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Q: Advance Thermodynamics ( No Answer,   1 Comment )
Question  
Subject: Advance Thermodynamics
Category: Science > Physics
Asked by: kran-ga
List Price: $4.50
Posted: 04 Dec 2004 17:54 PST
Expires: 03 Jan 2005 17:54 PST
Question ID: 438185
Determine the value of f/P for nitrogen at 250K and (1) 200 bars and
(2) 200 bars by the following methods:
a> the gas is an ideal gas
b> the gas follows the equation of state, Pv=RT+bP, where
b=39.5-1X10(exp4)/T-1.084X10(exp6)/T(exp2), jT is in Kelvins and b is
in cm(exp3)/gmol
c> the two parameters generalized fugacity coefficient chart
d> the Lee-Kessler three parameter data
Answer  
There is no answer at this time.

Comments  
Subject: Re: Advance Thermodynamics
From: hfshaw-ga on 08 Dec 2004 16:30 PST
 
You obviously made a mistake in copying the question;  parts (1) and
(2) both ask for the fugacity coefficient (f/P) at 200 bars.

To answer parts c) and d), one needs information you haven't given us,
namely the "generalized fugacity coefficient chart" referred to in
part c), and the Lee-Kessler equation of state parameters needed for
part d).

Part a) is trivial.  By definition for an ideal gas, the fugacity is
equal to the partial pressure.  f/P = 1

Part b) is asking for the fugacity coefficient for a nonideal gas that
obeys the van der Waals equation of state when the size parameter (b
parameter) dominates, and the intermolecular forces are negligible (a
parameter = 0).

Let g = f/P (g is the fugacity coefficient, usually written as a lower
case Greek gamma).  For any real gas,

     (d ln(g)/dp)_T = v/RT - 1/p,   [look this up in your textbook]

where (d ln(g)/dp)_T means the partial derivative of the ln of the
fugacity coefficient with respect to pressure at constant temperature.

Integrating this, we get:

     ln(g) - ln(g0) = Integral from p=0 to P { (v/RT -- 1/p) dp},

where g0 is the fugacity coefficient at P=0 bars.  But the fugacity is
defined to be equal to the pressure as P-> 0, so ln(g0) =ln(P/P) = 0.

We are given that v = RT/p + b.  Substituting this into the integrand
above, you'll find that the RT/p terms cancel out, and we are left
with:

    ln(g) = Integral from p = 0 to P {b/RT dp}
           = b/RT*(P-0) = bP/RT

exponentiating both sides yields the desired result:

    g = f/P = exp(bP/RT)

or, noting that from the equation of state that P/RT = 1/(V-b)

    g = f/P = exp(b/(V-b))

You need to plug in the particular values of b, T, and P (or V)
corresponding to the conditions asked about in the problem.  Note that
b is a function of T.

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