Flue-gas averaging: A flue gas contains 0.15 kmol CO2, 0.05 kmol O2, and 0.80 kmol N2 per kmol of mixture. What is the average molecular weight of the gas mixture?
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A28.6
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B30.0
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C30.6
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D32.6
Answer
Correct Answer: 30.6
Explanation
Introduction / Context:Determining a gas mixture’s average molecular weight is a routine step in combustion calculations, gas density estimates, and compressor sizing. The average is a mole-fraction-weighted sum of component molecular weights.
Given Data / Assumptions:
- Mole fractions: yCO2 = 0.15, yO2 = 0.05, yN2 = 0.80 (sum = 1.00).
- Molecular weights: MCO2 = 44, MO2 = 32, MN2 = 28 kg/kmol.
- Ideal mixing assumed.
Concept / Approach:The average molecular weight Mmix is given by Mmix = Σ yi * Mi. This uses mole fractions as weights because each kmol of mixture contains yi kmol of species i.
Step-by-Step Solution:
Step 1: Compute contributions: 0.15 * 44 = 6.60.Step 2: Add oxygen term: 0.05 * 32 = 1.60.Step 3: Add nitrogen term: 0.80 * 28 = 22.40.Step 4: Sum = 6.60 + 1.60 + 22.40 = 30.60 kg/kmol.Verification / Alternative check:As a reasonableness check, the mixture is mostly N2, so the result should be close to 28 but pulled upward by CO2 (44) and slightly by O2 (32). A value around 30 is sensible; 30.6 matches the calculation.
Why Other Options Are Wrong:
- 28.6: Too low; underweights CO2 contribution.
- 30.0: Rounded down; not the precise weighted sum.
- 32.6: Too high; would require more CO2/O2 than present.
Common Pitfalls:Using mass fractions by mistake; the correct weighting for Mmix is mole fraction. Also, forgetting to ensure that fractions sum to unity before applying the formula.
Final Answer:30.6