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### Topic: isothermal Joule-Thomson coefficient  (Read 21734 times)

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#### vdemas

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##### isothermal Joule-Thomson coefficient
« on: May 13, 2008, 01:21:21 AM »
OK, so here's the question:

a) Given that mu = 0.25 K/atm for Nitrogen, calculate the value of   its isothermal Joule-Thomson coefficient.

b) Calculate the energy that must be supplied as heat to maintain constant temperature when 15,0 mol N2 flows through a throttle in an isothermal Joule-thomson experiment and the pressure drop is 75atm.

This is what I got so far :

muT = - Cp . mu

Now, the value of mu is given in the Question.
What I would like to know is how do I calculate Cp ?
Can I assume that Cp = 5/2 R ? Since nitrogen is a monoatomic gas.
So, muT = - ( 5/2 ) (8.314510 J/K.mol) x (0.25K/atm)
= - 5,1966J/atm.mol

But at the back of the book the answer is -7,2 J/atm.mol.
How did they get to this?
« Last Edit: May 13, 2008, 01:41:35 AM by vdemas »

#### leo1

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##### Re: isothermal Joule-Thomson coefficient
« Reply #1 on: May 14, 2008, 12:49:46 PM »
You should note that nitrogen is a diatomic gas (N2) and hence cp=7/2 *R.

I hope this helps.

#### vdemas

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##### Re: isothermal Joule-Thomson coefficient
« Reply #2 on: May 21, 2008, 08:03:17 AM »
Thanx, I figured that out just after I posted the question.
I feel so stupid.