Question

According to the ideal gas law, a 9.776 mol sample of xenon gas in a 0.8177 L container at 499.7 K should exert a pressure of 490.2 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Xe gas, a = 4.194 L2atm/mol2 and b = 5.105×10-2 L/mol.

Answer 1

The ideal gas equation is

PV = nRT where V = 0.8177 L; n = 9.776 moles; R = 0.082 L-atm/mol.K and T = 499.7 K. Plug in the values to obtain P as

P = nRT/V = (9.776 moles)*(0.082 L-atm/mol.K)*(499.7 K)/(0.8177 L) = 489.88 atm;

the value you have reported is 490.2 atm and I shall use that value.

The Vander Waal’s equation is

(P + n²a/V²)(V – nb) = nRT

===> P = nRT/(V – nb) – n²a/V²

Given, a = 4.194 L²-atm/mol² and b = 5.105*10^-2 L/mol.

Plug in values.

P = [(9.776 mole)*(0.082 L-atm/mol.K)*(499.7 K)/{0.8177 L – (9.776 mol)*(5.105*10^-2 L/mol)}] – [(9.776 mole)²*(4.194 L²-atm/mol²)/(0.8177 L)²] =

(400.5755 L-atm)/(0.3186 L) – 599.4635 atm = 1257.2991 atm – 599.4635 atm = 657.8356 atm.

The precent error = ǀP_V.W - P_idealǀ/(P_ideal)*100 = (657.8356 atm – 490.2 atm)/(490.2 atm)*100 = 34.197 ≈ 34.2

Ans: The pressure obtained from Vander Waal’s equation is 34.2% higher than that obtained from the ideal gas law.