14 replies to this topic

[James_student]

Hi guys

I need to calculate the density of a mixture of gases, but am struggling.

I know the Mole%, % weight & molecular weight, temp and pressure

but I don't know how to turn that into a density. Its a mixture of different gases that is confusing me.

I assume that I have to employ PV=nRT is some way to calcultae the density but am becomming confused again by the mixture of gases. Can anyone help me out?

James

[Zauberberg]

The easiest way to get a quick (and relatively good) estimate is to divide molecular weight (kg/kmol) with Avogadro's volume (22.414 m3/kmol) and this will give you the density at normal conditions. I have seen this approach being used in many gas plants' material balances.

There are other procedures available in the literature as well, you might want to try with Prausnitz or Campbell's methods (if I recall correctly).

[James_student]

Thanks for the reply!

So if I had say 12 different gases, it would be the sum of the weight / 22.4?

For isntance I have

Nitrogen @ 0.084Mole%, 0.119% weight and a molecular weight of 28.13
Carbon Dioxide @ 1.139Mole%, 2.515% weight and a molecular weight of 44.010

How would I some up these values?

James

[Qalander (Chem)]

Dear James_Student,
My friend Zauberberg has guided you to very good and conceptually accurate/helping(I believe) approach to solve your problem.

[Zauberberg]

The sum of component's mole % should be equal to 100; otherwise you have to normalize the composition.
Molecular weight of gas mixture is defined as:

M = X1M1 + X2M2 + X3M3 +... + XnMn,

where Xi and Mi represent individual component mole fraction and molecular weight, respectively.

Once when you calculate the average mole weight of a gas mixture, simply divide it by 22.414 to get density (in kg/m3) at normal conditions. Where is the problem with that?

[James_student]

Hi

So it would simply be molecular weight of nitrogen = 28.013/22.4 to give the density of that gas?

[Zauberberg]

You have something else on your mind?
Remember, it is at normal conditions (0C, 1atm).

[Patrician]

It's best to get it from Clapeyron equation:

pV=nRT where V=m/d and n=m/M divie by m
p/d=RT/M put it to ^-1 and you get

d=pM/RT where M is a mole mass of gas mixture

with pressure higher than 20 atm you must include Compressibility Factor which is function of reducted parameters and can be find on web.

d=pM/zRT

Regards,

[Zauberberg]

This yields the same result as dividing mole weight by 22.414, since in Clapeyron equation RT/P equals 22.414 at normal conditions (8.314*273.15/101.325).

As long as we are speaking about ideal gases, all those approaches will yield the same results.
Best regards,

[Qalander (Chem)]

Dears Since for any calculating estimations some benchmark conditions are mandatory as such for 'Normal" conditions and as usual for Ideal Gas law applications what 'Zauberberg' said is fine,I still believe

[James_student]

Im still struggling, I can appreciate that this must be fairly basic stuff for you guys, but for some reason I cant get my head round it

[Zauberberg]

James,

You have the full procedure explained above. So where is the problem?

[James_student]

I have the Mole%, the weight% & molecular weight of each gas. But what is the mole fraction? I have the density of all the gases combined which is 0.6901, but I need to know how to get to this. I'm just finding thsi a bit confusing, sorry

[Zauberberg]

Mole fraction = Mole % divided by 100. Do this for every single component in your gas mixture, and the sum of all components' mole fractions should be equal to 1.

Now, when you have mole fractions and you know each component's mole weight, you can calculate the mole weight of gas mixture, by applying the abovementioned formula:

M = X1M1 + X2M2 +... + X12M12

Divide M by 22.414 and that is your normal density in kg/m3.
I hope it is clear now, because it is sleeping time in Malabo.

Best regards,

[James_student]

Got it! Thanks, sorry for being a pain!