Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose 120.00 kg of \(\mathrm{N}_{2}(g)\) is stored in a \(1100.0-\mathrm{L}\) metal cylinder at \(280^{\circ} \mathrm{C}\) (a) Calculate the pressure of the gas, assuming ideal-gas behavior. (b) By using the data in Table \(10.3,\) calculate the pressure of the gas according to the van der Waals equation. (c) Under the conditions of this problem, which correction dominates, the one for finite volume of gas molecules or the one for attractive interactions?

First, we need to find the number of moles of nitrogen gas. We will use the molar mass of nitrogen, which is 28.02 g/mol. Moles of \(\mathrm{N}_{2} = \frac{120.00 \times 10^3 \, g}{28.02 \, g/mol} = 4284.80 \, mol\) Next, we will convert the temperature from Celsius to Kelvin: \(T = 280^{\circ}\mathrm{C} + 273.15 = 553.15 \, K\) Now, we can use the Ideal Gas Law, which is: \(PV = nRT\) Where, P = pressure V = volume n = number of moles R = gas constant, which equals \(8.314 \, J \cdot K^{-1} \cdot mol^{-1}\) for this case T = temperature We need to solve for P: \(P = \frac{nRT}{V} = \frac{4284.80 \, mol \times 8.314 \, J \cdot K^{-1} \cdot mol^{-1} \times 553.15 \, K}{1100.0 \times 10^{-3} \, m^3} = 2399521.87 \, Pa\)

The van der Waals equation for real gases is: \(\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\) Where, a and b are van der Waals constants, which for nitrogen gas are: a = 1.390 L²·atm/mol² b = 0.0391 L/mol We need to convert the units of the van der Waals constants to match the other units: a = 1.390 L²·atm/mol² × (101325 Pa/atm) × (0.001 m³/L)² = 0.142 \(\frac{m^6 Pa}{mol^2}\) b = 0.0391 L/mol × (0.001 m³/L) = 0.0000391 \(\frac{m^3}{mol}\) Now, we can plug in the values and rearrange the equation to solve for P: \(P = \frac{nRT}{(V - nb)} - \frac{an^2}{V^2}\) \(P = \frac{4284.80 \, mol \times 8.314 \, J \cdot K^{-1} \cdot mol^{-1} \times 553.15 \, K}{1100.0 \times 10^{-3} \, m^3 - 4284.80 \, mol \times 0.0000391 \, m^3/mol} - \frac{0.142 \, m^6 Pa/mol^2 \times (4284.80 \, mol)^2}{(1100.0 \times 10^{-3} \, m^3)^2} = 2376262.39 \, Pa\)

To find which correction dominates, we need to compare the finite volume correction and the attractive forces correction terms: Finite volume correction (FVC) = nb Attractive forces correction (AFC) = \(\frac{an^2}{V^2}\) FVC = 4284.80 mol × 0.0000391 \(\frac{m^3}{mol}\) = 0.167 \, m^3 AFC = \(\frac{0.142 \, m^6 Pa/mol^2 \times (4284.80 \, mol)^2}{(1100.0 \times 10^{-3} \, m^3)^2}\) = 22459.48 \, Pa Now, let's calculate the ideal pressure without any corrections, which is: Ideal pressure (IP) = nRT/V IP = \(4284.80 \, mol \times 8.314 \, J \cdot K^{-1} \cdot mol^{-1} \times 553.15 \, K / 1100.0 \times 10^{-3} \, m^3 = 2399521.87 \, Pa\) Now, let's compare these corrections to the ideal pressure (IP): Difference due to FVC = \(\frac{FVC}{V} \times 100 \% \) = \(\frac{0.167 \, m^3}{1100.0 \times 10^{-3} \, m^3} \times 100 \% = 15.18 \% \) Difference due to AFC = \(\frac{AFC}{IP} \times 100 \% \) = \(\frac{22459.48 \, Pa}{2399521.87 \, Pa} \times 100 \% = 0.94 \% \) As we can see, the correction for finite volume (15.18%) is more significant than the correction for attractive forces (0.94%). Therefore, the finite volume correction dominates.