finding the concentration of a gas (thermodynamics)


by raul_l
Tags: concentration, thermodynamics
raul_l
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#1
Jan16-07, 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.
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GCT
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Jan16-07, 07:54 PM
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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
raul_l
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#3
Jan17-07, 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.

Gokul43201
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Jan17-07, 04:12 AM
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finding the concentration of a gas (thermodynamics)


Quote Quote by raul_l View Post
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]
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).
GCT
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Jan17-07, 07:01 PM
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(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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