Enthalpy and temperature

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I know enthalpy

H = U + PV

and for ideal gas, PV = nRT so it becomes

H = U + nRT

However, U = 3NkT/2 = 3nRT/2 for ideal gas as well.

So why is it incorrect to put

H = 3nRT/2 + nRT ??

Thanks
 

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  • #2
Andrew Mason
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I know enthalpy

H = U + PV

and for ideal gas, PV = nRT so it becomes

H = U + nRT

However, U = 3NkT/2 = 3nRT/2 for ideal gas as well.

So why is it incorrect to put

H = 3nRT/2 + nRT ??

Thanks
It is correct for a monatomic ideal gas (only).

AM
 
  • #3
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Thanks Andrew, but what about using H = 5nRT/2 + nRT or H = 7nRT/2 + nRT etc. for ideal gases made up of molecules???
 
  • #4
Andrew Mason
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Thanks Andrew, but what about using H = 5nRT/2 + nRT or H = 7nRT/2 + nRT etc. for ideal gases made up of molecules???
For non-monatomic ideal gas, the specific heat will depend upon temperature. A diatomic ideal gas will have a Cv of 5R/2 at temperatures below around 1000K but as temperatures increase the Cv increases and eventually reaches 7R/2

AM
 
  • #5
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But why does the graphs of Enthalpy usually depends on both T and P? Is it solely because the gas is not an ideal gas??

The Enthalpy graph I found:
http://www.nt.ntnu.no/users/haugwarb/Phase_diagrams_and_thermodynamic_tables/PhaseDiagrams/R134a.pdf [Broken]


I also found this example:
http://www.learnthermo.com/examples/ch05/p-5c-3.php

In EQN 4, the example applied the equation of state for ideal gas, so I suppose it is treating the gas as an ideal gas.
But when I try to calculate the enthalpy using H = (3 or 5 or 7)nRT/2 + nRT, the values mismatch the value of H1 = 87.41 kJ/kg and H2 = 239.78 kJ/kg in the example, even after converting kJ/mol into kJ/kg using the provided molar mass of 29g/mole...


Please help me on this, this issue has been troubling me for long, thank you very much
 
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  • #6
Bill_K
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The Enthalpy graph I found:
This graph is for water liquid/vapor. Clearly not an ideal gas! :smile:
I also found this example:
How do your answers for H compare to the values given in http://www.wiley.com/college/moran/CL_0471465704_S/user/tables/TABLE5S/table5sframe.html" [Broken] table for air? (Note that everything in the table is in Kelvin except for the first column!
 
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  • #7
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I know enthalpy

H = U + PV

and for ideal gas, PV = nRT so it becomes

H = U + nRT

However, U = 3NkT/2 = 3nRT/2 for ideal gas as well.

So why is it incorrect to put

H = 3nRT/2 + nRT ??

Thanks
For an ideal gas,

R = cp - cv

This is entirely consistent with writing

u = cvT

and

h = cpT

although you should be cautious to use these formulas only in the regime where the specific heats are constant. In the more general case, cv = du/dT and cp = dh/dT. If the specific heats are not constant, you have to integrate from some reference temperature.

The formula you wrote with cv=(3/2)R is only appropriate for monatomic ideal gases. As you point out, for diatomic ideal gases, the coefficient is 5/2 because of the extra degrees of freedom available at room temperature, although as the temperature rises above ~1E3 K for gases like O2 and N2, additional modes become unfrozen and cv/R rises. At some point the ideal gas model doesn't work all that well. I would point out that complex gases like R-134 aren't all that well-modeled by the ideal gas law.

BBB
 
  • #8
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This graph is for water liquid/vapor. Clearly not an ideal gas! :smile:

How do your answers for H compare to the values given in http://www.wiley.com/college/moran/CL_0471465704_S/user/tables/TABLE5S/table5sframe.html" [Broken] table for air? (Note that everything in the table is in Kelvin except for the first column!
Thanks alot! Now I am clear about the concept.

I tried to calculate h using the table you provided, and the values matched well with the formula H = 5nRT/2 + nRT

For an ideal gas,

R = cp - cv

This is entirely consistent with writing

u = cvT

and

h = cpT

although you should........
Thanks for your detailed explanations, which furthers my understandings on this topic. And now I am sure that R-134a can't be modeled as an ideal gas.
 
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