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Topic: Thermodynamics: Work on an isothermal non-ideal gas  (Read 10038 times)

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Offline kidechizen

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Thermodynamics: Work on an isothermal non-ideal gas
« on: October 27, 2007, 05:29:43 PM »
Say I have a non ideal gas benzene and I want to calculate the integral of w(reversible) = int(vdp). How do I go about finding the relationship between v and p? Can I still use P1V2^n = P2V2^n as a relationship? I don't think I can use PV^n = constant, can I? Or is there a way to use compressibility factor z in PV=ZRT to form some sort of relationship? Help please.

Also, when I want to compare this result with an ideal gas, would I simply calculate benzene using PV=ZRT or PV=NRT or something else?

Thanks

Offline Yggdrasil

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #1 on: October 27, 2007, 06:19:17 PM »
Do you know the equation of state for your non-ideal gas?  For example, if you are using the van der Waals equation of state, this equation will give you P in terms of V.

Offline kidechizen

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #2 on: October 27, 2007, 07:56:49 PM »
I'm only given that Z = PV / (RT) = 1 + BPP + CP P2 + …………… With B and C as coefficients instead. I'm told that I have a steady isothermal flow of non-ideal gas of benzene. The pressures that I'm given are reduced pressures Pr. And I'm asked to integrate Wrev = int(vdp). I can't seem to find a relationship between p and v to integrate the work reversible.

Offline kidechizen

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #3 on: October 29, 2007, 03:15:05 AM »
I'm actually given compressibility factor as a function of pr: Z = A0+A1(pr)+A2(pr)2+A3(pr)3...and w_rev = -int(vdp), would I use pv=Zrt and thus pvn=C as my relationship to integrate the work?

Offline Yggdrasil

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #4 on: October 29, 2007, 10:58:52 AM »
Isn't w = int(-pdV)?  Also, what do you mean by reduced pressures?  I've normally seen Z expanded as the sum of powers of the density (ρ [rho]) or 1/Vm (one over the molar volume).

Anyway, if what you are given is correct, you could find V as a function of P by rearranging the equation:

PV/(RT) = 1 + BP + CP2

to get

V = RT/P + BRTN + CRTN2/P

(assuming reduced pressure means Pr = P/N)

[edit: fixing tags]
« Last Edit: October 29, 2007, 07:24:23 PM by Yggdrasil »

Offline kidechizen

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #5 on: October 29, 2007, 04:23:47 PM »
I'm given that reduced pressure is Pr = P/Pc with Pc as the critical pressure of benzene. And I'm actually given Z = A0+A1(pr)+A2(pr)2+A3(pr)3...all the way to the 5th power and w_rev = -int(vdp) (I know it's strange). So from what you told me, I should just have PV/RT = A....and just find them as a function of v and just integrate that really long integration to find the work? But when I'm comparing that to ideal work, should I use PV=nRT as my equation? I believe this to be correct, but I want to ascertain the idea. Thanks.

Offline Yggdrasil

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #6 on: October 29, 2007, 07:23:49 PM »
Yup.  That sounds like the right approach to me.

Offline kidechizen

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #7 on: November 04, 2007, 05:24:06 AM »
Sorry Yggdrasil, I'm confused again. Here is the actual statement of the question: Develop a program to calculate the reversible work per unit mass to compress benzene (not ideal) in an isothermal steady flow process. Consider a temp of 600 K and an initial p = critical p. The compressibility factor Z, at this temp, as a function of the reduced pressure pr is given by Z = A0 + A1(pr) + A2(pr)2 +A3(pr)3...to A5(pr)5. If the reversible work/kg is given by w_rev = - int(vdp), calculate w_rev for final reduced pressures = 1.5, 2.0, 2.5...5.0.

What you told me before seemed very logical, but when I tried to apply the idea, I got confused. Z = Pv/RT or (pr)(vr')/(Tr) with pr = p/pc, pc = critical p,   Tr = T/Tc, Tc = critical T. After I made pr in terms of vr', I don't know where to go from there. Since w_rev = - int(vdp), my v could be made in terms of pr, but can I integrate pr with respect to dp? or should I calculate pr in terms of p/pc? Also, I don't know what to do with the Tr afterwards. I don't think I can take it out of the integral since it isn't a constant.


Offline Yggdrasil

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Re: Thermodynamics: Work on an isothermal non-ideal gas
« Reply #8 on: November 04, 2007, 11:37:24 AM »
So, the way I see it is:

Z = A0 + A1p + A2p2 + ...
pV/RT = A0 + A1p + A2p2 + ...
V = (A0 + A1p + A2p2 + ...) RT/p
V = A0RT/p + A1RT + A2pRT + ...

which you can plug into the integral w_rev = -int(Vdp).  Since your process is isothermal, T is constant.

Since your process is isothermal, T should be constant.

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