1) Graphical method
2) General method
3)From equation of state.
1) (l) GRAPHICAL METHOD :-
The fugacity of a real gas is equal to pressure of very low P.
dG = vdp and dG. RTlnf.
Vdp : RTlnf →
(or)
∂lnf , v
∂p RT
A function '∝' is introduced in the above equation which is defined as measure of deviation of real gas from ideal behaviour.
∝ = RT − V ( ∴ v = RT − ∝ )
P P
RTdlnf = ( RT − ∝ ) dp
p
(or)
dlnf = dp − ∝ . dp.
p RT
dlnf = dlnp − ∝ . dp
RT
dlnf - dlnp = - ∝ dp
RT
dlnf = − ∝ .dp.
p RT
upon ∫ "
∫ dln i − ∝ ∫₀p dp
p RT
lnf = lnp − ∫₀p ∝ . dp
RT
f∕p →1 as p→0
The value of integral of the above equation can be evaluated by plotting a graph between p = 0 and p = p gives the value of integral
Then the value of 'f' is can be calculated.
(ll) APPROXIATE CALCULATION METHOD :
∝ = RT − v
P
′ ∝ ′ only constant over a range of moderate P,
Then,
lnf = lnp - ∝ P
RT
(or)
ln f = − ∝. P → .........(1)
P RT
At moderate presuure f/p = 1, temperature
lnf = f - 1. [lnx ≈ x − 1 when x →1]
P P
From equation (1)
f −1 = − ∝ p
P RT
Substituting the value of ∝, we get
f = P² V
RT
This method becomes more accurate at low P and High T.
2) GENERAL METHOD :-
Real gas
∝ = RT ( 1− PV ) → (1)
P RT
The compressibility factor ' k' of gases is defined as,
k = PV
RT
So equation (1) becomes,
∝ = RT (1− k )
P
Inserting the value of ∝ in the previous equation,
lnf = lnp - ∫₀p ∝ . dp.
RT
Replacing the ' p' terms in the integrated by corresponding values of reduced P (π)
We get,
lnf = lnp + ∫₀p ( k − 1 . dp)
π
Where π = p/p₀
f/p is plotted against reduced pressure.
The compressibility factor for all goes are
∆G equal at average reduced T and P
3) FROM EQUATION OF STATE :-
if f and f₀ represent the fugacities of a gas at pressure P and P₀ where P₀ is very low value, Then on ∫ .
lnf = 1 ₚ₀ᵖ∫ᵖ vdp. → (1) f₀ -RT
Integrating by parts,
ₚ₀∫ ᴾ vdp = [ PV]ᵥ₀ⱽ − ᵥₒ∫ ᵛ pdv
= PV − P₀ V₀ − ᵥₒ∫ ᵛ pdv
Where 'V' terms represent the molar volume. At very low P, P₀V₀ = RT. Therefore,
lnf/f₀ = 1/RT ( Pv − RT− ᵥₒ∫ ᵛ pdv) → (2)
At very low P
lnf/f₀ = ln f/P₀
From equation (2)
ln f/f₀ = 1/RT ( PV − RT− ᵥₒ∫ ᵛ pdv )
lnf = lnP₀ 1/RT ( PV −RT−ᵥₒ∫ ᵛ pdv ) → ( 3)
By means of equation of state, say vaander waal's equation, we express 'P' as a function of V at constant T and the ∫ pdv can be easily calculated from vaander waal's equation of state
( P + a/v²) ( v− b) = RT
P = RT − a → (4)
v − b v²
Multiply by dv
Pdv = ( RT − a ) dv v−b v² →(5)
ᵥₒ∫ ᵛ pdv = ᵥₒ∫ ᵛ ( RT ) dv − ∫ a . dv
v− b v² = RTln v− b + a/v − a/v⁰ → (6) v₀ −b
since v₀ is very large as compared to b.
So, v₀ − b ≈ v₀
v₀ is equal to RT/P₀ ( P₀ v₀ = RT, v₀ = RT/P₀
a/v₀ can be rejected ,being very small. Therefore equation (6) becomes,
ᵥₒ∫ ᵛ pdv = RTln v- b + a/v v₀
H* − H̄ is the increase in partial molal enthalpy accompanying the isothermal expansion of the gas from a pressure ' P' into vacuum.
H̄* − H̄ = − ₀∫ ᴾ (∂H ) T. dp ∂P
Since (∂H ) = − μJ.T . Cp.
∂P
From equation........................... (6)
RT² [ ∂lnf ]ₚ = ₀∫ ᵖ μJ,T. cp. dp ∂T
If Cp is treated as a constant and μJ, T is known as a function of P. The integral of the above equation can be evaluated.
Good
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