Problem 13-10E (Intro to Thermo 7E)

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The discussion revolves around a thermodynamics problem involving the calculation of mole fractions and apparent molecular weight of a gas mixture. The user is confused about the term "pound mole" and its relation to molar mass, comparing it to "lb/mol." They provide calculations for mole fractions of various gases but find discrepancies with the instructor's answers. Clarification is offered that "pound mole" refers to a unit where the mass in pounds equals the grams per mole values, and the problem's wording was misleading. The calculations themselves are confirmed as correct, emphasizing the importance of understanding unit conversions in thermodynamics.
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Hey I was given this problem:

The moles of components of a gas mixture are given. The mole fractions and the apparent molecular weight are to be determined.

The given properties are:

The molar masses of He, O2, N2, and H2O are 4.0, 32.0, 28.0 and 18.0 lbm/lbmol respectively and there are 3 lbmol He, 1.5 lbmol O2, 0.3 lbmol H2O, 2.5 lbmol N2

I am a little bit confused on what a pound mol is I think, in my old chem class it was lb/mol. Anyways the analysis seems simple enough but I can't get the same answer as my instructor.
 
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Hey just so you know it is 25lbmol N2
 
Thanks that was it. Does anybody know how to move this to the homework help section
 
I've been reading up on this site and I didn't do the correct style so I will finish it here.

Relevant equations:

mf=mi/mt

Attempt at a Solution
mf o2 = 1.5 /7.3 = .206

mf h20=.3 /7.3 = .041

mf he = 3/7.3 = .41

mf n2 = 2.5/7.3 = .342

This was not the answer that was given by my instructor
 
Mschi said:
Hey just so you know it is 25lbmol N2

Please do not give solutions to schoolwork questions on the PF.
 
Heeb said:
I've been reading up on this site and I didn't do the correct style so I will finish it here.

Relevant equations:

mf=mi/mt

Attempt at a Solution
mf o2 = 1.5 /7.3 = .206

mf h20=.3 /7.3 = .041

mf he = 3/7.3 = .41

mf n2 = 2.5/7.3 = .342

This was not the answer that was given by my instructor
Your calculations look correct to me. What answer was given by the instructor? Also, is it supposed to be 2.5 or 25 lbmol of N2?

p.s. I've never run across the term "pound mole" before, but apparently it's the number of molecules such that the mass in pounds is equal to the usual grams/mole values we normally associate with molecules: Eg., Helium is 4.0 lbm per pound-mole. So a pound-mole is not Avogadro's number, but some other large number. Fortunately we don't have to actually know that number to solve this problem.
 
It was an incorrectly worded problem statement it was 25lbmol N2, but the method was fine.

Thanks for the explanation of pound mole that makes a little bit more sense
 
Similar unit is kgmole.
 

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