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?

Short answer

(a) Using the ideal gas law, the pressure of the nitrogen gas is \(22,843,078.5 \mathrm{~Pa}\). (b) Using the van der Waals equation, the pressure of the nitrogen gas is \(3.091 \times 10^{9} \mathrm{~Pa}\). (c) The correction for attractive forces dominates the correction for finite volume of gas molecules.

Step-by-step solution

Calculate the pressure using the ideal gas law.

To calculate the pressure using the ideal gas law, we will use the equation \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of the gas, \(R\) is the universal gas constant, and \(T\) is the absolute temperature. First, we need to convert the given mass of nitrogen gas (\(120.00 \mathrm{~kg}\)) to moles. We know that the molar mass of nitrogen, \(M_N\), is \(28.0134 \mathrm{~g/mol}\). Therefore, we can find the number of moles (\(n\)) of nitrogen gas as follows: \[ n = \frac{120,000 \mathrm{~g}}{28.0134 \mathrm{~g/mol}} = 4284.95 \mathrm{~mol} \] Next, we convert the given temperature (\(280^{\circ}\mathrm{C}\)) to the absolute temperature (in Kelvin): \[T = 280 + 273.15 = 553.15 \mathrm{K}\] Now we can solve for the pressure using the ideal gas law: \[P = \frac{nRT}{V}\]

Plug in the values and solve for the pressure.

Plug in the values into the ideal gas law equation: \[P = \frac{4284.95 \mathrm{~mol} \times 8.314 \mathrm{~J/mol\cdot K} \times 553.15\mathrm{K}}{1100.0 \mathrm{~L}}\] Now, solve for the pressure: \[ P = 22,843,078.5 \mathrm{~Pa}\]

Calculate the pressure using the van der Waals equation.

The van der Waals equation is: \[\left(P + a\left(\frac{n^2}{V^2}\right)\right)\left(V - nb\right) = nRT\] Using Table 10.3, we find the values for the constants \(a\) and \(b\), which are specific for nitrogen gas: \[ a = 1.390 \times 10^{-2} \mathrm{L^2\cdot atm/mol^2} \] \[ b = 3.9 \times 10^{-2} \mathrm{L/mol} \] First, we need to consider the units: - Convert \(V\) from liters to cubic meters (\(1\mathrm{L} = 10^{-3} \mathrm{m^3}\)): \(V = 1.10\mathrm{m^3}\) - Convert the universal gas constant, \(R\), to \( \mathrm{J/mol\cdot K} \): \(R = 0.0821 \mathrm{L\cdot atm/K \cdot mol}\) We now rewrite the van der Waals equation for pressure: \[P = \frac{nRT}{(V - nb)} - a\left(\frac{n^2}{V^2}\right)\]

Plug in the values and solve the pressure using the van der Waals equation.

Plug the values into the van der Waals equation: \[P = \frac{4284.95 \mathrm{~mol} \times 0.0821 \mathrm{~L\cdot atm/K \cdot mol} \times 553.15\mathrm{K}}{(1100.0 \mathrm{~L} - 4284.95 \mathrm{~mol} \times 3.9 \times 10^{-2} \mathrm{L/mol})} - 1.390 \times 10^{-2} \mathrm{L^2\cdot atm/mol^2} \times \frac{(4284.95 \mathrm{~mol})^2}{(1100.0 \mathrm{~L})^2}\] Solve for the pressure: \[ P = 30512.9 \mathrm{~atm}\] Convert the pressure to Pascals: \[ P = 30512.9 \mathrm{~atm} \times \frac{101325 \mathrm{~Pa}}{1 \mathrm{~atm}} = 3.091 \times 10^{9} \mathrm{~Pa}\]

Compare corrections and determine the dominant one.

Since the pressure calculated using the van der Waals equation is smaller than the pressure calculated using the ideal gas law, it means the pressure is being reduced by the attractive forces between the gas molecules. This means the correction for attractive forces dominates the correction for finite volume of gas molecules. So, the results are: (a) The pressure using the ideal gas law is \(22,843,078.5 \mathrm{~Pa}\). (b) The pressure using the van der Waals equation is \(3.091 \times 10^{9} \mathrm{~Pa}\). (c) The correction for attractive forces dominates the correction for finite volume of gas molecules.

Key concepts

van der Waals equation

The van der Waals equation is a modification of the ideal gas law, designed to account for the actual behavior of gases. It introduces two main corrections to the equation:

• Correcting for the finite volume of gas molecules: In an ideal gas law, gas molecules are treated as point particles with no volume. However, real gas molecules occupy space, and this must be incorporated into calculations.
• Correcting for intermolecular forces: In reality, gas molecules experience forces of attraction and repulsion. The van der Waals equation factors in these attractive forces, especially significant at high pressures.

The full form of the equation is: \[ \left(P + a\left(\frac{n^2}{V^2}\right)\right)\left(V - nb\right) = nRT \]where:

• \(P\) is the pressure,
• \(n\) is the number of moles,
• \(V\) is the volume,
• \(R\) is the gas constant,
• \(T\) is temperature,
• \(a\) and \(b\) are constants specific to each gas, representing the strength of intermolecular forces and the actual size of molecules, respectively.

These adjustments make the van der Waals equation particularly useful for gases at high pressure and low temperature.

pressure calculation

Calculating the pressure of a gas under different conditions is essential for understanding its behavior. The ideal gas law is the simplest formula used, expressed as \(PV = nRT\).This equation assumes that the gas particles occupy no space and do not interact with each other. However, in reality, the physical characteristics of gases lead to deviations from ideal behavior. This is where the van der Waals equation plays a crucial role.
When calculating pressure using the van der Waals equation, it involves additional steps to consider the corrections for volume and intermolecular attractions. One should:

• Insert the values for the constants \(a\) and \(b\), specific to the gas you're dealing with.
• Modify the volume \(V\) and pressure \(P\) in the equation, accommodating the finite size and intermolecular forces.

This results in a more accurate representation of the gas's state at conditions far from ideal.

nitrogen gas

Nitrogen gas, chemical symbol \(N_2\), is prevalent both in industrial applications and the atmosphere, where it makes up approximately 78%. It is fundamental in the production of ammonia, primarily used in fertilizers.
Nitrogen molecules consist of two nitrogen atoms, forming a stable diatomic molecule. Due to its triple covalent bond, \(N_2\) is inert and doesn't easily react under normal conditions. However, at elevated temperatures and pressures, such as in industrial settings, its reactivity increases.

• For calculations like the ones involving the ideal gas law or van der Waals equation, the molar mass of \(N_2\) must be considered, which is approximately 28.0134 g/mol.
• In the context of gas law problems, understanding the properties of nitrogen can help predict its behavior, particularly when the conditions deviate from ideality.

Given its significant role in industrial processes, accurately calculating the state of nitrogen gas is crucial.

molar conversion

Molar conversion involves converting between the mass of a substance and the number of moles, using the material's molar mass. It is a foundational skill for solving stoichiometric calculations.
For nitrogen gas, with a molar mass of approximately 28.0134 g/mol, the conversion from grams to moles is straightforward:

• Convert the mass from kilograms to grams for ease of calculation, as the molar mass is in g/mol.
• Use the formula \( n = \frac{m}{M} \) where \(m\) is mass in grams and \(M\) is the molar mass in g/mol, to find the number of moles.

This step is crucial when setting up gas law problems, as it determines \(n\), the quantity required for further calculations. It allows you to bridge the gap between the macroscopic mass and the molecular quantity, enabling you to apply gas laws effectively.