Using the van der waals equation, the pressure in a 22.4 l vessel containing 1.50 mol of chlorine gas at 0.00 °c is atm. (a = 6.49 l2-atm/mol2, b = 0.0562 l/mol)

Answer is: the pressure in a vessel is 1.48 atm.

V(Cl₂) = 22.4 L; pressure of chlorine gas.

n(Cl₂) = 1.50 mol; amount of chlorine gas.

T = 0.00°C = 273.15 K; temperature.

a = 6.49 L²·atm/mol²; the constant a provides a correction for the intermolecular forces.

b = 0.0562 L/mol; value is the volume of one mole of the chlorine gas.

R = 0.08206 L·atm/mol·K, universal gas constant.

Van de Waals equation: (P + an² / V²)(V - nb) = nRT.

(P +  6.49 L²·atm/mol² · (1.5 mol)² / (22.4 L)²) · (22.4 L - 1.5 mol·0.0562 L/mol) = 1.5 mol · 0.08206 L·atm/mol·K · 273.15 K.

(P +  6.49 L²·atm/mol² · (1.5 mol)² / (22.4 L)²) = (1.5 mol · 0.08206 L·atm/mol·K · 273.15 K) ÷ (22.4 L - 1.5 mol · 0.0562 L/mol).

P + 0.029 atm = 33.62 L·atm ÷ 22.31 L.

P = 1.507 atm - 0.029 atm.

P = 1.48 atm; the pressure.