The virial form of van der Waal's gas equation is PV = RT (1 + B/V + C/V^2 + ... ) = RT (1 + B'P + C β States of Matter and Gaseous State Chemistry Question
Question
The virial form of van der Waal's gas equation is PV = RT (1 + B/V + C/V^2 + ... ) = RT (1 + B'P + C'P^2 + ... ). The second virial coefficient of argon gas at 262.5 K is -1 l mol^-1. What is the density of argon gas at 262.5 K and 1 atm? Neglect all the terms after second term in the virial forms, under these condition. (R = 0.08 L-atm/K-mol, Ar = 40)
Answer: A
π‘ Solution & Explanation
virial equation: pv = rt (1 + b/vm). given b = -1 l/mol, p = 1 atm, t = 262.5 k, r = 0.08. rt = 21. vm = 21 Γ (1 - 1/vm) β vm^2 - 21vm + 21 = 0. solving: vm = (21 + β(441 - 84)) / 2 β 19.95 l/mol. density d = m / vm = 40 / 19.95 = 2.0 g/l.
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