
#1
Jan1607, 09:29 AM

P: 105

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. 



#2
Jan1607, 07:54 PM

Sci Advisor
HW Helper
P: 1,769

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 



#3
Jan1707, 04:01 AM

P: 105

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.




#4
Jan1707, 04:12 AM

Emeritus
Sci Advisor
PF Gold
P: 11,154

finding the concentration of a gas (thermodynamics)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). 



#5
Jan1707, 07:01 PM

Sci Advisor
HW Helper
P: 1,769

(n/v)=20 kPa(1000 Pa/1 kPa)(1.3806503 × 1023 m2 kg s2 K1)3/[8.314472 m^3 · Pa · K1 · mol1(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. I'm going to need to see if the units cancel out exactly....... 


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