Problem 120 - Chemistry The Central Science

Question

Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose \(120.00 \mathrm{~kg}\) 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 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?

Step-by-Step Solution

Answer

In this problem, we calculated the pressure of the nitrogen gas using the ideal gas law and the van der Waals equation. The pressure according to the ideal gas law was approximately \(3.69 \times 10^4\,\frac{\text{N}}{\text{m}^2}\), whereas by the van der Waals equation was approximately \(3.68 \times 10^4\,\frac{\text{N}}{\text{m}^2}\). Comparing the correction terms in the van der Waals equation, it was found that the correction for attractive interactions dominates under the given conditions, as it contributed a much larger difference between the two pressure calculations compared to the correction for finite volume of gas molecules.

Step 1: Gather given data

We have the following data available: Mass of nitrogen gas: \(120.00\,\text{kg}\) Volume of the cylinder: \(1100.0\,\text{L}\) Temperature: \(280^\circ \text{C}\) First, we need to convert the given temperature to Kelvin scale: \(T = 280 + 273.15 = 553.15\,\text{K}\)

Step 2: Calculate pressure using the Ideal Gas Law

The ideal gas law is given by the formula: \(PV=nRT\) To find the pressure, first, we need to find the number of moles of nitrogen gas. We can do this using its molar mass: Molar mass of \(N_2 = 28.02\,\text{g/mol}\) Number of moles of nitrogen gas: \(n = \frac{120.00\,\text{kg}}{28.02\,\text{g/mol} \cdot 1000\,\text{g/kg}} = 4285.95\,\text{mol}\) Now, let's calculate the pressure using the ideal gas law: \(P = \frac{nRT}{V} = \frac{4285.95\,\text{mol} \times 8.314\,\frac{\text{J}}{\text{mol}\cdot\text{K}} \times 553.15\,\text{K}}{1100.0\,\text{L} \cdot 0.001\,\frac{\text{m}^3}{\text{L}}}= 36852.30\,\frac{\text{N}}{\text{m}^2}\) The pressure of the nitrogen gas using the ideal gas law is approximately \(3.69 \times 10^4\,\frac{\text{N}}{\text{m}^2}\) or Pascals (Pa).

Step 3: Calculate pressure using the van der Waals Equation

The van der Waals equation is given by: \[\left(P + \frac{an^2}{V^2}\right)(V-nb)=nRT\] For Nitrogen gas (\(N_2\)), the van der Waals constants are: \(a = 1.39\,\frac{\text{L}^2\,\text{bar}}{\text{mol}^2}\) and \(b = 0.0391\,\frac{\text{L}}{\text{mol}}\) We need to solve the equation for pressure (P). Converting the volume from liters to cubic meters: \(V = 1100.0\,\text{L} \times 0.001\frac{\text{m}^3}{\text{L}} = 1.100\,\text{m}^3\) Now, we can isolate the pressure (P): \(P = \frac{nRT}{V-nb} - \frac{an^2}{V^2}\) \(P = \frac{4285.95\,\text{mol} \times 8.314\,\frac{\text{J}}{\text{mol}\cdot\text{K}} \times 553.15\,\text{K}}{1.10\,\text{m}^3 - 4285.95\,\text{mol}\times 0.0391\,\frac{\text{L}}{\text{mol}} \cdot 0.001\,\frac{\text{m}^3}{\text{L}}} - \frac{1.39\,\frac{\text{L}^2\,\text{bar}}{\text{mol}^2} \times 100\,\frac{\text{N}}{\text{m}^2\cdot\text{bar}} \times (4285.95\,\text{mol})^2}{(1.100\,\text{m}^3)^2}\) \(P = 36827.00\,\frac{\text{N}}{\text{m}^2}\) The pressure of the nitrogen gas using the van der Waals equation is approximately \(3.68 \times 10^4\,\frac{\text{N}}{\text{m}^2}\) or Pascals (Pa).

Step 4: Identify which correction dominates

In determining which correction dominates the van der Waals equation, we need to compare the two correction terms: \(\frac{an^2}{V^2}\) which accounts for the attractive interactions and \(nb\) which accounts for the finite volume of gas molecules. Attractive interactions term: \(\frac{an^2}{V^2} = \frac{1.39\,\frac{\text{L}^2\,\text{bar}}{\text{mol}^2} \times 100\,\frac{\text{N}}{\text{m}^2\cdot\text{bar}} \times (4285.95\,\text{mol})^2}{(1.100\,\text{m}^3)^2} = 2151.88\,\frac{\text{N}}{\text{m}^2}\) Finite volume of gas molecules term: \(nb = 4285.95\,\text{mol} \times 0.0391\,\frac{\text{L}}{\text{mol}} \times 0.001\,\frac{\text{m}^3}{\text{L}} = 0.16775\, \text{m}^3\) The difference between ideal gas law pressure and van der Waals pressure is mainly due to the attractive interactions term since the finite volume term contributes just a small difference to the total volume. Thus, under the given conditions, the correction for attractive interactions dominates.

Key Concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental relationship that allows us to relate the volume, pressure, temperature, and amount of moles of a gas. It is mathematically represented as \( PV=nRT \), where:

• \(P\) stands for the pressure of the gas,
• \(V\) denotes the volume occupied by the gas,
• \(n\) is the number of moles of gas,
• \(R\) represents the ideal gas constant, and
• \(T\) is the temperature of the gas in Kelvin.

By manipulating this equation, we can solve for any one of the variables if the others are known. When applied to real-life situations, such as storing nitrogen gas for fertilizer production, we typically assume that the gases behave ideally. However, ideal behavior is a simplification that ignores intermolecular forces and the volume occupied by gas molecules. Despite these simplifications, the ideal gas law provides a good approximation for gases under many conditions, and it's especially useful for calculating pressures in situations where high accuracy is not critical.

Pressure Calculation with the Ideal Gas Law

Pressure calculation in the context of the ideal gas law involves rearranging the law to isolate the pressure variable: \( P = \frac{nRT}{V} \). To calculate pressure, there are key steps we must follow:

• Convert all given measurements to the appropriate units, such as temperature to Kelvin and volume to cubic meters if necessary.
• Calculate the number of moles of gas using its mass and molar mass.
• Substitute these values, along with the ideal gas constant, into the equation to find the pressure.

In our nitrogen gas example, the calculation shows the pressure to be approximately \(3.69 \times 10^4\,\frac{\text{N}}{\text{m}^2}\) Pascals. This step gives us a baseline of the pressure assuming ideal behavior, which can then be refined by including non-ideal gas behavior through the van der Waals equation.

Attractive Interaction Correction in the van der Waals Equation

The van der Waals equation modifies the ideal gas law to account for non-ideal gas behavior by introducing corrections for gas particles' finite volume and intermolecular forces. The equation is expressed as \[\left(P + \frac{an^2}{V^2}\right)(V-nb) = nRT\], where the constants \(a\) and \(b\) are specific to each gas and provide the necessary corrections:

• \(a\) corrects for the attractive forces between the gas molecules, and
• \(b\) accounts for the volume occupied by the gas molecules themselves.

The attractive interactions term \(\frac{an^2}{V^2}\) has a significant impact on the pressure calculation under certain conditions. In the case of our nitrogen storage, this term outweighs the volume correction term, indicating that the attractive forces between nitrogen molecules have a more pronounced effect on the pressure of the gas. This degree of interaction can have critical implications for industrial processes, and understanding the extent of these forces is essential for accurate pressure predictions in non-ideal gas systems.