A vessel divided into two parts by a partition, contains 4 moles of nitrogen gas at 75 degC and 30 bar on one side and 2.5 moles of argon gas at 130 degC and 20 bar on the other. If the partition is removed and the gases mix adiabatically and completely, what is the change in entropy ? Assume nitrogen to be an ideal gas with Cv = (5/2)R and argon to be an ideal gas with Cv = (3/2)R
Hints :
Calculate ΔS for each step of a hypothetical process path.
First convert N2 from the initial to the final state. Then, change Ar from the initial to the final state. Finally, mix the N2 and the Ar at the final T and P.
Ans.: ΔS ≈ 38 J/K
2 comments:
Hi, I just have an unrelated conceptual question. The Gibb's free NRG equation becomes a function of moles or concentration when the system becomes open, right? Where before we strictly dealt with closed systems and pure components?
SpongeBob 5:05 PM
Gbar / RT is used for mixtures. The bar means it is a partial molar property (PMP). PMPs in general and this one in particular are used to describe the behavior of mixtures.
So, yes, Gbar / RT depends on the mole fractions of all the chemical species present in the mixture.
G~ / RT is generally used for pure systems and therefore it wouldn't usually depend on the compositions...because it is pure.
But none of this really depends on whether the system is open or closed !
The interesting point is that there are no mixture problems on HW #3 !!
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