
 
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  • #12
Chestermiller said:
Yes, that's fine (811K, but who's counting?). Chet.

Eheh thank you so much Chet, sorry for any incovinience.

Best Regards
 

FAQ: Calculation of ΔH for Ideal Gas Processes

1. What is thermodynamics enthalpy?

Thermodynamics enthalpy is a thermodynamic property of a system that represents the total amount of energy in a system, including both internal energy and the energy required to create or maintain the system's pressure and volume.

2. How is enthalpy different from internal energy?

Enthalpy takes into account the energy required to create or maintain a system's pressure and volume, while internal energy only considers the total amount of energy in a system.

3. What is the significance of enthalpy in thermodynamics?

Enthalpy is used to quantify the energy transfer in a thermodynamic process, making it a crucial concept in understanding and analyzing various thermodynamic systems and processes.

4. How is enthalpy measured?

Enthalpy is typically measured in units of joules (J) or kilojoules (kJ). It can be measured experimentally using calorimetry or calculated using mathematical equations based on the properties of a system.

5. Can enthalpy be negative?

Yes, enthalpy can be negative. A negative enthalpy value indicates that energy has been released from the system, while a positive value indicates that energy has been absorbed by the system.

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