Question

# A mixture of CH4 & O2 is used as an optimal fuel if O2 is present in thrice the amount required theoretically for combustion of CH4. Calculate number of effusions steps required to convert a mixture containing 1 part of CH4 in 193 parts mixture (parts by volume). If calorific value (heat evolved when 1 mole is burnt) of CH4 is 100 cal/mole & if after each effusion 80% of CH4 is collected. Find out what initial mole of each gas in initial mixture required for producing 1000 cal of energy after processing.[Given (0.9)5=0.6]

A
9Steps,14.32molCH4,2886.7molO2
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B
10Steps,13.89molCH4,2666.65molO2
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C
10Steps,27.78molCH4,5333.3molO2
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D
9Steps,28.64molCH4,5773.4molO2
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Solution

## The correct option is C 10Steps,27.78molCH4,5333.3molO2CH4+2O2→CO2+H2O3 times O2 theoretical is required for optimal fuel For optimal fuel nCH4nO2=12×3=16We know, rCH4rO2=nCH4nO2×√MO2MCH416=1192×(√2)n; 1926=(√2)n32=(√2)nn= 10 stepsLet initial moles be nA,nBAfer 1 effusionnAnb=nAinBi×√2And 90% 0f nA is removenAnO=nAinBi×√2×0.9After 10 steps We know 1 mole CH4 produces 100 - foursto produces 13=nAinBi×(√2x)16=nAinBi×32xAlso o produce 1000 cal,10 moles of CH4 is begin10=nCH4×0.910nCH4=27.78 moleMOles of O2=2778×3×36×32=5333.33 moles

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