Chemical Engineering - Chemical Engineering Thermodynamics - Discussion

Discussion Forum : Chemical Engineering Thermodynamics - Section 2 (Q.No. 17)
17.
The number of degrees of freedom for an azeotropic mixture in a two component vapour-liquid equilibria is/are
zero
one
two
three
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Aamina rabab said:   4 years ago
Yes, you are right, Thanks @Monika Agarwal.

Pranav said:   5 years ago
The Correct answer is Option B.

F = C-P + 1 for two components system.

Ref : KV Narayan.
(1)

Yuvaraj said:   7 years ago
Yes, right @Monika Agarwal.

Vimal said:   8 years ago
Which one is right? Please explain the answer.

Monika Agarwal said:   8 years ago
We will use equation F = C-P+2-r-s.
r = No. of reaction in system
s = Specialty constant

Here s=1 because of azeotropic solution,
So F = 2 - 2 + 2-1,
F = 1.
(6)

Gihi said:   8 years ago
The rule applies to non-reactive multi-component heterogeneous systems in thermodynamic equilibrium and is given by the equality.

F = C - P + 2.

Anmol said:   9 years ago
@Rameshwer, which equation you have used can you tell me please?

Rameshwer said:   1 decade ago
A binary system containing an azeotropic mixture in equilibrium with its vapor has two species, two phases, and one relation among intensive variables: xA=yA.

The number of degrees of freedom is then F = 2+s^\'r^\'P = 2+2^\'1^\'2 = 1. Hence, degree of freedom = 1.

Bajwa said:   1 decade ago
No @Krunal, correct answer is B.

For one component systems, we have.

F = C-P+2.

But for two component systems, we have.

F = C-P+1.

=> F = 1 for the given problem.

Krunal said:   1 decade ago
Correct answer is 2.
F = C -P+2.

Here num.of compo.C = 2.

No.of phases = 2.
So, F=2-2+2 = 2.

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