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The ideal gas equation $PV=nRT$ can be converted into real gas equation by compression factor $Z$ i.e $PV=Z~ nRT)$. My question is what is $Z$ and how does it arise? Is $PV/nRT$ a compression ratio of any gas? How does $Z$ adjust the ideal gas assumptions and allow for calculations with a real gas?

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2 Answers 2

You can define $Z$ phenomenologically as follows: calculate the ratio $PV/(nRT)$ for each PV. Call the ratio as a function of $PV$ the compression ratio, and assume it's independent of temperature and mole number. Then $PV=Z(nRT)$.

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sorry dude, I didn't get my answer. –  newera Apr 19 '13 at 3:21
@kiranadhikari I'll put my money that there's no better answer to your question. Typically the people who come up with these corrections give some weak justification, but there's a reason they call it a fudge factor! –  Douglas B. Staple Apr 19 '13 at 3:45
ok, just answer , How is PV/nRT a compression ratio?? I am really puzzled here.. I found no answer in my course book too. –  newera Apr 19 '13 at 14:40
@kiranadhikari Is the problem that you're trying to interpret $Z$ as a compression ratio like the kind defined for engines? These quantities have nothing to do with one another. You can tell this, because you can compress an ideal gas, e.g. to $V^\prime = V/2$, which would give a "compression ratio" of 2, but still have Z=1 in your notation, as long as the gas is still well described as an ideal gas. –  Douglas B. Staple Apr 19 '13 at 14:45

Compressibility factor comes from the virial expansion, any (monoatomic) gas can be study as an ideal gas with Z=1 but it's obviously just an approximation. The problem is that for the ideal gas law you assume that the particles (atoms) are punctiform without a proper volume. In the real gas model we have to correct volume and pressure because of the finite dimension of particles, and this correction introduce successive element of the virial expansion. I give you just an idea of the problem.

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