Finding the concentration of a gas (thermodynamics)

by raul_l
Tags: concentration, thermodynamics
 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 $$E_{kin}=\frac{m \overline{v}^2}{2}=\frac{3}{2}kT$$ 3. The attempt at a solution $$m \overline{v}^2=3kT \Rightarrow T= \frac{m \overline{v}^2}{3k}$$ $$PV=nR \frac{m \overline{v}^2}{3k} \Rightarrow n=\frac{3kPV}{Rm \overline{v}^2}$$ where $$m=2 \times 16 \times 1.66 \times 10^{-27} kg$$ I set V=1m^3 and get n=8.137mol and therefore $$\frac{n \times n_{a}}{V}=4.899 m^{-3}$$ where $$n_{a}=6.02 \times 10^{23}$$ Is this correct? I'm sure there's a simpler way to do this.
 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
 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.
Emeritus
PF Gold
P: 11,155
Finding the concentration of a gas (thermodynamics)

 Quote by raul_l 3. The attempt at a solution $$m \overline{v}^2=3kT \Rightarrow T= \frac{m \overline{v}^2}{3k}$$ $$PV=nR \frac{m \overline{v}^2}{3k} \Rightarrow n=\frac{3kPV}{Rm \overline{v}^2}$$ where $$m=2 \times 16 \times 1.66 \times 10^{-27} kg$$ I set V=1m^3 and get n=8.137mol and therefore $$\frac{n \times n_{a}}{V}=4.899 m^{-3}$$ where $$n_{a}=6.02 \times 10^{23}$$
You're missing a factor of 10^{24} in that final bit.

 Is this correct? I'm sure there's a simpler way to do this.
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).
 Sci Advisor HW Helper P: 1,769 (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. I'm going to need to see if the units cancel out exactly.......

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