The critical pressure and temperature of a van der Waal gas are 41.05 atm and 128 K, respectively. T — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

critical volume v_c = 3b = 3 × (r tc / 8 pc). substituting the values: v_c = 3 × 0.0821 × 128 / (8 × 41.05) = 31.5264 / 328.4 = 0.096 l/mol.