Understanding Enthalpy and Temperature in Ideal Gas Systems

[Rizer]

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

 

[Andrew Mason]

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

 

[Rizer]

Thanks Andrew, but what about using H = 5nRT/2 + nRT or H = 7nRT/2 + nRT etc. for ideal gases made up of molecules?

 

[Andrew Mason]

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

 

[Rizer]

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


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

 

[Bill_K]

The Enthalpy graph I found:

This graph is for water liquid/vapor. Clearly not an ideal gas!

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" table for air? (Note that everything in the table is in Kelvin except for the first column!

 

[bbbeard]

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 = c_p - c_v

This is entirely consistent with writing

u = c_vT

and

h = c_pT

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

The formula you wrote with c_v=(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 c_v/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

 

[Rizer]

This graph is for water liquid/vapor. Clearly not an ideal gas!

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" table for air? (Note that everything in the table is in Kelvin except for the first column!

Thanks a lot! 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 = c_p - c_v

This is entirely consistent with writing

u = c_vT

and

h = c_pT

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.