Gas mixture at 25 °C: Equal masses of oxygen (O2) and methane (CH4) are placed in an empty vessel. What fraction of the total pressure is due to oxygen?

Introduction / Context:When different gases occupy the same container at the same temperature, Dalton’s law of partial pressures states that each gas exerts a pressure proportional to its mole fraction. Therefore, converting given mass information into moles is the key step to finding the pressure fraction of a component in a gas mixture.

Given Data / Assumptions:

• Equal masses of O2 and CH4.
• Ideal-gas behavior at 25 °C.
• Molar masses: O2 = 32 g/mol, CH4 = 16 g/mol.

Concept / Approach:Partial pressure fraction equals mole fraction for ideal gases at a common T and V. With equal masses, species with lower molar mass contributes more moles. Compute the mole ratio n_O2 : n_CH4 from the mass ratio and molar masses, then find the oxygen mole fraction x_O2 and hence the fraction of total pressure exerted by oxygen.

Step-by-Step Solution:

Verification / Alternative check:Assume m = 32 g. Then n_O2 = 1 mol, n_CH4 = 2 mol, total = 3 mol. Oxygen’s fraction is 1/3 regardless of absolute pressure or temperature (as long as both gases share T and V and behave ideally).

Why Other Options Are Wrong:

• 2/3 corresponds to methane’s mole fraction in this setup, not oxygen’s.
• 1/2 would require equal moles, which is not true for equal masses here.
• 1/4 or 3/5 do not match the stoichiometric mole split derived from the molar masses.

Common Pitfalls:Using mass fractions instead of mole fractions in gas-mixture partial pressure calculations; forgetting that lighter gases contribute more moles for the same mass.

Final Answer:1/3