For a real gas (van der Waal's gas), behaving ideally, the pressure may be — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

A van der Waals gas behaves ideally when it is at its Boyle temperature, $T_B = \frac{a}{Rb}$. Under these conditions, it approximately follows the ideal gas law $P = \frac{RT_B}{V_m}$. Substituting $T_B$ gives $P = \frac{R(a/Rb)}{V_m} = \frac{a}{b V_m}$. Therefore, correct answer is C.