At a given temperature T,gases Ne,Ar,Xe,and K | vedclass

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(C) The given equation of state is $P(V - b) = RT$.Expanding this,we get $PV - Pb = RT$.Dividing by $RT$,we obtain $\frac{PV}{RT} - \frac{Pb}{RT} = 1$

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$Xe$

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(C) The given equation of state is $P(V - b) = RT$.Expanding this,we get $PV - Pb = RT$.Dividing by $RT$,we obtain $\frac{PV}{RT} - \frac{Pb}{RT} = 1$.Since the compression factor $Z = \frac{PV}{RT}$,the equation becomes $Z = 1 + \frac{Pb}{RT}$.This shows that $Z$ increases linearly with $P$ with a slope of $\frac{b}{RT}$.Therefore,the gas with the largest van der Waals constant $b$ will exhibit the steepest increase in the plot of $Z$ versus $P$.The magnitude of $b$ depends on the size of the gas molecules,which follows the order $Ne Thus,$Xe$ has the largest value of $b$ and will show the steepest increase.

$Gas$	$a \ (atm \ L^2 \ mol^{-2})$	$b \ (L \ mol^{-1})$
$W$	$4.0$	$0.027$
$X$	$8.0$	$0.030$
$Y$	$6.0$	$0.032$
$Z$	$12.0$	$0.027$

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$Gas$ $a \ (atm \ L^2 \ mol^{-2})$ $b \ (L \ mol^{-1})$

$W$ $4.0$ $0.027$

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$Y$ $6.0$ $0.032$

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