Problem 118 - 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 \(\mathrm{N}_{2}(\mathrm{~g})\) is stored in a \(1100.0\) - L metal cylinder at \(280^{\circ} \mathrm{C}\). (a) Calculate the pressure of the gas, assuming idealgas 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

In summary, the pressure of the nitrogen gas using the ideal gas equation is \(2.21 \times 10^6\ \text{Pa}\), while the pressure using the Van der Waals equation is \(2.25 \times 10^6\ \text{Pa}\). Comparing the magnitudes of the corrections in the Van der Waals equation, the attractive interaction correction (\(1.47 \times 10^5\ \text{Pa}\)) dominates over the finite volume of gas molecules correction (\(0.168\ \text{m}^3\)).

Step 1: Convert the given information to appropriate units

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: \ \[T = 280 + 273.15 = 553.15\ K \]

Step 2: Calculate the pressure using the ideal gas equation

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

Step 3: Calculate van der Waals constants and pressure

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: \[P = \frac{n^2a'}{V^2} + \frac{nRT}{V - nb'}\] \[P = \frac{(4282.66\ \text{mol})^2 \times 1.41 \times 10^{-4} \frac{\text{m}^6\text{Pa}}{\text{mol}^2}}{(1.100\ \text{m}^3)^2} + \frac{4282.66\ \text{mol} \times 8.314 \frac{\text{J}}{\text{mol}\cdot\text{K}} \times 553.15\ \text{K}}{1.100 \text{ m}^3 - 4282.66 \text{mol} \times 0.00003913\ \text{m}^3/\text{mol}}\] \[P = 2.25 \times 10^6\ \text{Pa}\]

Step 4: Determine which correction dominates

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 (\(0.168\ \text{m}^3\)).

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in the study of gases. It relates several physical properties of an ideal gas, namely pressure (P), volume (V), temperature (T), and the amount of gas in moles (n). The relationship is described by the equation
\[ PV = nRT \]
Here, R is the ideal gas constant and has a value of 8.314 J/(mol·K) in SI units. The ideal gas law assumes no interactions between gas molecules and that these molecules occupy no volume. It works well under conditions of high temperature and low pressure but fails to account for real gas behavior where attractive forces and finite molecule sizes become significant.

Gas Constants

Gas constants (R) appear in various forms depending on the units for pressure, volume, and temperature used in a given situation. The most commonly used constant is the ideal gas constant, which, in SI units, is 8.314 J/(mol·K). Other forms include 0.0821 Latm/(mol·K) if pressure is in atmospheres and volume is in liters. It's crucial to match the units of R with those of the other variables in the ideal gas law to perform accurate calculations.

Pressure Calculation

Calculating the pressure of a gas involves rearranging the ideal gas law equation to solve for P (pressure). The equation becomes
\[ P = \frac{nRT}{V} \]
where n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and V is the volume in cubic meters. The result gives the pressure in pascals (Pa). This calculation assumes the gas behaves ideally, which means there are no attractions between molecules and the molecules do not have their own volume.

Molar Mass Conversion

When performing calculations with gases, it's essential to use the proper units for mass, volume, and temperature. The conversions involve changing mass from grams to moles using the substance's molar mass, volume from liters to cubic meters, and temperature from Celsius to Kelvin. These conversions ensure that the units match those of the gas constant in the ideal gas law.