Problem 6 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

A vessel of volume \(50.0 \mathrm{dm}^{3}\) contains \(2.50 \mathrm{mol}\) of argon and \(1.20 \mathrm{mol}\) of nitrogen at \(273.15 \mathrm{K}\) (i) Calculate the partial pressure in bar of each gas. (ii) Calculate the total pressure in bar. (iii) How many additional moles of nitrogen must be pumped into the vessel in order to raise the pressure to 5 bar? (Sections \(8.2 \text { and } 8.3)\)

Step-by-Step Solution

Answer

(i) Argon: 1.134 bar, Nitrogen: 0.5445 bar. (ii) Total pressure: 1.6785 bar. (iii) Additional 7.32 moles of nitrogen required.

Step 1: Calculate the Partial Pressure of Argon

Use the Ideal Gas Law formula to calculate the partial pressure of argon. The formula is \(P = \frac{nRT}{V}\), where \(n\) is the number of moles, \(R\) is the ideal gas constant \(0.08314 \, \mathrm{bar} \, \mathrm{dm}^3\, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}\), \(T\) is the temperature in Kelvin, and \(V\) is the volume in dm³.For argon: \(P_{\text{Ar}} = \frac{2.50 \, \mathrm{mol} \times 0.08314 \, \mathrm{bar} \, \mathrm{dm}^3 \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \times 273.15 \, \mathrm{K}}{50.0 \, \mathrm{dm}^3}\)Calculate the value: \(P_{\text{Ar}} = \frac{56.69875}{50.0} = 1.134 \mathrm{bar}\). The partial pressure of argon is 1.134 bar.

Step 2: Calculate the Partial Pressure of Nitrogen

Use the same formula to calculate the partial pressure for nitrogen.\(P_{\text{N}_2} = \frac{1.20 \, \mathrm{mol} \times 0.08314 \, \mathrm{bar} \, \mathrm{dm}^3 \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \times 273.15 \, \mathrm{K}}{50.0 \, \mathrm{dm}^3}\)Calculate the value: \(P_{\text{N}_2} = \frac{27.22668}{50.0} = 0.5445 \mathrm{bar}\). The partial pressure of nitrogen is 0.5445 bar.

Step 3: Calculate the Total Pressure

Add the partial pressures of argon and nitrogen to find the total pressure.\(P_{\text{total}} = P_{\text{Ar}} + P_{\text{N}_2} = 1.134 \, \mathrm{bar} + 0.5445 \, \mathrm{bar} = 1.6785 \, \mathrm{bar}\). The total initial pressure in the vessel is 1.6785 bar.

Step 4: Determine Additional Moles of Nitrogen Required

To find out how many more moles of nitrogen are required to reach a total pressure of 5 bar, use the Ideal Gas Law for the additional pressure needed. First, calculate the required increase in pressure:\(\Delta P = 5 \, \mathrm{bar} - 1.6785 \, \mathrm{bar} = 3.3215 \, \mathrm{bar}\).We will use the same formula: \(n = \frac{PV}{RT}\).\(n_{\text{additional}} = \frac{3.3215 \, \mathrm{bar} \times 50.0 \, \mathrm{dm}^3}{0.08314 \, \mathrm{bar} \, \mathrm{dm}^3 \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \times 273.15 \, \mathrm{K}}\)Calculate the value: \(n_{\text{additional}} = \frac{166.075}{22.68381} \approx 7.32 \, \mathrm{mol}\).Therefore, approximately 7.32 additional moles of nitrogen are needed.

Key Concepts

Partial Pressure

Partial pressure is a concept used to describe the pressure that a gas would exert if it occupied the entire volume of its container on its own. This pressure is part of a larger system where multiple gases are present. Understanding partial pressure is crucial in physical chemistry, especially when dealing with gas mixtures. The partial pressure of a gas can be calculated using the Ideal Gas Law, which relates pressure, volume, moles, temperature, and the gas constant.

In our exercise, we used the formula \(P = \frac{nRT}{V}\) to find the partial pressures of both argon and nitrogen. Here's how it works:

• \(n\): Number of moles of the gas.
• \(R\): Ideal Gas Constant.
• \(T\): Temperature in Kelvin.
• \(V\): Volume of the gas.

The partial pressure of a gas is directly proportional to the temperature and number of moles but inversely proportional to the volume. By calculating the partial pressure of each gas in the vessel, we can determine how much pressure it contributes to the total pressure of the system.

Moles Calculation

Moles describe the amount of substance. It is a fundamental concept in chemistry that helps to quantify how much of a material is present. When solving gas-related problems, it's essential to calculate the correct number of moles to proceed accurately with other steps, such as calculating pressure or volume changes.

In our scenario, both argon and nitrogen have known quantities in moles – 2.50 moles for argon and 1.20 moles for nitrogen. To make calculations involving the Ideal Gas Law, having the accurate number of moles is non-negotiable. Furthermore, when additional moles are needed, as in raising the pressure from 1.6785 bar to 5 bar, the moles calculation becomes the key to determining how much more of a gas is needed.

To find the additional required moles of nitrogen to reach the desired pressure, we used the gas law formula rearranged to \(n = \frac{PV}{RT}\), highlighting the careful balance between the existing and additional gas quantities in response to changes in pressure.

Gas Constant

The Gas Constant \(R\) is a fundamental constant in the Ideal Gas Law equation, symbolizing the proportionality between the amount of gas, temperature, and pressure. In different units, it takes various values, but when dealing with pressures in bar and volumes in cubic decimeters, its value is \(R = 0.08314 \,\text{bar} \,\text{dm}^3\,\text{mol}^{-1}\,\text{K}^{-1}\). This specific constant also neatly ties the moles of gas to the Kelvin temperature scale, providing a clear bridge to calculate pressures.

For exercises like ours, using the correct gas constant is vital because we'll use it to determine partial pressures and the adjustments necessary to change pressure states in our vessel. It ensures our calculations align with the practical settings and units used for the vessel's volume. The constant helps uphold the laws of chemistry in a quantifiable, repeatable manner.

Temperature in Kelvin

Temperature in Kelvin is the standard unit of temperature used in chemistry, especially in relation to the Ideal Gas Law. Unlike Celsius, the Kelvin scale places absolute zero at 0 K, where molecular motion stops. This makes it an absolute measure of temperature, crucial for any gas law calculation.

In our application, the vessel's temperature is given as 273.15 K, which corresponds to 0°C. Using Kelvin allows for more accurate and direct calculations, as the Ideal Gas Law requires temperature units to be in Kelvin. It prevents any negative temperature values that would be nonsensical in calculations involving pressure and volume.

When the temperature changes, it affects the overall energy in the gas system, which in turn influences the pressure a gas exerts. In any comparison, consistency in units is crucial, and hence working with Kelvin is non-negotiable when dealing with gas laws.