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Question

(B) The Van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$. At high pressure,the term $\frac{a}{V^2}$ is very small compared t

Vedclass · Accepted Answer

$1 + \frac{Pb}{RT}$

Vedclass · Answer

(B) The Van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$. At high pressure,the term $\frac{a}{V^2}$ is very small compared to $P$ and can be neglected. Thus,the equation simplifies to $P(V - b) = RT$. Expanding this,we get $PV - Pb = RT$. Dividing both sides by $RT$,we get $\frac{PV}{RT} - \frac{Pb}{RT} = 1$. Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z - \frac{Pb}{RT} = 1$. Therefore,$Z = 1 + \frac{Pb}{RT}$. Hence,the correct option is $B$.

Gas	$Ar$	$Ne$	$Kr$	$Xe$
$a \ (atm \ dm^6 \ mol^{-2})$	$1.3$	$0.2$	$5.1$	$4.1$
$b \ (10^{-2} \ dm^3 \ mol^{-1})$	$3.2$	$1.7$	$1.0$	$5.0$

The compressibility factor for a real gas at high pressure is :-

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