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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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3 Answers 3

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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An ideal gas is treated as a set of $N$ indistinguishable particles with no interactions, which means the partition function of a single particle is simply,

$$Z = \int d^3p \, d^3 q \, \exp \left\{ -\beta\frac{p^2}{2m}\right\}$$

and the partition function of the entire system is $Z^N/N!$. From this, we can obtain the ideal gas law,

$$PV = N k_B T$$

Of course, this equation has many limitations due to the simplifying assumptions, but it is a good approximation if $N/V$ is small. Otherwise, there are higher order corrections, namely,

$$\frac{P}{k_B T} = \frac{N}{V} + B_2(T) \frac{N^2}{V^2} + B_3 \frac{N^3}{V^3} + \dots$$

which is known as the virial expansion, and $B_n$ are Virial coefficients. These correspond to higher order computations of the partition function if we expand the exponential. In fact, it is quite similar to quantum field theory in that one can assign diagrams to these terms. Now the compressibility factor $Z$ (not to be confused with the partition function) is just an experimental way to take into account that one is omitting these higher order corrections that describe the real behaviour of the gas. One possible potential is to use a hard-core potential, namely,

$$U(r) = \left\{ \begin{array}{lr} \infty, & r < r_0 \\ -U_0 \frac{r_0^6}{r^6}, & r \geq r_0\end{array} \right.$$

It takes into account van der Waals interactions, but only at a certain point, hence the name 'hard core' since the particles cannot get closer than some distance $r_0$ to each other.

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