Heat capaticy of isobaric process with real gas

  • Thread starter lakmus
  • Start date
  • #1
23
1

Main Question or Discussion Point

Hey,
during last lecture of thermodynamics we did polytropic processes, and with no
discuss teacher said, that isobaric process is in general polytropic. If we had ideal
gas, then it is clear, bud what if the gas isn't ideal?
I tried to proof it, but I stuck and don't know how to continue.
If I write the definition of polytropic process
[itex]c \mathrm{d}T = \mathrm{\delta}Q[/itex], where [itex]c[/itex] must be constant. For isobaric process we have
[itex]c_P = \left(\frac{\mathrm{d}Q}{\mathrm{d}T}\right)_P[/itex], where index right down the derivation notes what state variable is constant. To define the state in thermodynamics, we
need just two state variable, so
[itex]c_P = \left(\frac{\mathrm{d}U(P,T)}{\mathrm{d}T}\right)_P + \left(\frac{P\mathrm{d}V
(P,T)}{\mathrm{d}T}\right)_P[/itex]. And know I don't know how to continue. I tried said that
[itex]U(T,V(P,T))[/itex] and chain rule, but it didn't look better . . .
Thanks for any help . . .
 
Last edited:

Answers and Replies

  • #2
20,244
4,265
In an isobaric process, $$Q=\Delta H = C_p \Delta T$$. Also, for a polytropic process, $$PV^n=constant$$For an isobaric process, n = 0.
 

Related Threads on Heat capaticy of isobaric process with real gas

  • Last Post
Replies
2
Views
1K
  • Last Post
3
Replies
61
Views
3K
Replies
5
Views
4K
Replies
14
Views
4K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
3
Views
2K
Top