Graph of compressibility factor vs pressure when real gas is assigned Z=1

**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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> [![enter image description here](https://i.stack.imgur.com/GwOz2.png)](https://i.stack.imgur.com/GwOz2.png)
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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?




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 $Z= 1-Pb/(RT)\implies Z<1$.




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


 2

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.
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> 
> 


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.


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