The van der Waal's constants and for a gas of molar mass 164.2 g/mol are 4.105 and 0.1 L/mol, respec — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

From the van der Waals equation $(P + a/V_m^2)(V_m - b) = RT$, substituting values gives $(2 + 4.105/V_m^2)(V_m - 0.1) = 0.0821 \times 500 = 41.05$. By checking reasonable integer densities, $d = M/V_m$. If $d = 8\text{ kg/m}^3 = 8\text{ g/L}$, then $V_m = 164.2 / 8 = 20.525\text{ L/mol}$. Plugging this in: $2 + 4.105 / (20.525)^2 \approx 2.0097$ and $20.525 - 0.1 = 20.425$. The product is $2.0097 \times 20.425 \approx 41.05$, perfectly matching $RT$. Therefore, correct answer is 8.