Average molecular weight?


by tag16
Tags: average, molecular, weight
tag16
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#19
Oct30-10, 06:03 PM
P: 98
I just assumed what I originally thought the volume was must be wrong or you wouldn't asked about the volume so I just came up with something random because I didn't know what else it could be, hence the 250 cm^3.

So I have to do the calculations for each of the three bulbs seperately as well as for each temperature, correct?
Borek
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#20
Oct30-10, 06:22 PM
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To quote the document you have read:

Treat each bulb separately, do not average the values from the two bulbs.
tag16
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#21
Oct31-10, 12:11 PM
P: 98
So the volume of the bulbs using V= mass (w/water)/ water density

bulb 1: 397.122
bulb 2: 362.841
bulb 3: 398.928

Which brings me back to my problem of finding the average molecular weight of the gas mixture at each temperature. I originally thought I was suppose to use this equation: M(mix)=x(NO2)M(NO2)+x(N2O4)M(N2O4), is that right?
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#22
Oct31-10, 12:31 PM
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These are not volumes, volumes should be much closer to 250. You forgot to subtract mass of empty bulb.

Calculate volumes, calculate masses of gas mixtures. Knowing mass, volume, temperature and pressure you should be able to calculate average molar mass, later you can use this number to calculate molar fractions of both gases.
tag16
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#23
Oct31-10, 12:43 PM
P: 98
opps thats what I did originally but I also averaged them all together which was wrong. This is minus the mass of the bulb:

bulb1:254.882
bulb2:263.277
bulb3:284.049
tag16
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#24
Oct31-10, 04:16 PM
P: 98
I'm not really sure what to do next, I think I'm suppose to use this equation: M(mix)=x(NO2)M(NO2)+x(N2O4)M(N2O4), if so I'm not sure how to get the mole fractions from the data I have or the degree of dissociation.
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#25
Oct31-10, 04:38 PM
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Calculate volumes, calculate masses of gas mixtures. Knowing mass, volume, temperature and pressure you should be able to calculate average molar mass, later you can use this number to calculate molar fractions of both gases.
tag16
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#26
Oct31-10, 06:47 PM
P: 98
so I used M=DRT/P to find the average molecular mass and got this:

For bulb 1: P=717.41 Torr= 0.94396 atm
V= 254.882 cm^3
T=303.15 K
mass= 141.9780 g


M=(141.9780 g/254.882 cm^3)(0.0821)(303.15 K)/(0.94396 atm)
ave. molecular mass of bulb 1: 14.686

Am I going about this the right way or am I completely off?
Borek
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#27
Oct31-10, 07:04 PM
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You are off. Perhaps not completely, but seriously.

First - 141.8780 g can't be mass of the gas. I guess you forgot to subtract bulb mass again.

Second - your volume units are inconsistent with R units.
tag16
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#28
Oct31-10, 07:18 PM
P: 98
For bulb 1: P=717.41 Torr= 0.94396 atm
V= 254.882 cm^3
T=303.15 K
mass= 141.9780 g(bulb w/gas)-141.229 g(bulb empty)= 0.749 g


M=(0.749g/254.882 cm^3)(8.314)(303.15 K)/(0.94396 atm)
ave. molecular mass of bulb 1: 7.846g
tag16
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#29
Oct31-10, 07:29 PM
P: 98
opps I think I used the wrong gas constant again:

M=(0.749g/254.882 cm^3)(82.057)(303.15 K)/(0.94396 atm)
ave. molecular mass of bulb 1: 77.4395 g
Borek
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#30
Nov1-10, 04:14 AM
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Quote Quote by tag16 View Post
ave. molecular mass of bulb 1: 77.4395 g
That looks OK.

Now, write the equation for average molar mass using x & y for molar fractions of both gases. You know that x+y=1, that means you have two equations and two unknowns. Solve and you will know how much of each gas is present.
tag16
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#31
Nov1-10, 04:45 PM
P: 98
x + y = 1
92.011 x + 46.0055 y = 77.4395

x= -y + 1
92.011 (-y +1) + 46.0055 y = 77.4395
-92.011 y + 92.011 + 46.0055 y = 77.4395
-46.0055 y + 92.011= 77.4395
-46.0055 y = -14.5715
y = 0.316734
1- y = x
x = 0.683266

To find the degree of dissociation I did this:

n(N2O4)=77.4395/92=0.841 mol

degree dissociation: x(N2O4)/n(N2O4)= 0.683266/0.841= 0.812445= 81.24%


To find Kp I did:

n(N2O4)=77.4395/92=0.841 mol
n(NO2)= 77.4395/46=1.683 mol

Kp= n(N2O4)/(n(NO2))^2= 0.841/(1.683)^2= 0.296912

I think to find delta G standard I'd want to use this equation: delta G= -RTlnKp

and for delta S: delta S= delta G/T

Did I go about any of that right or is that not what I'm suppose to do?
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#32
Nov1-10, 04:56 PM
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Quote Quote by tag16 View Post
y = 0.316734
1- y = x
x = 0.683266
Looks OK.

To find the degree of dissociation I did this:

n(N2O4)=77.4395/92=0.841 mol

degree dissociation: x(N2O4)/n(N2O4)= 0.683266/0.841= 0.812445= 81.24%
So you have calculated composition of the mixture only to ignore it now and not use the information in following calculations?
tag16
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#33
Nov1-10, 05:14 PM
P: 98
Is this right? If not I don't know how to find it.

PV = nRT --> n =PV/RT--> n = (1 − α)n + 2αn = PV/RT

α = (PV/nRT)− 1
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#34
Nov1-10, 05:41 PM
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Quote Quote by tag16 View Post
n = (1 − α)n + 2αn
You have two different n in this equation - on the left, total number of moles after reaction took place, on the right - initial number of moles.

You know molar fractions of both gases present. Calculate numbers of moles of both. Use stoichiometry of the reaction to calculate initial number of moles of gas. Use definition of dissociation degree.


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