Why ∆u=Cv ∆T for isochoric transformation of non-ideal gases?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 5K views
maCrobo
Messages
51
Reaction score
1
I simply report what I read:
"For an ideal gas, but for every kind of transformation ∆u=Cv ∆T, while for every kind of material in the thermodynamic system, but only for isochoric transformation ∆u=Cv ∆T."

Where does this second statement come from?
Everything is clear about ideal gases, but I don't figure out how to prove the second part of this statement.
 
Physics news on Phys.org
That is how Cv is defined.

Cv is the limit when ∆T goes to zero of ∆u/∆T. The first statement is simply a consequence of internal energy being dependent only of T for an ideal gas.
 
maCrobo said:
I simply report what I read:
"For an ideal gas, but for every kind of transformation ∆u=Cv ∆T, while for every kind of material in the thermodynamic system, but only for isochoric transformation ∆u=Cv ∆T."

Where does this second statement come from?
Everything is clear about ideal gases, but I don't figure out how to prove the second part of this statement.
It follows from the first principle of thermodynamics. In isochoric transformations the work (compression-expansion work) is zero so Δu=Qv.