Question

0.6 kg of a gas mixture of N₂ and O2 is inside a rigid tank at 1.4 bar, 70°C with an initial composition of 20% O₂ by mole. O₂ is added such that the final mass analysis of O2 is 32%. How much O₂ was added? Express your answer in kg.

Answer 1

To determine the amount of O₂ added to the gas mixture, we can use the mass analysis of O₂ and the given initial and final compositions.

Given:

Initial mass of gas mixture = 0.6 kg

Initial mole fraction of O₂ = 20% = 0.2

Final mole fraction of O₂ = 32% = 0.32

Let's assume the mass of O₂ added is m kg.

The initial mass of O₂ in the gas mixture is:

m_initial_O2 = 0.2 * 0.6 kg

The final mass of O₂ in the gas mixture is:

m_final_O2 = (0.2 * 0.6 + m) kg

Since the final mole fraction of O₂ is 0.32, we can write:

m_final_O2 / (0.6 + m) = 0.32

Solving the equation for m, we can find the amount of O₂ added in kg.

Alternatively, we can rearrange the equation and solve for m_final_O2 directly:

m_final_O2 = 0.32 * (0.6 + m) kg

By substituting the given values and solving the equation, we can determine the amount of O₂ added to the gas mixture in kg.