Question

For a Vander Waals gas with$b \approx 0,a \ne 0$:
A. $Z = 0.375$
B. $Z < 1$
C. $Z > 1$
D. $Z = 1$

Answer 1

We know that the der waals equation is required for special cases such as non ideal gases which are used to calculate an actual value in which during collision in gaseous molecule, the gas molecules are in contact with the container, not each other. We can write the Vander Waals gas equation as,
$P + \left( {\dfrac{{{n^2}a}}{{{V^2}}}} \right)\left( {v - nb} \right) = nRT$
Where,
V refers to the volume of the gas.
n refers to the moles of gas.
The specific value of a particular gas is denoted as a.
b is the volume of gas molecules.
R is the gas constant.

Complete step by step answer: As we know that, critical compressibility factor (Z) is given by-
$Z = \dfrac{{PV}}{{RT}}$
We know that $P = \dfrac{a}{{27{b^2}}}$
$V = 3b$
$T = \dfrac{{8a}}{{27Rb}}$
We can substitute the known values we get,
$\Rightarrow$ $Z = \dfrac{{\left( {\dfrac{a}{{27{b^2}}}} \right)3b}}{{R\left( {\dfrac{{8a}}{{27Rb}}} \right)}}$
On simplification we get,
$\Rightarrow$ $Z = 0.375$
The critical compressibility factor (Z) is $0.375$.
So, the correct answer is “Option A”.

Note:
Van der Waal added two more assumptions and those they can account for deviation of real gases from ideal behavior. They are:
Volume occupied by the gas molecules.
The attractive forces between the molecules.
When $n$ mole of gas is placed during a container of volume$V$, the quantity during which the molecules are liberal to move is adequate to V as long as the quantity occupied by the molecules themselves is negligible. The presence of molecules of this non vanishing size means there's a particular volume called excluded volume, which isn't available for the molecules to maneuver in.
If this extended volume of one mole of a gas sample is represented by$b$ , then we can write a replacement corrected volume as $\left( {V{\text{ /}}nb} \right).$ the second correction term is obtained from the attractive forces among the molecules. This will be accounted for by pressure also because of the attractive forces among the molecules. Thus the effective pressure is going to be but the particular pressure required to confine the gas molecules, since the attractive forces also will help keep the gas molecules together. So, if there is $n$ mole of gas in volume$V$ , then the amount of molecules that are near any given molecules is proportional to $(n/V)$i.e. number of moles per unit volume. At an equivalent time, the opposite neighboring molecules also will attract its neighbors. Thus, the entire pull of gas thanks to these molecular interactions is proportional$(n/V)$.