At very high pressure, the compressibility factor of 1 mole of a van der Waal's gas can be given as — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

At very high pressures, the volume is so small that the pressure is extremely high, rendering the $a/V^2$ intermolecular attraction term negligible compared to $P$. The van der Waals equation simplifies to $P(V-b) = RT$. Thus, $PV - Pb = RT \implies PV = RT + Pb \implies Z = PV/RT = 1 + Pb/RT$. Alternatively, from $P = RT/(V-b)$, multiplying by $V/RT$ gives $Z = PV/RT = V/(V-b)$. Therefore, correct answer is A,B.