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Problem:

According to the real gas equation, $Z = 1$ for an ideal gas and $Z$ is variable for a real gas. Suppose, in order to easy our calculations, we fixed $Z=1$ for a real gas and for ideal gas, $Z$ will become variable. $Z$ vs $P$ for an ideal gas will be similar to:

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Attempt:
I imagined the answer by thinking the normal graph of $Z$ vs $P$ for a real gas and transposed each point on the locus of the real gas curve on $Z=1$ line. I then imagined where the corresponding point of the ideal gas curve would land.

This gave option (a) as the correct answer, but the book says the answer is (b). I believe then the van der Waals equation must be manipulated to get the answer, but can someone tell me how? Or is the answer given in the book wrong?

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user45838
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2 Answers2

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For a real gas, $(P+a/V^2)(V-b) = RT$

Now, since we are considering $Z=1$ for real gas, all the real gas equations will be reversed for the ideal gas. Meaning, we had to subtract $b$ for volume correction of real gas, for correction of ideal gas we need to add $b$. Same, with pressure.

The equation then comes out to be: $(P-a/V^2)(V+b) = RT$

Now, for the graph, we can calculate the approximate values of $Z$ (relative to unity) for very low pressure, low pressure and high pressure.

Case 1: Very low pressure: $P \downarrow \downarrow \implies V \uparrow \uparrow $

$(P-a/V^2) \approx P$ and $(V-b)\approx V \implies Z \approx 1$

Case 2: Low pressure $P \downarrow \implies V \uparrow$

$(V+b)\approx V$. On solving, we get $Z= 1+a/(RTV)\implies Z>1$.

Case 3: Very high pressure $P \uparrow \uparrow \implies V^2\downarrow \downarrow$

$(P - a/V^2) \approx P$. On solving, we get $Z= 1-Pb/(RT)\implies Z<1$.

The graph would thus be (A), because it is the only one satisfying all these conditions.

Iceberry
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  • I fail to see how the inversion of the sign of the a- and b-terms can be physically justified. To me, this seems actually to be a kind of coordinate transformation. But then, how would it be compatible with the usual definition of pressure an volume? – aventurin Apr 22 '18 at 15:44
  • I thought that if Z=1 is assigned to real gas, we need to see ideal gas relative to real gas. If we see real gas relative to ideal gas, we need to add 'b' to compensate our volume. But, if we assign, real gas Z=1, that's other way round. We need to subtract 'b' from the ideal gas volume to make it equal w.r.t real gas. – Iceberry Apr 22 '18 at 16:17
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The compressibility factor $Z$ of a real gas is defined by the equation

$$p ~ v = Z ~ R ~ T,\tag{1}$$

where $p$ is the pressure, $v$ the molar volume, $R$ the gas constant, and $T$ the thermodynamic temperature. It describes the deviation of a real gas from ideal gas behavior

$$p ~ v = R ~ T.\tag{2}$$

For low pressure real gases behave like an ideal gas, thus $Z = 1$.

For medium pressures the attractive forces between the molecules of the real gas make the volume and thus $p ~ v$ smaller than for an ideal gas. Therefore $Z < 1$.

For high pressures the size the of molecules of the real gas comes into play, $v$ converges to a constant, and lets $Z \propto p$ go to infinity.

Suppose, in order to easy our calculations, we fixed $Z = 1$ for a real gas and for ideal gas, $Z$ will become variable.

If we want to measure the compressibility factor $Z$ in terms of the compressibility factor $Z_R$ of a certain real gas, the resulting scaled compressibility factor $Z_s$ is

$$Z_s = \frac{Z}{Z_R}. \tag{3}$$

In this units the compressibility factor of the real gas becomes $1$ and the compressibility factor of an ideal gas becomes

$$Z_s = \frac{1}{Z_R}, \tag{4}$$

i.e. $Z_s$ is in reciprocal proportion to the compressibility factor of the real gas. This means for an ideal gas we have $Z_s = 1$ for low pressures, $Z_s > 1$ for medium pressures, and $Z_s \to 0$ for high pressures.

With this interpretation answer a is correct.

aventurin
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