⎛ ∂ ln f ⎞ ⎛ ∂μ ⎞ ⎜ ⎟ = RT ⎜ ⎟ =V ⎝ ∂P ⎠T ⎝ ∂μ ⎠T 1 ⎛ ∂ ln f ⎞ ⎜ ⎟ =V RT ⎝ ∂P ⎠T The fugacity is not the same as the pressure but it is closely related to it. Multiplying by P gives the relation to the compressibility, z: PV ⎛ ∂ ln f ⎞ P⎜ =z ⎟ = ⎝ ∂P ⎠T RT This shows that the fugacity is not the pressure of the real gas, which is P, but follows it and is related to it by the compressibility. The need for the fugacity is to have a form for the chemical potential that looks like that of an ideal gas but still obeys the rule that: ⎛ ∂μ ⎞ ⎜ ⎟ =V ⎝ ∂P ⎠T We are now in a position to use the partial derivative to obtain the fugacity coefficient, which is the (hopefully, small) correction to the actual (real gas) pressure, P . To obtain the fugacity, f . f =γP

ln γ = ln

f = ln f − ln P P 1

We use the usual integral rule of calculus: γ (P) ⎛ ∂ ln γ ⎞ ⎛ ∂ ln f ∂ ln P ⎞ ln γ γ ( P ) = ∫ ⎜ dP = ∫ ⎜ − ⎟ ⎟ dP o ∂P ⎠T ⎝ ∂P ⎠T ⎝ ∂P

⎛ z 1⎞ ⎛ z −1⎞ = ∫ ⎜ − ⎟ dP = ∫ ⎜ ⎟ dP P ⎠T ⎝ P P ⎠T Po ⎝ P

The range of integration can go to Po = 0 at which point the activity coefficient is one. All gasses in the limit of zero pressure behave as ideal gasses. With this boundary condition on the integral, the final form for the fugacity coefficient is: ⎛ z −1 ⎞ ln γ = ∫ ⎜ ⎟ dP P ⎠T 0⎝ P

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Fugacity for vdW gas for b > 0 and a = 0 One way to compare fugacity with actual pressure is to do a few examples. The first such example is the vdW case where the a = 0 . PV V b z= z −1 = = RT V − b V −b Putting this into the integral gives: P ⎛ ⎞ 1 ln γ = b ∫ ⎜ ⎟ dP − P V b ( ) 0⎝ ⎠T Notice that b can come out of the integral as it is a constant, but P and V are not constants and they must remain inside the integral. The temperature is considered to be held constant so if we can rewrite the integrand in terms of P and T we can do the integral. For this EoS P (V − b ) = RT , so the integral is quite simple: P

P

bP ⎛ 1 ⎞ ⎛ b ⎞ ln γ = b ∫ ⎜ ⎟ dP = ⎜ ⎟ ∫ dP = RT ⎠T RT ⎝ RT ⎠ 0 0⎝ Notice that the log of the activity coefficient looks like the ideal gas EoS where the volume is replaced by the true molar volume b, which is generally 1/1000 of a typical volume of gas. So this is a positive number and small as b 0 . Now the situation is reversed, and the vdW based pressure is less than the ideal gas pressure: RT a − P= V V2 (a) PV = 1 − RT z= RT V The fugacity (i.e. activity) coefficient can be determined for this gas as well. To do the integral we must write the activity coefficient as a function of {P, T } P a RT = 1 − cP z = 1− c= 2 VP z ( RT ) RT Rearrange and solve the quadratic function with the understanding that when P goes to zero, z goes to one. z 2 − z + cP = 0

( RTa )

1 + 1 − 4cP 1 − 4cP − 1 z −1 = 2 2 u ≡ 1 − 4cP 2z = 1 + u This gives an integrand that can be directly integrated, because the constant c (containing only the temperature and a ) can be taken out of the integral. P ⎛ 1 − 4cP − 1 ⎞ ln γ = ∫ ⎜ ⎟ dP 2 P 0⎝ ⎠T The integral can be done, and we will do it next, but it is important to know that the integrand is well behaved at low pressures. So the way to determine that is to expand the numerator in a Taylor series and keep the first few powers of the pressure, to see that it is indeed well behaved. So the Taylor series for the above integrand is: 1 + x = 1 + x 12 + x 2 2!1 12 −21 + x 3 3!1 12 −21 −23 +" From this series, and substituting in for x = −4cP , 2 3 cP ) 42 1 1 ( cP ) 43 1 1 3 ( 1 − 4cP − 1 = −c − − 2P P 2⋅2 2 2 P 2⋅3! 2 2 2 2 3 ⎛ cP ) cP ) ⎞ ( ( = −⎜c + +2 ⎟ ⎜ ⎟ P P ⎝ ⎠ Putting these first three terms into the integral, and evaluating the integral: z=

5

2 3 ⎛ cP ) cP ) ⎞ ( ( ln γ = − ∫ ⎜ c + +2 ⎟ dP ⎜ P P ⎟⎠ 0⎝ T P

(

= − cP + 12 ( cP ) + 23 ( cP ) 2

3

)

