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Topic: Compressibility factor... (UPDATE: see second post)  (Read 3933 times)

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Byrne

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Compressibility factor... (UPDATE: see second post)
« on: January 13, 2006, 10:53:33 PM »
Can someone explain to me how the compressibility factor, Z (PV/nRT = Z), is the ratio of the volume of an actual gas to the ratio of the volume of an ideal gas.

I'm having trouble understanding exactly how the compressibility factor works.
« Last Edit: January 14, 2006, 03:58:59 PM by Byrne »

Byrne

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Re:Compressibility factor...
« Reply #1 on: January 14, 2006, 03:58:13 PM »
Okay, I guess what I'm trying to ask, is that the compressibility factor is equal to PV/RT, but, V is equal to the MOLAR VOLUME.  How do you determine the molar volume of a gas?  After doing a quick search, all I'm finding is that the molar volume for all gases at STP is 22.4 L, which I understand because I have learned this already.  However, what about when I'm not at ideal conditions?

Say I have samples of gas at 300 K as in the first graph at the following link:

http://itl.chem.ufl.edu/2041_f97/lectures/lec_e.html

Another question, the compressibility factor can be expressed as either PV/RT (where V is the molar volume) = Z or PV/nRT = Z... does this mean molar volume is equal to V/n?

Offline Donaldson Tan

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Re:Compressibility factor... (UPDATE: see second post)
« Reply #2 on: January 19, 2006, 11:47:29 AM »
Z=PVm/RT

Zideal = 1

if Z > Zideal then repulsive intermolecular forces dominate over attractive intermolecular forces.

if Z < Zideal then attractive intermolecular forces dominate over repulsive intermolecular forces.

at STP, all real gas behaves almost ideally, therefore the molar volume for all gases is the same (22.4L) at STP.

in the equation for compressibility, Vm refers to the molar volume for the real gas. The molar volume is obtained experimentally.
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