Question

Use the van der Waals equation and the ideal gas equation to calculate the pressure for 2.00 mol He gas in a 1.00 L container at 300.0 K. 1st attempt

Part 1 (5 points)

Ideal gas law pressure_____ atm

Part 2 (5 points)

Van der Waals pressure_____ atm

Answer 1

Given,

Number of moles of He = 2.00 mol

Volume of container(V) = 1.00 L

Temperature(T) = 300.0 K

Part 1)

We know, the ideal gas equation,

PV = nRT

Here, R =0.08206 L atm /mol.K

Rearranging the equation,

P = nRT /V

Substituting the known values,

P = (2.00 mol x 0.08206 L.atm /mol.K x 300.0 K) / 1.00 L

P = 49.2 atm

Thus, ideal gas law pressure is 49.2 atm

Part 2)

We know, Van der Waals equation,

(P - n²a/V²) (V- nb) = nRT

Here,

a (for He) = 0.0346 L² atm / mol²

b (for He) =0.0238 L/mol

Rearranging the van der Waals equation,

P = (nRT /V-nb) - (n²a / V²)

Substituting the known values,

P = [( 2.00 mol x 0.08206 L. atm/mol.K x 300.0 K) / (1.00 L - 2.00 mol x 0.0238 L/mol)] - [(2.00 mol)² x 0.0346 L² atm / mol²/(1.00 L)²]

P = 51.6968 - 0.1384

P = 51.6 atm

Thus, Van der Waals pressure is 51.6 atm