Problem 117 - 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 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?

Step-by-Step Solution

Answer

(a) The pressure of nitrogen gas, assuming ideal-gas behavior, is approximately \(2399521.87 \, Pa\). (b) Using the van der Waals equation, the pressure of nitrogen gas is approximately \(2376262.39 \, Pa\). (c) The finite volume correction (15.18%) is more significant than the correction for attractive forces (0.94%), indicating that the finite volume correction dominates.

Step 1: (a) Calculate the pressure using Ideal Gas Law

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: \(P = \frac{nRT}{V} = \frac{4284.80 \, mol \times 8.314 \, J \cdot K^{-1} \cdot mol^{-1} \times 553.15 \, K}{1100.0 \times 10^{-3} \, m^3} = 2399521.87 \, Pa\)

Step 2: (b) Calculate pressure using van der Waals equation

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: \(P = \frac{nRT}{(V - nb)} - \frac{an^2}{V^2}\) \(P = \frac{4284.80 \, mol \times 8.314 \, J \cdot K^{-1} \cdot mol^{-1} \times 553.15 \, K}{1100.0 \times 10^{-3} \, m^3 - 4284.80 \, mol \times 0.0000391 \, m^3/mol} - \frac{0.142 \, m^6 Pa/mol^2 \times (4284.80 \, mol)^2}{(1100.0 \times 10^{-3} \, m^3)^2} = 2376262.39 \, Pa\)

Step 3: (c) Determine which correction is more dominant

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.

Key Concepts

van der Waals equation

When gases behave ideally, they follow the Ideal Gas Law, characterized by the equation \(PV = nRT\). This equation assumes that gas particles have negligible volume and do not exert forces on each other. However, in reality, gas particles do occupy space and interact with each other. That's where the van der Waals equation comes into play.

The van der Waals equation is formulated as: \[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\] Here, the terms \(a\) and \(b\) are constants that are specific to each gas, known as van der Waals constants. These account for the real gas behavior:

• \(a\) represents the magnitude of attractive forces between particles.
• \(b\) accounts for the finite space occupied by the gas particles.

By introducing these corrections, the van der Waals equation modifies the ideal gas model to better fit real gas behavior, especially under conditions of high pressure and low temperature.

finite volume correction

In the real world, gas molecules occupy a definite volume, which affects how gases behave, particularly under high pressure. The finite volume correction within the van der Waals equation addresses the actual physical space occupied by gas particles. In the equation \((V - nb)\), the term \(nb\) accounts for the occupation of space:

• \(n\) is the number of moles of the gas.
• \(b\) is the constant unique to each gas that reflects the volume occupied by a mole of molecules.

When gas molecules are squeezed into a container, their volumes are non-negligible, reducing the available space for movement. This correction subtracts the volume occupied by the gas molecules from the total volume \(V\), ensuring a more accurate calculation of pressure, especially in conditions involving high gas densities or strong molecular interactions.

attractive forces correction

Gases do not only collide elastically like billiard balls but also exert forces on one another. The attractive forces correction in the van der Waals equation acknowledges these forces. Specifically, the term \(\frac{an^2}{V^2}\) compensates for the decrease in pressure caused by the gas molecules attracting each other.

This is how it works:

• The constant \(a\) quantifies the strength of attractive interactions between molecules. Different gases have unique values of \(a\), depending on their intermolecular forces.
• The term \(\frac{n^2}{V^2}\) suggests that these attractions become significant when the number of moles \(n\) is large compared to the volume \(V\), meaning particles are close together.

As a result, the effective pressure of a real gas is less than predicted by the ideal gas law because of these attractive forces, making this correction vital for accurate calculations under various conditions, particularly those involving gases with significant intermolecular attractions.