Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 8 (Q.No. 13)
13.
The number of degrees of freedom for a mixture of ice and water (liquid) are
Discussion:
4 comments Page 1 of 1.
TANMAY SHAH said:
3 years ago
Here ice and water [liquid] both have only one component i.e. WATER.
So, C=1; P=2.
F = C-P+2
F = 1-2+2
F = 1.
So, C=1; P=2.
F = C-P+2
F = 1-2+2
F = 1.
Vikas chaurasiya said:
3 years ago
F = c-p+2.
C = 2(ice+water)
P = 2(ice=solid , water=liquid)
Ans = F=2-2 + 2 = 2
The Correct answer is 2.
C = 2(ice+water)
P = 2(ice=solid , water=liquid)
Ans = F=2-2 + 2 = 2
The Correct answer is 2.
Monika said:
8 years ago
Answer should be 2.
Oojii said:
8 years ago
f = c-p+1.
f = 2-2+1.
f = 1.
c = 2 (ice+water).
p = 2(liquid+ solid).
f = 2-2+1.
f = 1.
c = 2 (ice+water).
p = 2(liquid+ solid).
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