Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 5 (Q.No. 42)
42.
What is the degree of freedom for two mis-cible (non-reacting) substances in vapor-liquid equilibrium forming an azeotrope ?
Discussion:
8 comments Page 1 of 1.
Chayanika said:
5 months ago
Azeotrope is always 1.
John said:
3 years ago
@Adnan.
Vapour and liquid are two phases.
Vapour and liquid are two phases.
Adnan said:
5 years ago
There is only one phase because they are miscible.
Abhishek said:
6 years ago
F=2-p+N-r-s where r= no of reaction and s= special constraint
So F= 2-2+2-0-1
F= 1.
So F= 2-2+2-0-1
F= 1.
Toyyeb said:
6 years ago
I think the answer is correct, because the P=1 since the two components are in equilibrium.
Jhhu said:
9 years ago
F = C - P + 2 - r - s.
Here c = 2, p = 2, r = 0 (non reacting) s = 1 (special condition) So the answer should be 1.
Here c = 2, p = 2, r = 0 (non reacting) s = 1 (special condition) So the answer should be 1.
Vikram said:
9 years ago
Hi, @Rajendar.
Why you take r = 1 whether it is non-reacting?
Why you take r = 1 whether it is non-reacting?
Rajendar said:
9 years ago
I think the answer will B because components are two, phases are two and it's forming an azeotrope.
F = C - P + 2 - r.
= 2 - 2 + 2 - 1 = 1.
F = C - P + 2 - r.
= 2 - 2 + 2 - 1 = 1.
(1)
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