Help With calculating molar volume?

your question.. Help With calculating molar volume? Calculate the molar volume of carbon dioxide at 400 K and 30 atm, given that the second virial coefficient (B) of CO2 is -0.0605 L mol-1. Compare your result with that obtained using the ideal-gas equation. *********virial equation of state.. PV / (nRT) = 1 + B x (n/V) + C x (n/V)² + D x (n/V)³ + … ∞ we want molar volume.. V/n… so let’s call that Vm and rewrite this as.. PVm / (RT) = 1 + B / Vm + C / Vm² + D / Vm³ + … ∞now in this case, we’re going to assume V/m is high enough (meaning the density is low) so that the C and D and the rest of the terms to ∞ drop out. leaving.. PVm / (RT) = 1 + B / Vmrearranging.. PVm² / (RT) = Vm + B.. PVm² = RTVm + RTB.. PVm² – RTVm = RTB.. Vm² – (RT/P) Vm = RTB/Pthis is a quadratic. let’s complete the square.. Vm² – (RT/P) Vm + (RT / 2P)² = RTB/P + (RT / 2P)²then.. (Vm – RT / 2P)² = RTB/P + (RT / 2P)²then taking sqrt and rearranging.. Vm = (RT / 2P) ±√(RTB/P + (RT / 2P)²)solvingVm = (0.08206Latm/molK)x(400K) / (2x30atm) ±√((0.08206 Latm/molK x 400K x -0.0605 L/mol / 30atm) + (0.08206Latm/molK x 400K / (2x30atm))²)reducing.. Vm = 0.547L/mol ± 0.483L/mol giving these 2 solutions.. Vm = 1.030 L/mol.. Vm = 0.064 L/molthe first is the correct onethe second is the trivial one that we’ll ignore.*******now if want to include that C term.. let C = -B² / 9and you get to recalc all that stuff…. AND.. you’ll end up with a cubic equation to solve and I’d recommend doing that graphically.and DO DOUBLE CHECK my work!.***** compared to.. PV = nRT.. V/n = RT/P = (0.08206Latm/molK) x (400K) / (30atm) = 1.09 L/mol******and the % difference… ’cause this chemistry class and we like to use % difference in chem class.. % difference = | 1.030 – 1.09 | / ( | 1.030 + 1.09 | / 2) x 100% = 5.7%and you can see that at high pressures and high temps, the deviation from ideal may be too large for the process you’re trying to design. *******btw… if you calc the molar volume of CO2 at 400K and 30atm using VDW, the answer turns out to be…. 1.024 L/mol*******that help?... Show More