Question

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 1

Using the van der Waals equation, the pressure in the 22.4 L vessel containing 1.50 mol of chlorine gas at 0 °C is approximately 43.1 atm.

Step-by-step explanation:

To calculate the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 °C using the van der Waals equation, we can use the formula:

P = (RT/(V - b)) - (a/V^2)

Where P is the pressure, R is the ideal gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin, V is the volume, a is a constant (6.49 L^2-atm/mol^2), and b is a constant (0.0562 L/mol).

Plugging in the values into the equation, we get:

P = ((0.0821 L·atm/(mol·K))(273 K))/(22.4 L - 0.0562 L/mol) - (6.49 L^2-atm/mol^2)/(22.4 L)^2

Simplifying the equation gives us:

P = (0.0821 * 273)/(22.4 - 0.0562) - (6.49)/(22.4^2)

Calculating this, we find that the pressure is approximately 43.1 atm.

Answer 2

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.