Question

Calculate the pressure exerted by 2.72 moles of CO₂ confined in a volume of 4.25L at 460K. Use both the van der Waals equation and the ideal gas equation to calculate two pressure values. The van der Waals constants for CO₂ are a = 3.59·atmL2mol2 and b = 0.0427Lmol. Round your answer to 3 significant digits.

Answer 1

Using the ideal gas equation, the pressure is calculated to be approximately 23.266 atm, while using the van der Waals equation, it is approximately 22.413 atm. The discrepancy arises from the non-ideal behavior of gas molecules accounted for by the van der Waals equation.

Step-by-step explanation:

To calculate the pressure exerted by 2.72 moles of CO₂ in a volume of 4.25L at 460K, we can use the ideal gas equation (PV=nRT) and the van der Waals equation [(P + (n²a)/V²)(V - nb) = nRT)]. The ideal gas constant R is 0.0821 L·atm/K·mol. Using the ideal gas equation:

P = (nRT)/(V)

P = (2.72 moles × 0.0821 L·atm/K·mol × 460K) / (4.25L)

The calculated pressure is approximately 23.266 atm.

Next, we apply the van der Waals equation:

P = [(nRT)/(V-nb)] - (n²a)/(V²)

P = [(2.72 moles × 0.0821 L·atm/K·mol × 460K) / (4.25L - 2.72 moles × 0.0427 L/mol)] - (2.72 moles² × 3.59 atm·L²/mol²) / (4.25L)²

Substitute the values to get the pressure adjusting for van der Waals corrections. The calculated pressure using the van der Waals equation is approximately 22.413 atm. The difference in the two values exists because the van der Waals equation accounts for the volume occupied by CO₂ molecules and the intermolecular attractions, which are not considered in the ideal gas equation.