At relatively high pressure,van der Waals' eq | vedclass

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Question

(C) The van der Waals equation for $1$ mole of gas is given by: $\left(P + \frac{a}{V^2}\right)(V - b) = RT$. At relatively high pressure,the volume $

Vedclass · Accepted Answer

$PV = RT + Pb$

Vedclass · Answer

(C) The van der Waals equation for $1$ mole of gas is given by: $\left(P + \frac{a}{V^2}\right)(V - b) = RT$. At relatively high pressure,the volume $V$ is small,making the pressure correction term $\frac{a}{V^2}$ negligible compared to $P$. However,the volume correction term $b$ remains significant. Thus,the equation simplifies to: $P(V - b) = RT$. Expanding this,we get: $PV - Pb = RT$,which rearranges to $PV = RT + Pb$. Therefore,option $C$ is correct.

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

At relatively high pressure,van der Waals' equation becomes

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

$W$ $4.0$ $0.027$

$X$ $8.0$ $0.030$

$Y$ $6.0$ $0.032$

$Z$ $12.0$ $0.027$

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