91Ó°ÊÓ

We are given mass, volume, and temperature. Before proceeding, let's convert these values to appropriate units: 1. Convert mass of nitrogen gas to moles: Using the molar mass of nitrogen (N2) \(\displaystyle M = 28.02 \frac{\text{g}}{\text{mol}}\), we get: \[n = \frac{m}{M} = \frac{120000 \text{ g}}{28.02 \frac{\text{g}}{\text{mol}}} = 4282.66 \text{ mol}\] 2. Convert volume to SI units: We have \(1100.0\) L, which can be converted to m³ as follows: \[V = 1100.0 \frac{\text{L}}{1} \times \frac{10^{-3} \text{ m}^3}{1\text{ L}} = 1.100 \text{ m}^3\] 3. Convert temperature to Kelvin: We have \(280 ^\circ\C) which can be converted to Kelvin as follows: \ \[T = 280 + 273.15 = 553.15\ K \]

We're given, \[PV = nRT\] Solve for pressure P: \[P = \frac{nRT}{V}\] Now plug in the values for \(n\), \(R\), \(T\), and \(V\): \[P = \frac{4282.66\ \text{mol} \times 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}} \times 553.15\ \text{K}}{1.100\ \text{m}^3}\] \[P = 2.21 \times 10^6\ \text{Pa}\]

For nitrogen gas, the van der Waals constants are given as \(a = 1.390 \frac{\text{L}^2\text{atm}}{\text{mol}^2}\) and \(b = 0.03913\ \text{L/mol}\). Convert these values into SI units: \[a' = 1.390 \frac{\text{L}^2\text{atm}}{\text{mol}^2} \times \frac{0.101325\ \text{Pa}}{1\ \text{atm}} \times \frac{10^{-3}\ \text{m}^3}{1\ \text{L}} = 1.41 \times 10^{-4} \frac{\text{m}^6\text{Pa}}{\text{mol}^2}\] \[b' = 0.03913\ \text{L/mol} \times \frac{10^{-3}\ \text{m}^3}{1\ \text{L}} = 0.00003913\ \text{m}^3/\text{mol}\] Now, plug these values into the Van der Waals equation: \[\left(P + \frac{n^2a'}{V^2}\right)(V - nb') = nRT\] Solve for pressure P: \[P = \frac{n^2a'}{V^2} + \frac{nRT}{V - nb'}\] \[P = \frac{(4282.66\ \text{mol})^2 \times 1.41 \times 10^{-4} \frac{\text{m}^6\text{Pa}}{\text{mol}^2}}{(1.100\ \text{m}^3)^2} + \frac{4282.66\ \text{mol} \times 8.314 \frac{\text{J}}{\text{mol}\cdot\text{K}} \times 553.15\ \text{K}}{1.100 \text{ m}^3 - 4282.66 \text{mol} \times 0.00003913\ \text{m}^3/\text{mol}}\] \[P = 2.25 \times 10^6\ \text{Pa}\]

From the ideal gas and Van der Waals pressure values, we can see that the pressure is slightly higher using the Van der Waals equation. The two corrections in the Van der Waals equation are the one for gas molecules' finite volume (\(V - nb\)) and the one for attractive interactions \(\left(\frac{n^2a}{V^2}\right)\). To determine which correction dominates, we can compare the magnitudes of the two terms: 1. In the term \(V - nb\), the volume correction is \(nb = 4282.66 \text{ mol} \times 0.00003913\ \text{m}^3/\text{mol} = 0.168\ \text{m}^3\). 2. In the term \(\displaystyle \frac{n^2a'}{V^2}\), the attractive interaction correction is \(\displaystyle \frac{n^2a'}{V^2} = \frac{(4282.66\ \text{mol})^2 \times 1.41 \times 10^{-4} \frac{\text{m}^6\text{Pa}}{\text{mol}^2}}{(1.100\ \text{m}^3)^2} = 1.47 \times 10^5\ \text{Pa}\). Comparing these values shows that the attractive interaction correction (\(1.47 \times 10^5\ \text{Pa}\)) dominates over the finite volume of gas molecules correction (\(0.168\ \text{m}^3\)).