How can I calculate the molar volume of a non ideal gas?
1 Answer
Well, it depends on what equation of state you WANT to use. The easiest one to use for REAL gases is the van der Waals equation of state (vdW EOS):
P = (RT)/(barV - b) - a/(barV^2)P=RT¯¯¯V−b−a¯¯¯V2 where
P,R,P,R, andTT are known from the ideal gas law,barV -= V/n¯¯¯V≡Vn is the molar volume of the vdW gas, andaa andbb are van der Waals constants accounting for the attractive intermolecular forces, and the excluded volume, respectively.
Solving for
P = (barV^2RT)/(barV^2(barV - b)) - (a(barV - b))/(barV^2(barV - b))P=¯¯¯V2RT¯¯¯V2(¯¯¯V−b)−a(¯¯¯V−b)¯¯¯V2(¯¯¯V−b)
PbarV^2(barV - b) = barV^2RT - a(barV - b)P¯¯¯V2(¯¯¯V−b)=¯¯¯V2RT−a(¯¯¯V−b)
PbarV^3 - bP barV^2 = barV^2RT - abarV + abP¯¯¯V3−bP¯¯¯V2=¯¯¯V2RT−a¯¯¯V+ab
PbarV^3 - (bP + RT)barV^2 + abarV - ab = 0P¯¯¯V3−(bP+RT)¯¯¯V2+a¯¯¯V−ab=0
This becomes a cubic equation for
barul|stackrel(" ")(" "barV^3 - (b + (RT)/P)barV^2 + a/PbarV - (ab)/P = 0" ")|
For this, we need
- specified pressure
P in"bar" , - temperature
T in"K" , R = "0.083145 L"cdot"bar/mol"cdot"K" ,- vdW constants
a in"L"^2"bar/mol"^2 andb in"L/mol" .
Then this can be solved iteratively via the Newton-Raphson method. Of course, you can use whatever method you want to solve this cubic.
To do the Newton-Raphson method, in your calculator, let:
b + (RT)/P = A
a/P = B
(ab)/P = C
Then we have:
barV^3 - AbarV^2 + BbarV - C = f(barV)
3barV^2 - 2AbarV + B = f'(barV)
Each iteration acquires
barV_(i+1) = barV_i - (f(barV_i))/(f'(barV_i))
In your TI calculator, let
"logical guess" -> X
Then if you believe you chose correctly, proceed to type the following:
(X - (X^3 - AX^2 + BX - C)/(3X^2 - 2AX + B)) -> X
This generates an iterative loop that triggers each time you press Enter. So, press Enter until the value you get stops changing.
That is ONE out of THREE molar volumes.
- One
barV is of the liquid. - One
barV is of the gas. - One
barV is a so-called spurious (i.e. UNPHYSICAL) solution.
To know what you have just gotten, compare with the other