States of Matter - Result Question 99
####3. At a given temperature $T$, gases $Ne, Ar, Xe$ and $Kr$ are found to deviate from ideal gas behaviour. Their equation of state is given as, $p=\frac{R T}{V-b}$ at $T$.
Here, $b$ is the van der Waals’ constant. Which gas will exhibit steepest increase in the plot of $Z$ (compression factor) $v s p$ ?
(2019 Main, 9 April II)
(a) $Xe$
(b) $Ar$
(c) $Kr$
(d) $Ne$
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Answer:
Correct Answer: 3. (d)
Solution:
- Noble gases such as $Ne, Ar, Xe$ and $Kr$ found to deviate from ideal gas behaviour.
Xe gas will exhibit steepest increase in plot of $Z$ vs $p$.
Equation of state is given as:
$$ p=\frac{R T}{(V-b)} \Rightarrow p(V-b)=R T $$
$$ \begin{aligned} & p V-p b=R T \Rightarrow p V=R T+p b \ & \frac{p V}{R T}=1+\frac{p b}{R T} \ & \text { As, } \quad Z=\frac{p V}{R T} \ & \text { so, } \quad Z=1+\frac{p b}{R T} \Rightarrow y=c+m x \end{aligned} $$
The plot of $z$ vs $p$ is found to be
The gas with high value of $b$ will be steepest as slope is directly proportional to $b . b$ is the van der Waals’ constant and is equal to four times the actual volume of the gas molecules. Xe gas possess the largest atomic volume among the given noble gases ( $Ne, Kr, Ar)$. Hence, it gives the steepest increase in the plot of $Z$ (compression factor) $v s p$.
