Estimate the molar volume of Argon at 100 and 100 atm by treating it is as ideal gas and a van der Waals gas. Give your conclusion based on the result. Given a/(atm dm6 mol-2) and b/(10-2 dm3 mol-1) for Argon are 1.337 and 3.20 respectively.
( P + "\\frac{a}{V_m^2}" ) (Vm -b) = RT
(100 + "\\frac{1.337}{V_m^2}" )( Vm - 3.20 ) = 0.0821 × 100
(100V"_m^2" + 1.337)(Vm - 3.20 ) = V"_m^2" 8.21
(100V"_m^3" + 1.337Vm - 320 V"_m^2" - 4.2784 - 8.21 V"_m^2" ) = 0
100V"_m^3" - 311.79 V"_m^2" + 1.337Vm - 4.2784 = 0
Vm = 3.118 L /mol
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