1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the concentration of a gas (thermodynamics)

  1. Jan 16, 2007 #1
    Hi.

    1. The problem statement, all variables and given/known data

    The root mean square velocity of oxygen molecules is 480m/s while the pressure is 20kPa.

    What is the concentration (particles/volume) of oxygen?

    2. Relevant equations

    PV=nRT
    [tex] E_{kin}=\frac{m \overline{v}^2}{2}=\frac{3}{2}kT [/tex]

    3. The attempt at a solution

    [tex] m \overline{v}^2=3kT \Rightarrow T= \frac{m \overline{v}^2}{3k} [/tex]

    [tex] PV=nR \frac{m \overline{v}^2}{3k} \Rightarrow n=\frac{3kPV}{Rm \overline{v}^2} [/tex] where [tex] m=2 \times 16 \times 1.66 \times 10^{-27} kg [/tex]

    I set V=1m^3 and get n=8.137mol and therefore [tex] \frac{n \times n_{a}}{V}=4.899 m^{-3} [/tex] where [tex] n_{a}=6.02 \times 10^{23} [/tex]

    Is this correct? I'm sure there's a simpler way to do this.
     
    Last edited: Jan 16, 2007
  2. jcsd
  3. Jan 16, 2007 #2

    GCT

    User Avatar
    Science Advisor
    Homework Helper

    Note that the mass refers to an individual oxygen molecule (from what I recall at the moment), (16 grams of Oxygen/mole of Oxygen)(1 mole/6.022 x 10^23 atoms)(1 kilogram/1000 grams)(2 atoms of Oxygen/1 diatomic molecule)=____.......the setup should be solved for n/V, this means that you need to incorporate a particular value of R, choose from the list on the page that can be linked to through the below text so that the final units for n/V is respect to moles/liter.



    http://en.wikipedia.org/wiki/Gas_constant
     
    Last edited: Jan 16, 2007
  4. Jan 17, 2007 #3
    Actually, since this is one of my physics class problems (not chemistry) concentration really does mean particles/volume in this case. Which means that I have chosen the right value for R, I think.
     
  5. Jan 17, 2007 #4

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You're missing a factor of 10^{24} in that final bit.

    The method is perfectly correct - I haven't checked the numbers, but I believe the final number looks close enough (I happen to know that the RMS speed of oxygen molecules at room temperature is about 500m/s, and at NTP, a mole of atoms occupies about 22.4 liters, so at a fifth of an atmosphere, the concentration would be roughly 6/(5*0.0224)*10^{23} per cubic meter, which is about 10% higher than your number, but this is very rough estimate.)

    As for a simpler way, I think this is as simple as it gets. Only, notice that since R=k*Na, and M=m*Na, your final expression simplifies to n=3PV/Mv^2 (in moles).
     
    Last edited: Jan 17, 2007
  6. Jan 17, 2007 #5

    GCT

    User Avatar
    Science Advisor
    Homework Helper

    (n/v)=20 kPa(1000 Pa/1 kPa)(1.3806503 × 10-23 m2 kg s-2 K-1)3/[8.314472 m^3 · Pa · K-1 · mol-1(16 grams of Oxygen/mole of Oxygen)(1 mole/6.022 x 10^23 atoms)(1 kilogram/1000 grams)(2 atoms of Oxygen/1 diatomic molecule)(480 m/s)^2]= 0.00813782909722 moles/m^3

    (0.00813782909722 moles Oxygen/L)(6.022x10^23 molecules Oxygen/mole)= 4.900600682 x 10^24 molecules/m^3

    So there's everything done in a perfunctory fashion for ya, I wanted to see what answer would result with the "chemist" method.:smile:

    I'm going to need to see if the units cancel out exactly.......
     
    Last edited: Jan 17, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Finding the concentration of a gas (thermodynamics)
Loading...