Question

What are some examples of molar volumes of a gases?

Answer 1

Some examples of molar volumes:

van der Waals
molar volume of Butane (g) at 370 K and 14.35 bar: 1.7417
molar volume of Butane (g) at 350 K and 13.35 bar: 1.7377
molar volume of Krypton (g) at 173.15 K and 15.00 bar: 0.8219
molar volume of Krypton (g) at 120 K and 1.00 bar: 9.7836

Redlich-Kwong
molar volume of Butane (g) at 370 K and 14.35 bar: 1.6453
molar volume of Butane (g) at 350 K and 13.35 bar: 1.6065
molar volume of Krypton (g) at 173.15 K and 15.00 bar: 0.7839
molar volume of Krypton (g) at 120 K and 1.00 bar: 9.6896

Ideal Gas Law
molar volume of Butane (g) at 370 K and 14.35 bar: 2.1438
molar volume of Butane (g) at 350 K and 13.35 bar: 2.1798
molar volume of Krypton (g) at 173.15 K and 15.00 bar: 0.9598
molar volume of Krypton (g) at 120 K and 1.00 bar: 9.9774

What this means is that Krypton is more ideal than Butane! What THAT means is that it does not compress that much relative to an ideal gas under the same conditions.

(`Z = (PV)/(nRT)` = 0.99768 vs. 0.96616, where 1 is ideal)

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Below is how you can calculate these molar volumes on your own (bear with me, it's long).

You can calculate molar volumes (albeit tediously) using the van der Waals or Redlich-Kwong equations of state and get reasonably accurate values (the ideal gas law is only accurate near the Boyle temperature and at a low enough pressure).

What you would need to do is solve these to get a cubic form with the molar volume as your variable.

Let `V` from this point on = molar volume.

van der Waals:
`P = (RT)/(V-b) - a/(V^2)`

Redlich-Kwong:
`P = (RT)/(V-B) - A/(sqrt(T)V(V+B))`

where `a` or `A` is proportional to the amount of attraction between particles, and `b` or `B` is proportional to the difference between the ideal volume and the non-ideal, smaller volume.

You can find these online, though in different units. I'm just using my Physical Chemistry textbook.
http://che31.weebly.com/uploads/3/7/4/3/3743741/handout_e-realgasconstants.pdf

Solving for cubic form of van der Waals :
`P = (RTV^2 - a(V-b))/((V-b)V^2)`
`P(V^3-bV^2) = RTV^2 - a(V - b)` (multiply denominator out)
`PV^3-bPV^2 = RTV^2 - aV + ab` (distribute)
`PV^3 - (RT+bP)V^2 + aV - ab = 0` (subtract stuff over)

`V^3 - (b+(RT)/P)V^2 + (a/P)V - (ab)/P = 0`

And when you do the same for the Redlich-Kwong , you get:

`V^3 - (RT)/PV^2 - (B^2 + (BRT)/P - A/(sqrt(T)P))V - (AB)/(sqrt(T)P) = 0`

For example, the van der Waals constants (`a` and `b`) for Butane are:
`a = 13.888 (dm^6*Bar)/(mol^2)`
`b = 0.11641 (dm^3)/(mol)`

and the A and B for Redlich-Kwong for butane are:
`A = 290.16 (dm^6*Bar*K^(1/2))/(mol^2)`
`B = 0.08068 (dm^3)/(mol)`

You would need the pressure, temperature, and a and b constants to calculate this (`R = 0.083145 (L*Bar)/(mol*K)`). What you could do afterwards is solve it using the Newton-Raphson approximation method, which is:

`V_(goal) = V_(guess) - f(V_(guess))/((deltaf(V_(guess)))/(deltaV_(guess)))_(T,P)`

What happens here is that you if you keep solving, you'll approach the answer from above from some guess you have. For example, if you calculate the molar volume of butane at 370 K and 14.35 bar using the van der Waals, if you guess 2, it should take 6 guesses before you get a molar volume that does not change within 9 decimal places. The highest molar volume is for the gas, the lowest one is for the liquid, and the middle one is physical nonsense. ;)

If you want to do this on your calculator for the van der Waals, you could store values into variables, and then write:

`(X - ((X^3 - AX^2 + BX - C)/(3X^2 - 2AX + B)))->` Store as `X`
where `X = V, A = b+(RT)/P, B = a/P, and C = (ab)/P`.
Then press Enter a lot. :)