Question

At lower pressures, the van der Waals equation written as $[P+\dfrac{a}{{{V}^{2}}}]V=RT$
The compressibility factor is then equal to:
(A) $[1 - \dfrac{a}{{RTV}}]$

(B) $[1-\dfrac{RTV}{a}]$

(C) $[1+\dfrac{a}{RTV}]$

(D) $[1+\dfrac{RTV}{a}]$

Answer 1

Hint: The compressibility factor (Z), also known as the compression factor or the gas deviation factor. It is simply defined as the ratio of the molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure.

Complete step by step answer:
> The van der waals equation or the real gas equation can be represented in many forms. One of those forms is given in the question.
> The equation is written as $[P+\dfrac{a}{{{V}^{2}}}]V=RT$ because the condition given is of
low pressure. At very low pressures the volume of the gas becomes too high.
> In larger volumes the contribution of the volume of individual gas molecules is neglected and hence the van der waals correction for volume in real gases is dropped.

Let's consider “Z” as a compressibility factor,
We know that,
    \[Z=\dfrac{PV}{RT}\]

We know the given van der Waals equation in the question is:
$[P+\dfrac{a}{{{V}^{2}}}]V=RT$
Now, we will multiply the V to the whole expression on the left hand side and will get the below equation:

\[PV+\dfrac{a}{V}=RT\]
Now we will divide the whole expression by RT and will get:
\[\dfrac{PV}{RT}+\dfrac{a}{VRT}=1\]

Now will transform the equation to get the value of $\dfrac{PV}{RT}$
\[\dfrac{PV}{RT}=1-\dfrac{a}{VRT}\]

Now, we know that $Z=\dfrac{PV}{RT}$ so, the value of Z will be:
\[Z=\left( 1-\dfrac{a}{VRT} \right)\]

According to the above-solved example, we can conclude that a correct answer is an option (A).

Note:
The compressibility factor (Z) is a useful thermodynamic property for modifying the ideal gas law to account for the behavior of real gases. It may be thought of as the ratio of the actual volume of a real gas to the volume predicted by the ideal gas at the same temperature and pressure as the actual volume.