- #1

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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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- Thread starter Rizer
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- #1

- 18

- 0

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

- #2

Andrew Mason

Science Advisor

Homework Helper

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It is correct for a

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

AM

- #3

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- 0

- #4

Andrew Mason

Science Advisor

Homework Helper

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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

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

Science Advisor

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This graph is for water liquid/vapor. Clearly not an ideal gas!The Enthalpy graph I found:

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!I also found this example:

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- #7

- 192

- 3

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

This is entirely consistent with writing

u = c

and

h = c

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

The formula you wrote with c

BBB

- #8

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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" [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 = c_{p}- c_{v}

This is entirely consistent with writing

u = c_{v}T

and

h = c_{p}T

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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