12.0 moles of gas are in a 6.00 tank at 20.3. Calculate the difference in pressu

Question

12.0 moles of gas are in a 6.00 tank at 20.3. Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300 L^2*atm/mol^2 and b=0.0430 L/mol .

Explanation / Answer

Instead of using the ideal gas equation pV = nRT the van der Waals Equation attempts to better model the behaviour of real gases. Ideal Gas model: pV = nRT {where p is gas pressure, V is volume, n is moles of gas, R is the gas constant, T is absolute temperature). ==> p = nRT/V = (15.0)(0.0826)(22.1+273.15)/(8.00) atm = 45.7 atm (to 3 s.f). Call this pressure p(ideal). van der Waals model (methane): {p + a(n/V)²} (V - bn) = nRT p, V, n, R and T have the same meaning as above. ==> p = nRT/(V - bn) - a(n/V)² {where a = 2.300 L^2 * atm/mol^2 & b=.0430 L/mol} p = (15.0)(0.0826)(22.1+273.15)/{8 - (0.0430)(15.0)} - (2.300){(15.0)/(8)}² = 41.7 atm (to 3 s.f.). Call this pressure p(vDW). Then p(ideal) - p(vdW) = (45.7 - 41.7) atm = 4.0 atm. The difference in pressure between methane and an ideal gas under these conditions is 4.0 atm (to 2 s.f).

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