The expansion is necessary to demonstrate that the integrand is stable as the pressure goes to zero, so the function is integrable, and well behaved. Also, as the pressure goes to zero the activity coefficient goes to one, as discussed above. This expansion shows that the log of the fugacity coefficient is negative (always). Therefore, the coefficient will be less than one -- it will be a fraction -- and cannot be less than zero. Therefore the fugacity will be less than the real vdW pressure, which is less than the ideal pressure. Because the Taylor series is terminated at a few terms, it is not obvious that the log of the fugacity coefficient is always negative. One can deduce this from the complete integrand. The numerator must always be negative and so the integral must always be negative. The pressure is constrained that 0 ≤ 4cP ≤ 1 . If the pressure exceeds this constraint then the compressibility becomes imaginary. If we revisit the complete integral, we can write it exactly as: z −1 = −

( RTa ) V

P ⎛ 1 ⎞ ⎛ 1 ⎞ a ln γ = − ( ) ∫ ⎜ ⎟ dP ⎟ dP = − ( RT ) ∫ ⎜ VP ⎠T RT − Va ⎠T 0⎝ 0⎝ If we work in the low pressure limit where RT >> Va , then the integrand can be simplified (drop the a term in the denominator of the integrand) and the result is first V order Taylor series answer: ln γ = − cP . So this simplification of the integrand gave a very simple answer. We can see how exact it is by comparing with the exact result below. The conclusion we will come to is: This result is very good, and only off by about 20% even at the highest possible pressures for the vdW gas. (See figure below). P

a RT

The integral can actually be done exactly by doing a change of variables in the integrand: u = 1 − 4 cP . P u u ⎛ 1 − 4cP − 1 ⎞ ⎛ u −1 ⎞ ⎛ u ⎞ ln γ = ∫ ⎜ dP udu = = ⎟ ⎜ 2 ⎟ ⎜ ⎟ du ∫ ∫ 2 P u 1 u 1 − + ⎝ ⎠ ⎝ ⎠T T 0⎝ 1 1 ⎠T Notice that at this point the integrand has no instability in it as P goes to zero (or u goes to one). This integral is known (G&R 2.213) and straightforward to verify: ⎛ u ⎛1+ u ⎞⎞ ln γ = u − ln(1 + u ) 1 = − ⎜ 1 − u + ln ⎜ ⎟⎟ ⎝ 2 ⎠⎠ ⎝ This analytic result is perhaps a bit unusual, but it is well behaved. As the pressure goes to zero the ln γ = − cP ; and as the pressure goes as high as it can (given the above constraint, cP = 14 ) the function goes to ln γ = −1 − ln ( 12 ) = −0.3 , instead of –0.25 for the linear result. 6

This figure shows how the fugacity coefficient changes with c ⋅ P for the Taylor series expansion (using the first 3 terms) and the complete answer, the lower (darker) line. The first order Taylor series answer is the dashed, straight line above the other two. Again to verify that the activity coefficient does give the correct chemical potential we evaluate: ⎧ 1 ⎛ ∂ ln γ ⎞ ⎫ ⎛ ∂μ ⎞ ⎛ ∂ ln f ⎞ ⎜ ⎟ = RT ⎜ ⎟ = RT ⎨ + ⎜ ⎟ ⎬ ⎝ ∂P ⎠T ⎝ ∂P ⎠T ⎩ P ⎝ ∂P ⎠T ⎭ to determine that we will indeed get V: ⎛ ∂ ln γ ⎞ ⎛ ∂u ⎞ ⎛ ∂ ln γ ⎞ u = 1 − 4cP ⎜ ⎟ =⎜ ⎟ ⎜ ⎟ ⎝ ∂P ⎠T ⎝ ∂P ⎠T ⎝ ∂u ⎠T −2 c ⎛1+ u ⎞ ⎛ ∂u ⎞ ln γ = ( u − 1) − ln ⎜ ⎟ ⎜ ⎟ = u ⎝ 2 ⎠ ⎝ ∂P ⎠T 1 u ⎛ ∂ ln γ ⎞ ⎛ ∂ ln γ ⎞ ⎛ 2 ⎞ = ⇒ ⎜ ⎜ ⎟ = 1− ⎟ = −c ⎜ ⎟ 1+ u 1+ u ⎝ ∂u ⎠T ⎝ ∂P ⎠T ⎝1+ u ⎠ Having evaluated the activity coefficient, which should be equivalent to the integrand we will determine the value of: ? ∂μ RT RT a ⎛ 2 ⎞ ⎛ ⎞ ⎛ ∂ ln γ ⎞ V =⎜ + RT ⎜ − ⎟ ⎟ = ⎟ = ⎜ P P RT ⎝ 1 + u ⎠ ⎝ ∂P ⎠T ⎝ ∂P ⎠T RT a vdW EoS V = − P PV a ⎛ 2 ⎞ a Therefore or 2 z = 1 + u ⎜ ⎟= RT ⎝ 1 + u ⎠ PV This is a correct statement, from above, so we found that this complete form of the fugacity coefficient does indeed guarantee that the fugacity is constructed so that ⎛ ∂μ ⎞ ⎜ ⎟ =V ⎝ ∂P ⎠T

7