91Ó°ÊÓ

Molar analysis helps us understand the composition of a gas mixture by analyzing the number of moles of each gas present. In the given problem, we start with nitrogen gas and later add oxygen to the container.
We begin by finding the initial number of moles of nitrogen using the ideal gas law equation \(\text{PV} = \text{nRT}\).
Given:

• Pressure (P): 180 kPa
• Volume (V): 2 m³
• Temperature (T): 50°C = 323 K
• Universal Gas Constant (R): 0.008314 kPa·m³/mol·K

Rearranging the ideal gas law to find moles, we get \(\text{n} = \frac{\text{PV}}{\text{RT}}\). Substituting the values, we find \(\text{n}_{\text{N}_2} \approx 134.15 \text{ moles}\).

Next, for oxygen, we use its molar mass \(\text{M}_{\text{O}_2}\ = 0.032 \text{ kg/mol}\). Given mass \(\text{m}_{\text{O}_2}\text{ is } 5 \text{ kg}\), the number of moles is calculated as \(\text{n}_{\text{O}_2} \approx 156.25 \text{ moles}\).
Total moles in the mixture become the sum of nitrogen and oxygen moles: \(\text{n}_{\text{total}}\text{ is } 290.4 \text{ moles}\).

To find the pressure of the gas mixture after oxygen is added, we need to apply the ideal gas law again. With constant temperature, volume, and new total moles of gas, we rearrange the ideal gas law to solve for pressure \(\text{P}_{\text{mixture}} = \frac{\text{n}_{\text{total}} \times \text{R} \times \text{T}}{\text{V}}\).

Given:

• Total moles (n): 290.4 moles
• Temperature (T): 323 K
• Volume (V): 2 m³
• Universal Gas Constant (R): 0.008314 kPa·m³/mol·K

Substituting into the equation, we find \(\text{P}_{\text{mixture}} \approx 388.68 \text{ kPa}\). Therefore, the pressure of the mixture increases significantly due to the addition of oxygen. This result helps illustrate the direct relationship between the number of moles of gas and the pressure, as per the Ideal Gas Law.

The mole fraction of a component in a gas mixture indicates the ratio of the moles of that component to the total moles of all components. This is important for determining the composition of the mixture.
For nitrogen \(\text{N}_2\), the mole fraction \(\text{X}_{\text{N}_2}\) is calculated as:

• \(\text{X}_{\text{N}_2} = \frac{\text{n}_{\text{N}_2}}{\text{n}_{\text{total}}} = \frac{134.15}{290.4} \approx 0.462\)

This indicates that nitrogen makes up approximately 46.2% of the total gas mixture.

For oxygen \(\text{O}_2\), the mole fraction \(\text{X}_{\text{O}_2}\) is calculated as:

• \(\text{X}_{\text{O}_2} = \frac{\text{n}_{\text{O}_2}}{\text{n}_{\text{total}}} = \frac{156.25}{290.4} \approx 0.538\)

This means oxygen makes up approximately 53.8% of the mixture.

Mole fractions are useful when dealing with partial pressures and gas compositions in chemistry and engineering applications.