Gas mixture partial pressures (Dalton’s law):\nAn empty rigid vessel at 40°C is charged with 40 g methane (CH4) and 40 g oxygen (O2). What fraction of the total pressure is exerted by oxygen?

Introduction / Context:
Dalton’s law states that each gas in a mixture exerts a partial pressure proportional to its mole fraction at a fixed temperature and volume. Converting masses to moles is crucial because pressure shares depend on moles, not mass.

Given Data / Assumptions:

• T = 40°C, rigid vessel (volume constant).
• m(CH4) = 40 g; m(O2) = 40 g.
• Ideal-gas behaviour and homogeneous mixture.

Concept / Approach:
Partial pressure fraction of a component equals its mole fraction: yi = ni / ntotal. Compute moles for both species, then take the ratio for oxygen.

Step-by-Step Solution:
Molar masses: M(CH4) = 16 g/mol; M(O2) = 32 g/mol.n(CH4) = 40 / 16 = 2.5 mol.n(O2) = 40 / 32 = 1.25 mol.n(total) = 2.5 + 1.25 = 3.75 mol.y(O2) = n(O2) / n(total) = 1.25 / 3.75 = 1/3.

Verification / Alternative check:
Because equal masses of CH4 and O2 were used, the lighter gas contributes more moles. Therefore, oxygen’s fraction must be less than methane’s, consistent with 1/3 vs 2/3.

Why Other Options Are Wrong:
1/2 assumes equal moles, not masses; 1/4 and 3/4 are inconsistent with the computed mole counts; 2/3 would be methane’s fraction, not oxygen’s.

Common Pitfalls:
Using mass fractions instead of mole fractions; forgetting to divide by the correct molar masses.

Final Answer:
1/3