At relatively high pressure,the van der Waals | vedclass

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Question

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

Vedclass · Accepted Answer

$PV_m = RT + Pb$

Vedclass · Answer

(C) The van der Waals equation for $1$ mole of gas is $\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT$.At high pressure,the molar volume $V_m$ is small,making the term $\frac{a}{V_m^2}$ negligible compared to $P$.Thus,the equation simplifies to $P(V_m - b) = RT$.Expanding this gives $PV_m - Pb = RT$,which rearranges to $PV_m = RT + Pb$.Therefore,option $C$ is correct.

At relatively high pressure,the van der Waals equation becomes?

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