# Homework Help: Problem 13-10E (Intro to Thermo 7E)

1. Sep 12, 2011

### Heeb

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.

2. Sep 12, 2011

### Mschi

Hey just so you know it is 25lbmol N2

3. Sep 12, 2011

### Heeb

Thanks that was it. Does anybody know how to move this to the homework help section

4. Sep 12, 2011

### Heeb

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

5. Sep 12, 2011

### Staff: Mentor

Please do not give solutions to schoolwork questions on the PF.

6. Sep 12, 2011

### Redbelly98

Staff Emeritus
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.

7. Sep 13, 2011

### Heeb

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

8. Sep 13, 2011

### Staff: Mentor

Similar unit is kgmole.