Question

A rigid vessel having a volume of \(3 \mathrm{~m}^{3}\) initially contains a mixture at \(21^{\circ} \mathrm{C}, 1\) bar consisting of \(79 \% \mathrm{~N}_{2}\) and \(21 \% \mathrm{O}_{2}\) on a molar basis. Helium is allowed to flow into the vessel until the pressure is 2 bar. If the final temperature of the mixture within the vessel is \(27^{\circ} \mathrm{C}\), determine the mass, in \(\mathrm{kg}\), of each component present.

Short answer

2.74 kg of \( \text{N}_2 \), 0.83 kg of \( \text{O}_2 \), and 0.47 kg of He.

Step-by-step solution

- Calculate initial number of moles using the Ideal Gas Law

The Ideal Gas Law is given by \[ PV = nRT \]Here, \( P = 1 \text{ bar} = 100000 \text{ Pa} \)\( V = 3 \text{ m}^3 \)\( R = 8.314 \text{ J/mol·K} \)\( T = 21 \degree \text{C} = 294.15 \text{ K} \)Substitute these values to solve for \( n \) (initial number of moles):\[ n = \frac{PV}{RT} = \frac{100000 \times 3}{8.314 \times 294.15} \approx 123.68 \text{ mol} \]

- Determine moles of individual gases initially

With the mixture consisting of 79% \( \text{N}_2 \) and 21% \( \text{O}_2 \), calculate the moles of each:\[ n_{\text{N}_2} = 0.79 \times 123.68 \approx 97.71 \text{ mol} \]\[ n_{\text{O}_2} = 0.21 \times 123.68 \approx 25.97 \text{ mol} \]

- Calculate final number of moles after helium addition

The new conditions are:\( P = 2 \text{ bar} = 200000 \text{ Pa} \)\( V = 3 \text{ m}^3 \)\( T = 27 \degree \text{C} = 300.15 \text{ K} \)Using the Ideal Gas Law again, solve for the new total moles \( n_f \):\[ n_f = \frac{PV}{RT} = \frac{200000 \times 3}{8.314 \times 300.15} \approx 241.17 \text{ mol} \]

- Determine moles of helium added

Subtracting the initial moles from the final moles provides the moles of helium added:\[ n_{\text{He}} = n_f - (n_{\text{N}_2} + n_{\text{O}_2}) = 241.17 - 123.68 \approx 117.49 \text{ mol} \]

- Convert moles to mass for each gas

Using the molar masses \( M \) of the gases:\( M_{\text{N}_2} = 28 \text{ g/mol} \)\( M_{\text{O}_2} = 32 \text{ g/mol} \)\( M_{\text{He}} = 4 \text{ g/mol} \)Convert moles to mass for each gas:\[ m_{\text{N}_2} = n_{\text{N}_2} \times M_{\text{N}_2} = 97.71 \times 28 \approx 2.74 \text{ kg} \]\[ m_{\text{O}_2} = n_{\text{O}_2} \times M_{\text{O}_2} = 25.97 \times 32 \approx 0.83 \text{ kg} \]\[ m_{\text{He}} = n_{\text{He}} \times M_{\text{He}} = 117.49 \times 4 \approx 0.47 \text{ kg} \]

Key concepts

Ideal Gas Law

The Ideal Gas Law is a crucial principle in understanding the behavior of gases. It's expressed as: \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) is volume, \( n \) represents the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvins. This law helps us to relate these variables and predict how changes in one affect the others. It assumes gases are perfect, without interactions between molecules.
In our problem, we first used the Ideal Gas Law to find the initial number of moles in the vessel. With the vessel's volume (3 m³), initial pressure (1 bar = 100000 Pa), and the initial temperature (21°C = 294.15 K), we substituted these values into the equation. This yielded an initial mole count of approximately 123.68 mol.
Later, when helium is added and conditions change to a pressure of 2 bar and temperature of 27°C, we applied the Ideal Gas Law again to find the new total moles in the vessel. This gave us around 241.17 mol.

Molecular Composition

Molecular composition describes the types and amounts of different molecules present in a mixture. In our problem, the vessel initially contains a mixture of nitrogen (\( \text{N}_2 \)) and oxygen (\( \text{O}_2 \)). The percentages provided (\text{79%} \( \text{N}_2 \) and \text{21%} \( \text{O}_2 \)) are on a molar basis, which means they refer to the number of moles of each gas relative to the total number of moles.
To break this down, we calculated the moles of each gas initially present. As calculated, there were approximately 97.71 mol of nitrogen and 25.97 mol of oxygen in the vessel. This was obtained by multiplying the total initial moles by their respective percentages (0.79 and 0.21).
When helium (\( \text{He} \)) was added, we calculated the change in the total number of moles and found that 117.49 moles of helium were added to the mixture, making this the third component of the final mixture.

Mass Calculation

The mass calculation involves converting the number of moles of each gas into their respective masses. This is done using the molar mass of each gas, which is the mass of one mole of that gas. The molar masses we used are: 28 g/mol for nitrogen (\( \text{N}_2 \)), 32 g/mol for oxygen (\( \text{O}_2 \)), and 4 g/mol for helium (\( \text{He} \)).
To find the mass, we simply multiply the number of moles of each gas by their molar masses. For example, for nitrogen: \( m_{ \text{N}_2 } = n_{ \text{N}_2 } \times M_{ \text{N}_2} = 97.71 \times 28 \text{ g (or 2.74 kg)} \). Similarly, for oxygen it's \( 0.83 \text{ kg} \), and for helium, \( 0.47 \text{ kg} \).
This process helps us translate the mole quantities into more tangible mass measurements, crucial for applications requiring accurate measurements of gas quantities.